Ioaa 2016 Problems

IOAA 2016 PROBLEMS

Table of Constants Page 1 of 1

Fundamental Constants 𝑐 ℎ

Speed of light in vacuum Planck Constant Boltzmann Constant Stefan-Boltzmann Constant Charge of electron Universal Gravitational Constant Universal Gas Constant Avogadro Constant Wien's displacement law Mass of electron Mass of proton Mass of neutron Atomic Mass Unit (a.m.u.)

𝑘B 𝜎 𝑒 𝐺 𝑅 𝑁A 𝜆m 𝑇 𝑚e 𝑚p 𝑚n 𝑢

= = = = = = = = = = = = =

2.998 × 108 m s −1 6.626 × 10−34 J s 1.381 × 10−23 J K −1 5.670 × 10−8 W m−2 K −4 1.602 × 10−19 C 6.674 × 10−11 N m2 kg −2 8.315 J mol−1 K −1 6.022 × 1023 mol−1 2.898 × 10−3 m K 9.109 × 10−31 kg 1.673 × 10−27 kg 1.675 × 10−27 kg 1.661 × 10−27 kg

= = = = = = = = = = = =

3.086 × 1016 m 1.496 × 1011 m 1.989 × 1030 kg 6.955 × 108 m 3.826 × 1026 W −26.72 mag 1366 W m−2 30′ 5.972 × 1024 kg 6.371 × 106 m 365.242 solar days 3.156 × 107 s

Astronomical Data 1 parsec (pc) 1 astronomical unit (AU) Solar Mass Solar Radius Solar Luminosity Apparent magnitude of the Sun at mid-day Solar Constant (at Earth) Apparent angular diameter of Sun Earth Mass Earth Radius 1 tropical year

𝑎⊕ 𝑀⊙ 𝑅⊙ 𝐿⊙ 𝑚⊙ 𝜃⊙ 𝑀⊕ 𝑅⊕

Theoretical Examination Page 1 of 1

Instructions to the Contestants Total time duration of this examination is five hours. (2) This examination consists of Question 1 to 5 - 10 marks each - 10 x 5 = 50 marks Question 6 to 10 - 20 marks each - 20 x 5 = 100 marks Question 11 to 13 - 50 marks each - 50 x 3 = 150 marks Total = 300 marks (3) Inside the envelope you will find a. Question paper in English b. Question paper in your native language (if applicable) c. Table of Constants in English d. Table of Constants in your native language (if applicable) e. A set of Summary Answersheets f. Cover Sheet (4) Use only black or blue pen for writing. For figures, you may use pencils. (5) There is no negative marking. (6) Some marks will be deducted if the final answer is given without detailed solution. (7) Some marks will be deducted if the final answers have inappropriate number of significant digits, no units or wrong units. (8) The necessary values of fundamental and astronomical constants should be taken from the Table of Constants provided to you. (9) Instructions regarding using the answersheets a. For each question a separate Summary Answersheet has been provided. Final answer(s) for each question / each part of the question must be written in the corresponding box in the Summary Answersheet. Write your contestant code and page number on each Summary Answersheet. b. You should also show detailed solution in the blank space provided on each Summary Answersheet. If necessary, you can ask for extra blank sheets from the invigilators. c. Use separate blank sheets for each question. d. Write your contestant code, question number and page number on FRONT side of each blank sheet. Write your contestant code and page number on the BACK side of each blank sheet. e. The page numbers should be continuous, i.e., if you use 20 sheets for the entire examination (including the Summary Answersheets), the page numbers should run from 1 to 40. f. Write only inside the boxed area. g. For the work that you do not want to be evaluated, cross that part out. (10) At the end of the examination a. On the Cover Sheet, clearly write page numbers for each question. b. Check that you have written your contestant code on all pages. c. Put the Cover Sheet, all Summary Answersheets, blank sheets and rough sheets inside the envelope. You may keep question paper and Table of Constants with you. (1)

Theoretical Examination Page 1 of 6 (T1) True or False Determine if each of the following statements is True or False. In the Summary Answersheet, tick the correct answer (TRUE / FALSE) for each statement. No justifications are necessary for this question. (T1.1) In a photograph of the clear sky on a Full Moon night with a sufficiently long exposure, the colour of the sky would appear blue as in daytime.

2

(T1.2) An astronomer at Bhubaneswar marks the position of the Sun on the sky at 05: 00 UT every day of the year. If the Earth's axis were perpendicular to its orbital plane, these positions would trace an arc of a great circle.

2

(T1.3) If the orbital period of a certain minor body around the Sun in the ecliptic plane is less than the orbital period of Uranus, then its orbit must necessarily be fully inside the orbit of Uranus.

2

(T1.4) The centre of mass of the solar system is inside the Sun at all times.

2

(T1.5) A photon is moving in free space. As the Universe expands, its momentum decreases.

2

(T2) Gases on Titan Gas particles in a planetary atmosphere have a wide distribution of speeds. If the r.m.s. (root mean square) thermal speed of particles of a particular gas exceeds 1/6 of the escape speed, then most of that gas will escape from the planet. What is the minimum atomic weight (relative atomic mass), 𝐴min , of an ideal monatomic gas so that it remains in the atmosphere of Titan?

10

Given, mass of Titan 𝑀T = 1.23 × 1023 kg, radius of Titan 𝑅T = 2575 km, surface temperature of Titan 𝑇T = 93.7 K. (T3) Early Universe Cosmological models indicate that radiation energy density, 𝜌r , in the Universe is proportional to (1 + 𝑧) 4 , and the matter energy density, 𝜌m , is proportional to (1 + 𝑧)3 , where 𝑧 is the redshift. The dimensionless density parameter, Ω, is given as Ω = 𝜌/𝜌c , where 𝜌c is the critical energy density of the Universe. In the present Universe, the density parameters corresponding to radiation and matter, are Ωr0 = 10 −4 and Ω m0 = 0.3, respectively. (T3.1) Calculate the redshift, 𝑧e , at which radiation and matter energy densities were equal.

3

(T3.2) Assuming that the radiation from the early Universe has a blackbody spectrum with a temperature of 2.732 K, estimate the temperature, 𝑇e , of the radiation at redshift 𝑧e .

4

(T3.3) Estimate the typical photon energy, 𝐸ν (in eV), of the radiation as emitted at redshift 𝑧e .

3

(T4) Shadows An observer in the northern hemisphere noticed that the length of the shortest shadow of a 1.000 m vertical stick on a day was 1.732 m. On the same day, the length of the longest shadow of the same vertical stick was measured to be 5.671 m.

10

Find the latitude, 𝜙, of the observer and declination of the Sun, 𝛿⊙, on that day. Assume the Sun to be a point source and ignore atmospheric refraction. (T5) GMRT beam transit Giant Metrewave Radio Telescope (GMRT), one of the world's largest radio telescopes at metre wavelengths, is located in western India (latitude: 19 ∘ 6′ N, longitude: 74∘ 3′ E). GMRT consists of 30 dish antennas, each with a diameter of 45.0 m. A single dish of GMRT was held fixed with its axis pointing at a zenith angle of 39 ∘ 42′ along the northern meridian such that a radio point source would pass along a diameter of the beam, when it is transiting the meridian. What is the duration 𝑇transit for which this source would be within the FWHM (full width at half maximum) of the beam of a single GMRT dish observing at 200 MHz? Hint: The FWHM size of the beam of a radio dish operating at a given frequency corresponds to the angular resolution of the dish. Assume uniform illumination.

10

Theoretical Examination Page 2 of 6 (T6) Cepheid Pulsations The star 𝛽-Doradus is a Cepheid variable star with a pulsation period of 9.84 days. We make a simplifying assumption that the star is brightest when it is most contracted (radius being 𝑅1) and it is faintest when it is most expanded (radius being 𝑅2 ). For simplicity, assume that the star maintains its spherical shape and behaves as a perfect black body at every instant during the entire cycle. The bolometric magnitude of the star varies from 3.46 to 4.08. From Doppler measurements, we know that during pulsation the stellar surface expands or contracts at an average radial speed of 12.8 km s −1 . Over the period of pulsation, the peak of thermal radiation (intrinsic) of the star varies from 531.0 nm to 649.1 nm. (T6.1) Find the ratio of radii of the star in its most contracted and most expanded states (𝑅1 /𝑅2 ).

7

(T6.2) Find the radii of the star (in metres) in its most contracted and most expanded states (𝑅1 and 𝑅2 ).

3

(T6.3) Calculate the flux of the star, 𝐹2 , when it is in its most expanded state.

5

(T6.4) Find the distance to the star, 𝐷star , in parsecs.

5

(T7) Telescope optics In a particular ideal refracting telescope of focal ratio 𝑓/5, the focal length of the objective lens is 100 cm and that of the eyepiece is 1 cm. (T7.1) What is the angular magnification, 𝑚0 , of the telescope? What is the length of the telescope, 𝐿 0, i.e. the distance between its objective and eyepiece?

4

An introduction of a concave lens (Barlow lens) between the objective lens and the prime focus is a common way to increase the magnification without a large increase in the length of the telescope. A Barlow lens of focal length 1 cm is now introduced between the objective and the eyepiece to double the magnification. (T7.2) At what distance, 𝑑B , from the prime focus must the Barlow lens be kept in order to obtain this desired double magnification?

6

(T7.3) What is the increase, Δ𝐿, in the length of the telescope?

4

A telescope is now constructed with the same objective lens and a CCD detector placed at the prime focus (without any Barlow lens or eyepiece). The size of each pixel of the CCD detector is 10 µm. (T7.4) What will be the distance in pixels between the centroids of the images of the two stars , 𝑛p , on the CCD, if they are 20′′ apart on the sky?

6

(T8) U-Band photometry A star has an apparent magnitude 𝑚U = 15.0 in the U-band. The U-band filter is ideal, i.e., it has perfect (100%) transmission within the band and is completely opaque (0% transmission) outside the band. The filter is centered at 360 nm, and has a width of 80 nm. It is assumed that the star also has a flat energy spectrum with respect to frequency. The conversion between magnitude, 𝑚, in any band and flux density, 𝑓, of a star in Jansky (1 Jy = 1 × 10 −26 W Hz −1 m−2 ) is given by 𝑓 = 3631 × 10−0.4𝑚 Jy (T8.1) Approximately how many U-band photons, 𝑁0 , from this star will be incident normally on a 1 m2 area at the top of the Earth's atmosphere every second?

8

This star is being observed in the U-band using a ground based telescope, whose primary mirror has a diameter of 2.0 m. Atmospheric extinction in U-band during the observation is 50%. You may assume that the seeing is diffraction limited. Average surface brightness of night sky in U-band was measured to be 22.0 mag/arcsec 2 . (T8.2) What is the ratio, 𝑅, of number of photons received per second from the star to that received from the sky, when measured over a circular aperture of diameter 2′′?

8

(T8.3) In practice, only 20% of U-band photons falling on the primary mirror are detected. How many photons, 𝑁t , from the star are detected per second?

4

Theoretical Examination Page 3 of 6 (T9) Mars Orbiter Mission India's Mars Orbiter Mission (MOM) was launched using the Polar Satellite Launch Vehicle (PSLV) on 5 November 2013. The dry mass of MOM (body + instruments) was 500 kg and it carried fuel of mass 852 kg. It was initially placed in an elliptical orbit around the Earth with perigee at a height of 264.1 km and apogee at a height of 23903.6 km, above the surface of the Earth. After raising the orbit six times, MOM was transferred to a trans-Mars injection orbit (Hohmann orbit). The first such orbit-raising was performed by firing the engines for a very short time near the perigee. The engines were fired to change the orbit without changing the plane of the orbit and without changing its perigee. This gave a net impulse of 1.73 × 105 kg m s −1 to the satellite. Ignore the change in mass due to burning of fuel. (T9.1) What is the height of the new apogee, ℎa above the surface of the Earth, after this engine burn? (T9.2) Find the eccentricity (𝑒) of the new orbit after the burn and the new orbital period (𝑃) of MOM in hours.

14 6

(T10) Gravitational Lensing Telescope Einstein's General Theory of Relativity predicts bending of light around massive bodies. For simplicit y, we assume that the bending of light happens at a single point for each light ray, as shown in the figure. The angle of bending, 𝜃b , is given by 2𝑅 𝜃b = sch 𝑟

where 𝑅sch is the Schwarzschild radius associated with that gravitational body. We call 𝑟, the distance of the incoming light ray from the parallel 𝑥-axis passing through the centre of the body, as the “impact parameter”.

A massive body thus behaves somewhat like a focusing lens. The light rays coming from infinite distance beyond a massive body, and having the same impact parameter 𝑟, converge at a point along the axis, at a distance 𝑓𝑟 from the centre of the massive body. An observer at that point will benefit from huge amplification due to this gravitational focusing. The massive body in this case is being used as a Gravitational Lensing Telescope for amplification of distant signals. (T10.1) Consider the possibility of our Sun as a gravitational lensing telescope. Calculate the shortest distance, 𝑓min , from the centre of the Sun (in A. U.) at which the light rays can get focused.

6

(T10.2) Consider a small circular detector of radius 𝑎, kept at a distance 𝑓min centered on the 𝑥-axis and perpendicular to it. Note that only the light rays which pass within a certain annulus (ring) of width ℎ (where ℎ ≪ 𝑅⊙ ) around the Sun would encounter the detector. The amplification factor at the detector is defined as the ratio of the intensity of the light incident on the detector in the presence of the Sun and the intensity in the absence of the Sun.

8

Express the amplification factor, 𝐴m , at the detector in terms of 𝑅 ⊙ and 𝑎. (T10.3) Consider a spherical mass distribution, such as dark matter in a galaxy cluster, through which light rays can pass while undergoing gravitational bending. Assume for simplicity that for the gravitational bending with impact parameter, 𝑟, only the mass 𝑀(𝑟) enclosed inside the radius 𝑟 is relevant. What should be the mass distribution, 𝑀(𝑟), such that the gravitational lens behaves like an ideal optical convex lens?

6

Theoretical Examination Page 4 of 6 (T11) Gravitational Waves The first signal of gravitational waves was observed by two advanced LIGO detectors at Hanford and Livingston, USA in September 2015. One of these measurements (strain vs time in seconds) is shown in the accompanying figure. In this problem, we will interpret this signal in terms of a small test mass 𝑚 orbiting around a large mass 𝑀 (i.e., 𝑚 ≪ 𝑀), by considering several models for the nature of the central mass.

The test mass loses energy due to the emission of gravitational waves. As a result the orbit keeps on shrinking, until the test mass reaches the surface of the object, or in the case of a black hole, the innermost stable circular orbit – ISCO – which is given by 𝑅ISCO = 3𝑅sch , where 𝑅sch is the Schwarzschild radius of the black hole. This is the “epoch of merger". At this point, the amplitude of the gravitational wave is maximum, and so is its frequency, which is always twice the orbital frequency. In this problem, we will only focus on the gravitational waves before the merger, when Kepler’s laws are assumed to be valid. After the merger, the form of gravitational waves will drastically change. (T11.1) Consider the observed gravitational waves shown in the figure above. Estimate the time period, 𝑇0 , and hence calculate the frequency, 𝑓0 , of gravitational waves just before the epoch of merger.

3

(T11.2) For any main sequence (MS) star, the radius of the star, 𝑅MS , and its mass, 𝑀MS , are related by 10 a power law given as, 𝑅MS ∝ (𝑀MS )𝛼 for 𝑀⊙ < 𝑀MS where 𝛼 = 0.8 for 0.08𝑀⊙ ≤ 𝑀MS ≤ 𝑀⊙ = 1.0 If the central object were a main sequence star, write an expression for the maximum frequency of gravitational waves, 𝑓MS , in terms of mass of the star in units of solar masses (𝑀MS /𝑀⊙ ) and 𝛼. (T11.3) Using the above result, determine the appropriate value of 𝛼 that will give the maximum possible frequency of gravitational waves, 𝑓MS,max for any main sequence star. Evaluate this frequency. (T11.4) White dwarf (WD) stars have a maximum mass of 1.44 𝑀⊙ (known as the Chandrasekhar limit) and obey the mass-radius relation 𝑅 ∝ 𝑀 −1/3 . The radius of a solar mass white dwarf is equal

9

8

Theoretical Examination Page 5 of 6 to 6000 km. Find the highest frequency of emitted gravitational waves, 𝑓WD,max , if the test mass is orbiting a white dwarf. (T11.5) Neutron stars (NS) are a peculiar type of compact objects which have masses between 1 and 3𝑀⊙ and radii in the range 10 − 15 km. Find the range of frequencies of emitted gravitational waves, 𝑓NS,min and 𝑓NS,max , if the test mass is orbiting a neutron star at a distance close to the neutron star radius.

8

(T11.6) If the test mass is orbiting a black hole (BH), write the expression for the frequency of emitted gravitational waves, 𝑓BH , in terms of mass of the black hole, 𝑀BH , and the solar mass 𝑀⊙ .

7

(T11.7) Based only on the time period (or frequency) of gravitational waves before the epoch of merger, determine whether the central object can be a main sequence star (MS), a white dwarf (WD), a neutron star (NS), or a black hole (BH). Tick the correct option in the Summary Answersheet. Estimate the mass of this object, 𝑀obj , in units of 𝑀⊙ .

5

(T12) AstroSat India astronomy satellite, AstroSat, launched in September 2015, has five different instruments.

In this question, we will discuss three of these instruments (SXT, LAXPC, CZTI), which point in the same direction and observe in X-ray wavelengths. The details of these instruments are given in the table below. Instrument SXT LAXPC

Band [keV] 0.3 – 80 3 – 80

Collecting Area [m2] 0.067 1.5

Effective Photon Detection Efficiency 60% 40%

CZTI

10 – 150

0.09

50%

Saturation level [counts] 15000 (total) 50000 (in any one counter) or 200000 (total) ---

No. of Pixels 512 x 512 ---

4 x 4096

You should note that LAXPC energy range is divided into 8 different energy band counters of equal bandwidth with no overlap. (T12.1) Some X-ray sources like Cas A have a prominent emission line at 0.01825 nm corresponding 13 to a radioactive transition of 44Ti. Suppose there exists a source which emits only one bright emission line corresponding to this transition. What should be the minimum relative velocity (𝑣) of the source, which will make the observed peak of this line to get registered in a different energy band counter of LAXPC as compared to a source at rest? These instruments were used to observe an X-ray source (assumed to be a point source), whose energy spectrum followed the power law, 𝐹 (𝐸 ) = 𝐾𝐸 −2⁄3

[in units of counts/keV/m2 /s ]

where 𝐸 is the energy in keV, 𝐾 is a constant and 𝐹(𝐸) is photon flux density at that energy. Photon flux density, by definition, is given for per unit collecting area (m2 ) per unit bandwidth (keV) and per unit

Theoretical Examination Page 6 of 6 time (seconds). From prior observations, we know that the source has a flux density of 10 counts/keV/m2 /s at 1 keV, when measured by a detector with 100% photon detection efficiency. The “counts” here mean the number of photons reported by the detector. As the source flux follows the power law given above, we know that for a given energy range from 𝐸1 (lower energy) to 𝐸2 (higher energy) the total photon flux (𝐹T ) will be given by 1 ⁄3 1 ⁄3 𝐹T = 3𝐾 (𝐸2 − 𝐸1 ) [in units of counts/m2 /s ] (T12.2) Estimate the incident flux density from the source at 1 keV, 5 keV, 40 keV and 100 keV. Also estimate what will be the total count per unit bandwidth recorded by each of the instruments at these energies for an exposure time of 200 seconds.

8

(T12.3) For this source, calculate the maximum exposure time (𝑡S ), without suffering from saturation, for the CCD of SXT.

4

(T12.4) If the source became 3500 times brighter, calculate the expected counts per second in LAXPC counter 1, counter 8 as well as total counts across the entire energy range. If we observe for longer period, will the counter saturate due to any individual counter or due to the total count? Tick the appropriate box in the Summary Answersheet.

8

(T12.5) Assume that the counts reported by CZTI due to random fluctuations in electronics are about 10 0.00014 counts per pixel per keV per second at all energy levels. Any source is considered as “detected” when the SNR (signal to noise ratio) is at least 3. What is minimum exposure time, 𝑡𝑐 , needed for the source above to be detected in CZTI? Note that the “noise” in a detector is equal to the square root of the counts due to random fluctuations. (T12.6) Let us consider the situation where the source shows variability in number flux, so that the factor 𝐾 increases by 20%. AstroSat observed this source for 1 second before the change and 1 second after this change in brightness. Calculate the counts measured by SXT, LAXPC and CZTI in both the observations. Which instrument is best suited to detect this change? Tick the appropriate box in the Summary Answersheet.

7

Theoretical Examination Page 1 of 27 (T1) True or False Determine if each of the following statements is True or False. In the Summary Answersheet, tick the correct answer (TRUE / FALSE) for each statement. No justifications are necessary for this question. (T1.1) In a photograph of the clear sky on a Full Moon night with a sufficiently long exposure, the colour of the sky would appear blue as in daytime. Solution: T The colour of the clear sky during night is the same as during daytime, since the spectrum of sunlight reflected by the Moon is almost the same as the spectrum of sunlight. Only the intensity is lower.

2

2.0

(T1.2) An astronomer at Bhubaneswar marks the position of the Sun on the sky at 05:00 UT every day of the year. If the Earth’s axis were perpendicular to its orbital plane, these positions would trace an arc of a great circle. Solution: T If the Earth’s axis were perpendicular to its orbital plane, the celestial equator will coincide with ecliptic and the Sun will remain along the celestial equator every day. However, as the Earth’s orbit is elliptical, the true sun would still lead or lag mean sun by a few minutes on different days of year.

2

2.0

(T1.3) If the orbital period of a certain minor body around the Sun in the ecliptic plane is less than the orbital period of Uranus, then its orbit must necessarily be fully inside the orbit of Uranus. Solution: F The semi-major axis of the orbit of the body will be less than that of Uranus. However the minor body’s orbit may have a high eccentricity, in which case it may go outside that of Uranus.

2

2.0

(T1.4) The centre of mass of the solar system is inside the Sun at all times. Solution: F The centre of mass of Sun-Jupiter pair is just outside the Sun. Thus, if all gas giants are on same side of the Sun, the centre of mass of Solar system is definitely outside the Sun.

2

2.0

(T1.5) A photon is moving in free space. As the Universe expands, its momentum decreases. Solution: T For photons the wavelength increases when the Universe expands.

2

2.0

Theoretical Examination Page 2 of 27 (T2) Gases on Titan Gas particles in a planetary atmosphere have a wide distribution of speeds. If the r.m.s. (root mean square) thermal speed of particles of a particular gas exceeds 1/6 of the escape speed, then most of that gas will escape from the planet. What is the minimum atomic weight (relative atomic mass), Amin , of an ideal monatomic gas so that it remains in the atmosphere of Titan?

10

Given, mass of Titan MT = 1.23 × 1023 kg, radius of Titan RT = 2575 km, surface temperature of Titan TT = 93.7 K. Solution: As the gas is monatomic, 3 1 2 kB TT ≈ mg vrms 2
2
Mg 2 v 3kB TT ≈ N rms sA 3kB NA TT ∴ vrms ≈ Mg

4.0

50% deduction if 3/2 pre-factor is not used and 1/2 or 1 are used instead. Full credit if students writes the relation for vrms directly. To remain in atmosphere, vesc 1 vrms < = 6 6 s r 3kB NA TT GMT < Mg 18RT

r

2GMT RT

54kB NA TT RT GMT 54 × 1.381 × 10−23 × 6.022 × 1023 × 93.7 × 2.575 × 106 > g 6.6741 × 10−11 × 1.23 × 1023 > 13.2 g

∴ Mg >

Thus, all gases with atomic weight more than Amin = 13.2 will be retained in the atmosphere of Titan. Half mark for understanding that atomic mass has no units.

4.0

1.5

0.5

Alternative solution 3 1 2 kB TT ≈ mg vrms 2
2
s 3kB TT ∴ vrms ≈ mg 54kB TT RT GMT mg > 2.19 × 10−26 kg

∴ mg >

∴ Amin =

mg 2.19 × 10−26 kg = atomic mass unit 1.66 × 10−27 kg

4.0 4.0 1.0

Theoretical Examination Page 3 of 27

Amin = 13.2

1.0

Answers between 13.0 and 13.4 are acceptable with full credit.

(T3) Early Universe Cosmological models indicate that radiation energy density, ρr , in the Universe is proportional to (1+z)4 , and the matter energy density, ρm , is proportional to (1+z)3 , where z is the redshift. The dimensionless density parameter, Ω, is given as Ω = ρ/ρc , where ρc is the critical energy density of the Universe. In the present Universe, the density parameters corresponding to radiation and matter, are Ωr0 = 10−4 and Ωm0 = 0.3, respectively. (T3.1) Calculate the redshift, ze , at which radiation and matter energy densities were equal.

3

Solution: Ω m0 ρm0 /ρc 0.3 = = −4 = 3000 ρr0 /ρc Ωr0 10 At ze , both matter density and radiation density were equal. ρr = ρm ∴ ρr0 (1 + ze )4 = ρm0 (1 + ze )3 ρm0 1 + ze = = 3000 ρr0 ∴ ze ' 3000

2.0 1.0

Only ze = 2999 and ze = 3000 are acceptable answers. (T3.2) Assuming that the radiation from the early Universe has a blackbody spectrum with a temperature of 2.732 K, estimate the temperature, Te , of the radiation at redshift ze . Solution: As the Universe behaves like an ideal black body, the radiation density will be proportional to the fourth power of the temperature (Stefan’s law).  4 Te ρr = e T0 ρr0 ρr (1 + ze )4 = 0 ρr0  4 Te = (1 + ze )4 2.732 Te = 1 + ze = 3000 2.732 Te = 3000 × 2.732 Te = 8200 K

4

2.0 1.0

1.0

8100 ≤ Te ≤ 8200 gives 1.0; 8200 < Te ≤ 9000 gives 0.5; else 0. (T3.3) Estimate the typical photon energy, Eν (in eV), of the radiation as emitted at redshift ze .

3

Theoretical Examination Page 4 of 27

Solution: Wien’s law: λmax = = Eν = = Eν =

0.002898 m K Te 0.002898 m = 354 nm 8200 hc λmax 6.62 × 10−34 × 3 × 108 5.62 × 10−19 J = eV 354 × 10−9 1.602 × 10−19 3.5 eV

1.0

1.0 1.0

Alternative solution: Eν = kB Te = 1.38 × 10−23 × 8200 J =

1.13 × 10−19 eV 1.602 × 10−19

2.0 1.0

Eν = 0.71 eV Use of either Wien’s law or E = kB T gets full credit. Eν = 3kB T /2 or similar gets no credit. Answers with Eν = 3kB T or Eν = 2.7kB T also get full credit.

(T4) Shadows An observer in the northern hemisphere noticed that the length of the shortest shadow of a 1.000 m vertical stick on a day was 1.732 m. On the same day, the length of the longest shadow of the same vertical stick was measured to be 5.671 m.

10

Find the latitude, φ, of the observer and declination of the Sun, δ , on that day. Assume the Sun to be a point source and ignore atmospheric refraction. Solution: As the longest shadow of the Sun on the given day is of finite length, the Sun is circumpolar for this observer on this day.

φ

90 − δ 90 − δ

A θ1

θ O

θ2 S

In the figure above, the left panel shows the shadow OS formed by stick OA (of length 1.000 m), and the right panel shows the Sun’s location in two cases.

2.0

Theoretical Examination Page 5 of 27

For an altitude θ of the Sun, OA 1.000 m tan θ = = OS OS ∴ cot θ = OS (in metres)

1.0

Let θ1 and θ2 be altitude in two extreme cases. θ1 = 180◦ − φ − (90◦ − δ ) = 90◦ − φ + δ

1.0

◦

cot(90 − φ + δ ) = 1.732 ∴ tan(φ − δ ) = 1.732 φ − δ = tan−1 (1.732) = 60◦ = 1.047 rad θ2 = φ − (90◦ − δ ) = φ − 90◦ + δ

1.5 1.0

◦

cot(φ − 90 + δ ) = 5.671 1 ∴ tan(φ − 90◦ + δ ) = 5.671  −1

φ + δ = tan

1 5.671



+ 90◦ = 100◦ = 1.745 rad

1.5

Solving, φ = 80◦ = 1.396 rad

1.0

δ = 20◦ = 0.349 rad

1.0

Given high accuracy of shadow length, only ±0.5◦ is allowed. One can also solve the question by manipulating tan(φ − δ ) and tan(φ + δ ), to get tan(φ) and tan(δ ).

(T5) GMRT beam transit Giant Metrewave Radio Telescope (GMRT), one of the world’s largest radio telescopes at metre wavelengths, is located in western India (latitude: 19◦ 60 N, longitude: 74◦ 30 E). GMRT consists of 30 dish antennas, each with a diameter of 45.0 m. A single dish of GMRT was held fixed with its axis pointing at a zenith angle of 39◦ 420 along the northern meridian such that a radio point source would pass along a diameter of the beam, when it is transiting the meridian.

10

What is the duration Ttransit for which this source would be within the FWHM (full width at half maximum) of the beam of a single GMRT dish observing at 200 MHz? Hint: The FWHM size of the beam of a radio dish operating at a given frequency corresponds to the angular resolution of the dish. Assume uniform illumination.

Solution: As the dish is pointed towards northern meridian at zenith angle of 39.7◦ , altitude of the centre of the beam is a = 90.00◦ − z = 90.00◦ − 39.70◦ = 50.30◦

1.0

Theoretical Examination Page 6 of 27

Thus, declination of the source should be, δ = 90.00 − a + φ = 90.00◦ − 50.30◦ + 19.10◦ = 58.80◦

2.0

Declination = ZA + Latitude also gets full credit. FWHM beam size (for uniform illumination) will be given by 1.22λ θ= D 1.22c 1.22 × 2.998 × 108 = = Dν 45.0 × 2 × 108 = 0.0406 rad θ = 2.33◦ θ × 3.99 min Ttransit = cos δ 2.33 × 3.99 min = cos 58.8 Ttransit = 17.9 min

1.5

1.5 3.0

1.0

• Use of 4 min per degree is also acceptable. • Missing cos δ gets a penalty of 2.0.

(T6) Cepheid Pulsation The star β-Doradus is a Cepheid variable star with a pulsation period of 9.84 days. We make a simplifying assumption that the star is brightest when it is most contracted (radius being R1 ) and it is faintest when it is most expanded (radius being R2 ). For simplicity, assume that the star maintains its spherical shape and behaves as a perfect black body at every instant during the entire cycle. The bolometric magnitude of the star varies from 3.46 to 4.08. From Doppler measurements, we know that during pulsation the stellar surface expands or contracts at an average radial speed of 12.8 km s−1 . Over the period of pulsation, the peak of thermal radiation (intrinsic) of the star varies from 531.0 nm to 649.1 nm. (T6.1) Find the ratio of radii of the star in its most contracted and most expanded states (R1 /R2 ). Solution: We first find flux ratio and then use Stefan’s law to compare the fluxes.   F1 m1 − m2 = −2.5 log F2 F1 ∴ = 10−0.4(m1 −m2 ) = 10−0.4(3.46−4.08) F2 = 1.77 Li = ∴ F1 =

4πRi2 σTi4 4πR12 σT14 4πD2

7

1.0

1.0 1.0

F2 =

4πR22 σT24 4πD2

1.0

Theoretical Examination Page 7 of 27

R2 T 4 F1 = 12 × 14 F2 R T2 r2  2 R1 T2 F1 = × R2 F2 T1 λ1 T2 = . From Wien’s displacement law, T1 λ2 r  2 R1 F1 λ1 ∴ = × R2 F2 λ2   √ 531.0 2 = 1.77 × 649.1

1.0 1.0

R1 = 0.890 R2

1.0

Acceptable range: ±0.010. (T6.2) Find the radii of the star (in metres) in its most contracted and most expanded states (R1 and R2 ).

3

Solution: R2 − R1 = v × P/2 R2 − R1 = 12.8 × 103 × 86 400 ×

2.0 9.84 m 2

(1 − 0.890)R2 = 5.441 × 109 m ∴ R2 = 4.95 × 1010 m

0.5

R1 = 4.41 × 1010 m

0.5

Acceptable range: ±0.02 × 1010 m for both. (T6.3) Calculate the flux of the star, F2 , when it is in its most expanded state.

5

Solution: To get the absolute value of flux (F2 ) we must compare it with observed flux of the Sun.   F2 m2 − m = −2.5 log F ∴ F2 = F 10−0.4(m2 −m ) L = × 10−0.4(4.08+26.72) 2 4πa⊕ =

3.0

3.826 × 1026 × 4.7863 × 10−13 W m−2 11 2 4π(1.496 × 10 )

F2 = 6.51 × 10−10 W m−2 Acceptable range: ±0.04 × 10−10 W m−2 .

2.0

Theoretical Examination Page 8 of 27 (T6.4) Find the distance to the star, Dstar , in parsecs.

5

Solution: r Dstar =

L2 = 4πF2

Wien’s law: T2 =

s

R22 σT24 = R2 T22 F2

r

σ F2

2.0

2.898 × 10−3 m K λ2

Dstar = 4.95 × 1010 ×



2.898 × 10−3 649.1 × 10−9

1.0 2 s

5.670 × 10−8 6.51 × 10−10

∴ Dstar = 9.208 × 1018 m = 298 pc

2.0

Acceptable range: 298 ± 2pc (depends on truncation).

(T7) Telescope optics In a particular ideal refracting telescope of focal ratio f /5, the focal length of the objective lens is 100 cm and that of the eyepiece is 1 cm. (T7.1) What is the angular magnification, m0 , of the telescope? What is the length of the telescope, L0 , i.e. the distance between its objective and eyepiece? Solution: The magnification will be given by, fo m0 = fe 100 = 100 = 1

4

1.0 1.0

The magnification is m0 = 100 Length of the telescope will be L0 = fo + fe = 100 + 1 = 101 cm

1.0 1.0

The telescope length will be L0 = 101 cm Exact answer required for credit. An introduction of a concave lens (Barlow lens) between the objective lens and the prime focus is a common way to increase the magnification without a large increase in the length of the telescope. A Barlow lens of focal length 1 cm is now introduced between the objective and the eyepiece to double the magnification. (T7.2) At what distance, dB , from the prime focus must the Barlow lens be kept in order to obtain this desired double magnification? Solution: We use the following sign convention. Lens is the origin. Direction along the direction

6

Theoretical Examination Page 9 of 27

1 1 1 = − (f is positive for convex f v u v lens and negative for concave lens). Magnification is m = . Solutions using other u sign conventions are acceptable. Let v be image distance from the Barlow lens. 1 1 1 = − fB v u Distance of Barlow lens, dB , before the prime focus is same as the object distance, u, in this case. of light is taken as positive. The lens formula is

1 1 1 = − fB v dB v v Also, mB = 2 = = u dB 2 1 = ∴ dB v 1 1 2 ∴ = − −1 v v −1 −1 = v v = 1 cm v 1 cm dB = = = 0.5 cm 2 2

1.0 1.0

0.5 1.0

2.5

The positive sign for dB indicates that the Barlow lens was introduced 0.5 cm before the prime focus. (T7.3) What is the increase, ∆L, in the length of the telescope?

4

Solution: The increase in the length will be, ∆L = v − dB = 1.0 − 0.5 = 0.5 cm

2.0 2.0

Thus, the length will be increased by ∆L = 0.5 cm

A telescope is now constructed with the same objective lens and a CCD detector placed at the prime focus (without any Barlow lens or eyepiece). The size of each pixel of the CCD detector is 10 µm. (T7.4) What will be the distance in pixels between the centroids of the images of the two stars, np , on the CCD, if they are 2000 apart on the sky? Solution: Plate scale at prime focus is given by, 1 1 rad s= = = 0.206 265 arcsec/µm fo 1m

6

2.0

Theoretical Examination Page 10 of 27

Since each pixel is 10 µm in size, sp = 10 × 0.206 arcsec/µm = 2.06 arcsec/pixel

2.0

Two stars will be separated by, 2000 pixels ' 10 pixels 2.0600 Acceptable range: 9.5 to 10.5 pixels. np =

2.0

(T8) U-band photometry A star has an apparent magnitude mU = 15.0 in the U -band. The U -band filter is ideal, i.e., it has perfect (100%) transmission within the band and is completely opaque (0% transmission) outside the band. The filter is centered at 360 nm, and has a width of 80 nm. It is assumed that the star also has a flat energy spectrum with respect to frequency. The conversion between magnitude, m, in any band and flux density, f , of a star in Jansky (1 Jy = 1 × 10−26 W Hz−1 m−2 ) is given by f = 3631 × 10−0.4m Jy (T8.1) Approximately how many U -band photons, N0 , from this star will be incident normally on a 1 m2 area at the top of the Earth’s atmosphere every second?

8

Solution: The U -band is defined as (360 ± 40) nm. Thus, the maximum, minimum and average frequencies of the band will be, c νmax = = 9.369 × 1014 Hz λmax νmin = 7.495 × 1014 Hz νavg = 8.432 × 1014 Hz ∆ν = νmax − νmin = 1.874 × 1014 Hz

2.0

fst1 = 3631 × 10−0.4×15 = 3.631 mJy = 3.631 × 10−29 W Hz−1 m−2 Now, N0 × hνavg = ∆ν × fst1 × A × ∆t

2.0 2.0

where, A = 1 m2 & ∆t = 1 s 1.874 × 1014 × 3.631 × 10−29 6.626 × 10−34 × 8.432 × 1014 ' 12180

∴ N0 =

Exact calculation including integration is accepted with full credit (exact answer: 12190). Accepted range: 12180 ± 200. Using flat spectrum for ∆λ instead of ∆ν is considered a major conceptual error, and will incur penalty of 2.0 marks. This star is being observed in the U -band using a ground based telescope, whose primary mirror has a diameter of 2.0 m. Atmospheric extinction in U -band during the observation is 50%. You may assume that the seeing is diffraction limited. Average surface brightness of night sky in U -band was measured to be 22.0 mag/arcsec2 .

2.0

Theoretical Examination Page 11 of 27 (T8.2) What is the ratio, R, of number of photons received per second from the star to that received from the sky, when measured over a circular aperture of diameter 200 ?

8

Solution: Let us call sky flux per square arcsec as Φ and total sky flux for the given aperture as φsky . Let total star flux be φst . φsky = AΦ = π × (1 arcsec)2 × Φ = πΦ   Φ ∴ msky = 22.0 + 2.5 log10 φsky   Φ  = 22.0 + 2.5 log10 π Φ = 22.0 − 2.5 log10 (π) msky = 20.76 mag

3.0 1.0

1.0

As extinction is 50% φst2 0.5φst1 R= = = 0.5 × 10(20.76−15)/2.5 φsky φsky ' 100

1.0

2.0

Accepted range: ±5. A student may also calculate number of photons incident per second per metre square in this case and then compare it with the answer in the first case to get the correct ratio. (T8.3) In practice, only 20% of U -band photons falling on the primary mirror are detected. How many photons, Nt , from the star are detected per second?

4

Solution: Nt × 1 m2 = N0 × 0.5 × 0.2 × At

2.0 

Nt = 12180 × 0.5 × 0.2 × π

2.0 2

2 = 1233π

Nt ' 3813

1.0 1.0

Accepted range (3813 ± 50).

(T9) Mars Orbiter Mission India’s Mars Orbiter Mission (MOM) was launched using the Polar Satellite Launch Vehicle (PSLV) on 5 November 2013. The dry mass of MOM (body + instruments) was 500 kg and it carried fuel of mass 852 kg. It was initially placed in an elliptical orbit around the Earth with perigee at a height of 264 km and apogee at a height of 23 904 km, above the surface of the Earth. After raising the orbit six times, MOM was transferred to a trans-Mars injection orbit (Hohmann orbit). The first such orbit-raising was performed by firing the engines for a very short time near the perigee. The engines were fired to change the orbit without changing the plane of the orbit and without changing its perigee. This gave a net impulse of 1.73 × 105 kg m s−1 to the satellite. Ignore the change in mass due to burning of fuel. (T9.1) What is the height of the new apogee, ha , above the surface of the Earth, after this

14

Theoretical Examination Page 12 of 27 engine burn?

Solution: Let apogee and perigee distances be ra and rp respectively. rp = R⊕ + hip = (6371 + 264) km = 6635 km

1.0

ra = R⊕ + hia = (6371 + 23904) km = 30 275 km

1.0

Conservation of energy and angular momentum gives GmM⊕ E=− rp + ra Total energy at perigee 1 GmM⊕ GmM⊕ =E=− mvp 2 − 2 rp rp + ra   rp 1 2 GM⊕ vp = 1− 2 rp rp + ra s GM⊕ ra ∴ vp = 2 ra + rp rp s 2 × 6.674 × 10−11 × 5.972 × 1024 × 3.0275 × 107 = 6.635 × 106 × (3.0275 + 6.635 × 106 ) = 9.929 km s−1 As the engine burn is just 41.6 s, we assume that the entire impulse is applied instantaneously at perigee. The impulse is J = 1.73 × 105 kg m s−1 . Note that the total mass of MOM must include the fuel, so we have to use m = 500 + 852 = 1352 kg. Change in velocity due to impulse at perigee is J 1.73 × 105 = = 128.0 m s−1 m 1352 The new velocity will be given by (we use 0 symbol to denote quantities after the first orbit-raising maneuvre)

2.0

1.0

1.0

∆v =

1.0

vp0 = vp + ∆v = 10.06 km s−1

1.0

The perigee remains unchanged. So we get rp0 = rp . Since the satellite is moving faster, the new apogee will be higher. s r0 vp0 = 2GM⊕ 0 0 a 0 rp (ra + rp ) ∴1+

rp0 2GM⊕ = 0 2 0 ra (vp ) × rp0 2 × 6.674 × 10−11 × 5.972 × 1024 = 1.188 (10.06 × 103 )2 × 6.635 × 106 6635 ra0 = = 35 380 km 0.188 ha = 35380 − 6371

1.0

2.0

=

= 29 009 km

2.0

1.0

Theoretical Examination Page 13 of 27

Acceptable range: ±150 km (T9.2) Find eccentricity (e) of the new orbit after the burn and new orbital period (P ) of MOM in hours.

6

Solution: As seen above, rp0 1−e = 0.188 = 0 ra 1+e 1 − 0.188 ∴e= = 0.683 1 + 0.188 Acceptable range: ±0.002 The new orbital semi-major axis and orbital period will be, ra0 + rp0 2 35380 + 6635 = = 20 933 km s 2

a0 =

P = 2π s = 2π

1.0 1.0

1.0 1.0

a0 3 GM⊕

1.0

(2.0933 × 107 )3 = 30 136 s = 8.37 h 6.674 × 10−11 × 5.972 × 1024

1.0

Acceptable range: ±0.1 h

(T10) Gravitational Lensing Telescope Einstein’s General Theory of Relativity predicts bending of light around massive bodies. For simplicity, we assume that the bending of light happens at a single point for each light ray, as shown in the figure. The angle of bending, θb , is given by 2Rsch θb = r where Rsch is the Schwarzschild radius associated with that gravitational body. We call r, the distance of the incoming light ray from the parallel x-axis passing through the centre of the body, as the “impact parameter”. θb r x-axis

A massive body thus behaves somewhat like a focusing lens. The light rays coming from infinite distance beyond a massive body, and having the same impact parameter r, converge at a point along the axis, at a distance fr from the centre of the massive body. An observer at that point

Theoretical Examination Page 14 of 27 will benefit from huge amplification due to this gravitational focusing. The massive body in this case is being used as a Gravitational Lensing Telescope for amplification of distant signals. (T10.1) Consider the possibility of our Sun as a gravitational lensing telescope. Calculate the shortest distance, fmin , from the centre of the Sun (in A.U.) at which the light rays can get focused.

6

Solution: The rays travelling closer to the gravitational body will bend more. Thus, we get shortest convergence point where the rays just grazing the solar surface will meet each other. θb R θb fmin

2Rsch R ' R fmin 2 R2 c2 R = = 2Rsch 4GM (6.955 × 108 × 2.998 × 108 )2 = m 4 × 6.674 × 10−11 × 1.989 × 1030 8.188 × 1013 AU = 8.188 × 1013 m = 1.496 × 1011 = 547.3 AU

θb = ∴ fmin

fmin

2.0 2.0

2.0

(T10.2) Consider a small circular detector of radius a, kept at a distance fmin centred on the x-axis and perpendicular to it. Note that only the light rays which pass within a certain annulus (ring) of width h (where h  R ) around the Sun would encounter the detector. The amplification factor at the detector is defined as the ratio of the intensity of the light incident on the detector in the presence of the Sun and the intensity in the absence of the Sun. Express the amplification factor, Am , at the detector in terms of R and a. Solution: The following figure needs to be drawn. h

8

1.0

θ2 Detector a

R fmin f2

Theoretical Examination Page 15 of 27

The light bending from the surface of the Sun (r = R ) will intersect the detector at its centre, as it is kept at fmin . The boundary of the detector will be intersected by a light ray with r = R + h. This ray will intersect the x-axis at a distance f2 . f2 =

(R + h)2 2Rsch

2.0

Same argument as in Part 1 For small angles, a = (f2 − fmin )θ2  2  R (R + h)2 2Rsch 2R h + h2 = − = 2Rsch 2Rsch (R + h) R + h ' 2h Let original intensity of the incoming radiation be I0 . The flux at the detector in the presence of Sun is Φ = I0 2π R h The flux at the detector in the absence of Sun is Φ0 = I0 π a2 The amplification is therefore Am =

I0 2πR h R Φ = = Φ0 I0 πa2 a

2.0 1.0 1.0

1.0

(T10.3) Consider a spherical mass distribution, such as a dark matter cluster, through which light rays can pass while undergoing gravitational bending. Assume for simplicity that for the gravitational bending with impact parameter, r, only the mass M (r) enclosed inside the radius r is relevant.

6

What should be the mass distribution, M (r), such that the gravitational lens behaves like an ideal optical convex lens ? Solution: θ2 θ1 r1 r2 f1 = f2

All rays should focus at the same spot. This should be evident from figure drawn on answersheet or otherwise. Let there be two rays with impact parameters r1 and r2 . The corresponding distances of focus will be ri2 ri2 c2 fi = = 2rschi 4GM (ri ) Same argument as in Part 1

2.0

2.0

Theoretical Examination Page 16 of 27

The rquirement f1 = f2 implies r12 M (r1 ) = M (r2 ) r22 The required mass distribution is: M (r) ∝ r2

1.0 1.0

(T11) Gravitational Waves The first signal of gravitational waves was observed by two advanced LIGO detectors at Hanford and Livingston, USA in September 2015. One of these measurements (strain vs time in seconds) is shown in the accompanying figure. In this problem, we will interpret this signal in terms of a small test mass m orbiting around a large mass M (i.e., m  M ), by considering several models for the nature of the central mass.

The test mass loses energy due to the emission of gravitational waves. As a result the orbit keeps on shrinking, until the test mass reaches the surface of the object, or in the case of a black hole, the innermost stable circular orbit – ISCO – which is given by RISCO = 3Rsch , where Rsch is the Schwarzschild radius of the black hole. This is the “epoch of merger”. At this point, the amplitude of the gravitational wave is maximum, and so is its frequency, which is always twice the orbital frequency. In this problem, we will only focus on the gravitational waves before the merger, when Kepler’s laws are assumed to be valid. After the merger, the form of gravitational waves will drastically change. (T11.1) Consider the observed gravitational waves shown in the figure above. Estimate the time period, T0 , and hence calculate the frequency, f0 , of gravitational waves just before the epoch of merger.

3

Theoretical Examination Page 17 of 27

Solution: From the graph, just before the peak of emission, the time period of gravitational waves is approximately (0.007 ± 0.004) s. T0 ≈ 0.007 s

2.0

Acceptable range: 0.003 to 0.011 s That is, the frequency of the gravitational waves is f0 ≈ 142.86 Hz . Acceptable range: 333.33 to 90.91 Hz Answer given in terms of angular frequency with correct value and units gains full credit. (T11.2) For any main sequence (MS) star, the radius of the star, RMS , and its mass, MMS , are related by a power law given as,

1.0

10

RMS ∝ (MMS )α where α = 0.8 = 1.0

for M < MMS for 0.08M ≤ MMS ≤ M

If the central object were a main sequence star, write an expression for the maximum frequency of gravitational waves, fMS , in terms of mass of the star in units of solar masses (MMS /M ) and α. Solution: Since m  M then, by Kepler’s third law r 1 GM forbital = 2π r3 Hence the frequency of the gravitational waves is r 1 GM fgrav = 2forbital = π r3 The frequency will be maximum when r = RMS . For main sequence stars,   RMS MMS α = R M   MMS α ∴ RMS = R M s   1 GMMS M 3α/2 ∴ fMS = 3 π MMS R s   1 GM M (3α−1)/2 = 3 π MMS R s   1 GM MMS (1−3α)/2 fMS = 3 π M R

(T11.3) Using the above result, determine the appropriate value of α that will give the maximum possible frequency of gravitational waves, fMS,max for any main sequence star. Evaluate

4.0

1.0 1.0

1.0

3.0

9

Theoretical Examination Page 18 of 27 this frequency. Solution: For possible values of α given in the question, the exponent 1−3α is negative. Thus, 2 if MMS > M , the frequency will be smaller. Thus, for highest possible frequency coming from a main sequence star, you should take lowest possible mass i.e. α = 1.0 4.0

fMS,max

 1−3×1 MMS ( 2 ) M

1 = π

s

GM 3 R

1 = π

s

GM M × 3 MMS R



2.0

The frequency of gravitational waves will be given by, s 1 6.674 × 10−11 × 1.989 × 1030 1 fMS,max = × 8 3 π 0.08 (6.955 × 10 ) fMS,max = 2.5 mHz

3.0

Answer given in terms of angular frequency with correct value and units gains full credit. (T11.4) White dwarf (WD) stars have a maximum mass of 1.44M (known as the Chandrasekhar limit) and obey the mass-radius relation R ∝ M −1/3 . The radius of a solar mass white dwarf is equal to 6000 km. Find the highest frequency of emitted gravitational waves, fWD,max , if the test mass is orbiting a white dwarf. Solution: The maximum frequency would be when r = RWD . We use the notation RW D for the radius of a solar mass white dwarf. Then for white dwarfs MW D 3 3 RW D = RW D M s 1 GMWD fWD = 3 π RW D s 1 GM MW D = 3 π RW D M For maximum frequency, we have to take highest white dwarf mass. s 1.989 × 1030 fWD,max = 2.600 × 10−6 × × 1.44 (6000 × 103 )3

8

1.0

1.0

2.0 2.0

= 2.600 × 10−6 × 95.96 × 103 × 1.44 fWD,max = 0.359 Hz

2.0

Answer given in terms of angular frequency with correct value and units gains full credit. (T11.5) Neutron stars (NS) are a peculiar type of compact objects which have masses between

8

Theoretical Examination Page 19 of 27 1 and 3M and radii in the range 10 – 15 km. Find the range of frequencies of emitted gravitational waves, fNS,min and fNS,max , if the test mass is orbiting a neutron star at a distance close to the neutron star radius. Solution:s 1 GMNS fgrav = 3 π RNS The lowest possible frequency is when MNS is lowest and RNS is the highest. For MNS = M and R = 15 km, we get fNS,min = 1.996 kHz Similarly, the largest possible frequency is when MNS is the largest and RNS is the smallest. For MNS = 3M and R = 10 km, we get fNS,max = 6.352 kHz

2.0 2.0 2.0 2.0

Answer given in terms of angular frequency with correct value and units gains full credit. (T11.6) If the test mass is orbiting a black hole (BH), write the expression for the frequency of emitted gravitational waves, fBH , in terms of mass of the black hole, MBH , and the solar mass M .

7

Solution: For black holes, we have to consider RISCO . Hence the equation will be, s 1 GM M fBH = × × 3 π MBH 27Rsch− fBH = 4.396 kHz ×

M MBH

2.0

2.0 3.0

Answer given in terms of angular frequency with correct value and units gains full credit. (T11.7) Based only on the time period (or frequency) of gravitational waves before the epoch of merger, determine whether the central object can be a main sequence star (MS), a white dwarf (WD), a neutron star (NS), or a black hole (BH). Tick the correct option in the Summary Answersheet. Estimate the mass of this object, Mobj , in units of M .

5

Solution: We found the frequency of the LIGO-detected wave to be 166.67 Hz just before merger. As per our analysis above, only black holes can lead to emission in this frequency range. Black Hole By using corresponding expression, 4396 Mobj = M ≈ 31M 142.86 Any answer between 13 to 50 will get full credit.

2.0 3.0

Theoretical Examination Page 20 of 27 (T12) Exoplanets Two major methods of detection of exoplanets (planets around stars other than the Sun) are the radial velocity (or so-called “wobble”) method and the transit method. In this problem, we find out how a combination of the results of these two methods can reveal a lot of information about an orbiting exoplanet and its host star. Throughout this problem, we consider the case of a planet of mass Mp and radius Rp moving in a circular orbit of radius a around a star of mass Ms (Ms  Mp ) and radius Rs . The normal to the orbital plane of the planet is inclined at angle i with respect to the line of sight (i = 90◦ would mean “edge on” orbit). We assume that there is no other planet orbiting the star and Rs  a. “Wobble” Method: When a planet and a star orbit each other around their barycentre, the star is seen to move slightly, or “wobble”, since the centre of mass of the star is not coincident with the barycentre of the star-planet system. As a result, the light received from the star undergoes a small Doppler shift related to the velocity of this wobble. The line of sight velocity, vl , of the star can be determined from the Doppler shift of a known spectral line, and its periodic variation with time, t, is shown in the schematic diagram below. In the diagram, the two measurable quantities in this method, namely, the orbital period P and maximum line of sight velocity v0 are shown. vl P

v0 t

(T12.1) Derive expressions for the orbital radius (a) and orbital speed (vp ) of the planet in terms of Ms and P . Solution: Kepler’s law:   GMs 2 1/3 a= P 4π 2 Gravitational force provides centripetal acceleration: r GMs vp = a   2πGMs 1/3 vp = P

3

1.0

0.5 1.5

Theoretical Examination Page 21 of 27 (T12.2) Obtain a lower limit on the mass of the planet, Mp, min in terms of Ms , v0 and vp .

4

Solution: Momentum conservation: Mp v p = M s v s

1.5

Observed quantity is v0 = vs sin i. Thus, Mp, min = Mp sin i =

Ms v 0 Ms vs sin i = vp vp

This is a lower limit on Mp .

2.5

Transit Method: As a planet orbits its host star, for orientations of the orbital plane that are close to “edge-on” (i ≈ 90◦ ), it will pass periodically, or “transit”, in front of the stellar disc as seen by the observer. This would cause a tiny decrease in the observed stellar flux which can be measured. The schematic diagram below (NOT drawn to scale) shows the situation from the observer’s perspective and the resulting transit light curve (normalised flux, f , vs time, t) for a uniformly bright stellar disc.

Rs

bRs 1

2

3

4

Rp f 1 ∆ tF tT t If the inclination angle i is exactly 90◦ , the planet would be seen to cross the stellar disc along a diameter. For other values of i, the transit occurs along a chord, whose centre lies at a distance bRs from the centre of the stellar disc, as shown. The no-transit flux is normalised to 1 and the maximum dip during the transit is given by ∆.

Theoretical Examination Page 22 of 27 The four significant points in the transit are the first, second, third and fourth contacts, marked by the positions 1 to 4, respectively, in the figure above. The time interval during the second and third contacts is denoted by tF , when the disc of the planet overlaps the stellar disc fully. The time interval between the first and fourth contacts is denoted by tT . These points are also marked in the schematic diagram below showing a “side-on” view of the orbit (NOT drawn to scale).

i

4 3

Observer

a 1

2

The measurable quantities in the transit method are P , tT , tF and ∆. (T12.3) Find the constraint on i in terms of Rs and a for the transit to be visible at all to the distant observer.

2

Solution: bRs = a cos i Therefore, for visibility, 0 ≤ b ≤ 1 ⇒ i ≥

1.0 cos−1 (Rs /a)

1.0

(T12.4) Express ∆ in terms of Rs and Rp . Solution: Blackbody ⇒ brightness is proportional to area. Since the observer is far away from the star-planet system, size of silhouette of planet on stellar disc is independent of a.  2 Rp ∆= Rs

1

1.0

(T12.5) Express tT and tF in terms of Rs , Rp , a, P and b. Solution: Circular orbit ⇒ uniform orbital speed aφ φ t = = ⇒ P 2πa 2π where φ is the angle subtended by the planet at the centre of the star during transit (over time t).

8

1.0

Theoretical Examination Page 23 of 27

4

φ14

Rs − Rp

Rs + Rp

a

l14

φ23

3 l23

bRs 1

2

3

l14 /2

4

1 2

l23 /2

(l23 /2)2 = (Rs − Rp )2 − (bRs )2 q l23 = 2Rs (1 − Rp /Rs )2 − b2 sin(φ23 /2) =

2.0

l23 /2 a

⇒ φ23 = 2 sin−1



2.0

l23 2a



 = 2 sin−1 

Rs a

s 1−

Rp Rs

2

 − b2 

 s   2 R P P R p s tF = φ23 = sin−1  1− − b2  2π π a Rs

1.0

1.0

Similarly,  tT =

Rs P sin−1  π a

s 1+

Rp Rs



2

− b2 

1.0

(T12.6) In the approximation of an orbit much larger than the stellar radius, show that the parameter b is given by   2 1/2 tF 1+  √ tT    b = 1 + ∆ − 2 ∆  2   tF  1− tT Solution: Since Rs  a, use sin−1 x ≈ x.  s    Rp 2 P  Rs 1+ − b2  tT ≈ π a Rs  tF ≈

P  Rs π a

s 1−

Rp Rs

2

 − b2 

5

2.0

Theoretical Examination Page 24 of 27 √ Dividing, and putting Rp /Rs = ∆, " #1/2 √ (1 − ∆)2 − b2 tF √ = tT (1 + ∆)2 − b2   (1 −  ⇒b= 

√

1.0

    2 1/2 √ 2 1/2 tF 2 tF (1 + ∆)  − 1+  √ tT tT      = 1 + ∆ − 2 ∆  2  2    tF tF  1− 1− tT tT

∆)2



2.0

Expressions lacking the use of approximation Rs /a  1, but otherwise correct will get a penalty of 2.0. Use of approximation with proper justification at a later stage than at the first step will get full credit. (T12.7) Use the result of part (T12.6) to obtain an expression for the ratio a/Rs in terms of measurable transit parameters, using a suitable approximation.

3

Solution:  tT =

P  Rs π a

s 1+

Rp Rs



2

− b2 

1.0

Either substitution of b or elimination of b gets 1.0. Substituting b and Rp /Rs ,  v  2  u u tF   u 1+ u √ √  P  Rs u tT  2 tT =  u(1 + ∆) − 1 − ∆ + 2 ∆  2  π a t tF   1− tT 1/2

 P Rs   ⇒ tT =  π a 

⇒

√ 4 ∆    2  tF  1− tT

a 2P ∆1/4 = Rs π(t2T − t2F )1/2

2.0

Expressions lacking the use of approximation Rs /a  1, but otherwise correct will get a penalty of 1.0. If penalty has already been imposed in part (T12.6), no further penalty for lack of approximation. (T12.8) Combine the results of the wobble method and the transit method to determine the Ms stellar mean density ρs ≡ in terms of tT , tF , ∆ and P . 4πRs3 /3

6

Theoretical Examination Page 25 of 27

Solution: From part (T12.7) a = Rs

2P ∆1/4 π(t2T − t2F )1/2

From part (T12.1)   GMs 2 1/3 a= P 4π 2 Combining, #3 " 2P ∆1/4 GMs 2 P = Rs 4π 2 π(t2T − t2F )1/2

3.0

Identifying the two equations for combining gets credit. No credit for writing only one equation or irrelevant equations. ⇒ ρs ≡

⇒ ρs =

Ms 3 4π 2 8P 3 ∆3/4 = 4πRs3 /3 4π P 2 G π 3 (t2T − t2F )3/2 24 P (∆)3/4 π 2 G (t2T − t2F )3/2

2.0

1.0

Rocky or gaseous: Let us consider an edge-on (i = 90◦ ) star-planet system (circular orbit for the planet), as seen from the Earth. It is known that the host star is of mass 1.00M . Transits are observed with a period (P ) of 50.0 days and total transit duration (tT ) of 1.00 hour. The transit depth (∆) is 0.0064. The same system is also observed in the wobble method to have a maximum line of sight velocity of 0.400 m s−1 . (T12.9) Find the orbital radius a of the planet in units of AU and in metres. Solution: From Kepler’s third law (with same mass of host star):   P 2/3 a = a⊕ P⊕  2/3 50.0 × 1 AU = 0.266 AU a= 365.242 = 0.266 × 1.496 × 1011 m = 3.97 × 1010 m

2

1.0 0.5 0.5

(T12.10) Find the ratio tF /tT of the system. Solution: Edge-on ⇒ b = 0 " #1/2 √ √ (1 − ∆)2 − b2 1− ∆ tF √ √ = 0.8519 = = tT (1 + ∆)2 − b2 1+ ∆

2

1.0 1.0

Theoretical Examination Page 26 of 27 (T12.11) Obtain the mass Mp and radius Rp of the planet in terms of the mass (M⊕ ) and radius (R⊕ ) of the Earth respectively. Is the composition of the planet likely to be rocky or gaseous? Tick the box for ROCKY or GASEOUS in the Summary Answersheet.

8

Solution: From parts (T12.1) and (T12.9) s r GM 6.674 × 10−11 × 1.989 × 1030 = = 57.798 km s−1 vp = a 3.97 × 1010

1.0

Assumption of small planet (Mp  Ms ) is valid because ∆ is very small; less dense planet would make the assumption stronger!

Mp =

Ms v 0 1.989 × 1030 × 0.400 = M⊕ = 2.30 M⊕ vp 5.7798 × 104 × 5.972 × 1024

2.0

From part (T12.4), √ Rp = Rs ∆ From part (T12.7), 2P ∆1/4 a 2P ∆1/4 = = Rs π(t2T − t2F )1/2 πtT (1 − (tF /tT )2 )1/2 ∴ Rs =

aπtT (1 − (tF /tT )2 )1/2 2P ∆1/4

Combining, aπtT (1 − (tF /tT )2 )1/2 ∆1/2 2P ∆1/4 aπtT (1 − (tF /tT )2 )1/2 ∆1/4 = 2P 1 10 3.97 × 10 × π × 24 × (1 − 0.85192 )1/2 × (0.0064)1/4 = R⊕ 2 × 50.0 × 6.371 × 106 = 1.21 R⊕

Rp =

2.0

1.0

Mean density ρp =

Mp 2.30 = ρ⊕ = 1.3ρ⊕ 3 4πRp /3 (1.21)3

Since mean density is higher than that of Earth, the planet is Rocky .

1.0 1.0

Transit light curves with starspots and limb darkening: (T12.12) Consider a planetary transit with i = 90◦ around a star which has a starspot on its equator, comparable to the size of the planet, Rp . The rotation period of the star is 2P . Draw schematic diagrams of the transit light curve for five successive transits of the planet (in the templates provided in the Summary Answersheet). The no-transit flux for each transit may be normalised to unity independently. Assume that the planet does not encounter the starspot on the first transit but does in the second.

4

Theoretical Examination Page 27 of 27

Solution: f

Transit no. 1

f

Transit no. 2

t

f

Transit no. 3

t

f

Transit no. 4

t

f

t

Transit no. 5

t

• No change in the first: 0.5 • Spike in second (width and phase of spike is arbitrary): 1.0 • Height of spike (almost) equal to maximum dip: 0.5 • No change in third: 0.5 • Spike again in fourth: 0.5 • Same phase of spike in second and fourth: 0.5 • No change in 5th: 0.5 (T12.13) Throughout the problem we have considered a uniformly bright stellar disc. However, real stellar discs have limb darkening. Draw a schematic transit light curve when limb darkening is present in the host star. Solution: f

t

Non-flat bottom with central minimum gets 2.0. Curvature of ingress and egress are tolerated.

2

Data Analysis Examination Page 1 of 4 (D1) Binary Pulsar Through systematic searches during the past decades, astronomers have found a large number of millisecond pulsars (spin period < 10 ms). Majority of these pulsars are found in binaries, with nearly circular orbits. For a pulsar in a binary orbit, the measured pulsar spin period (𝑃) and the measured line-of-sight acceleration (𝑎) both vary systematically due to orbital motion. For circular orbits, this variation can be described mathematically in terms of orbital phase 𝜙 (0 ≤ 𝜙 ≤ 2𝜋) as, 2𝜋𝑃0 𝑟 𝑃(𝜙 ) = 𝑃0 + 𝑃t 𝑐𝑜𝑠𝜙 where 𝑃𝑡 = 𝑐𝑃B

4𝜋 2 𝑟 𝑃B2 where 𝑃B is the orbital period of the binary, 𝑃0 is the intrinsic spin period of the pulsar and 𝑟 is the radius of the orbit. 𝑎 (𝜙) = −𝑎t 𝑠𝑖𝑛𝜙

where 𝑎t =

The following table gives one such set of measurements of 𝑃 and 𝑎 at different heliocentric epochs, 𝑇, expressed in truncated Modified Julian Days (tMJD), i.e. number of days since MJD = 2,440,000. No. 1 2 3 4 5 6 7 8 9

T (tMJD) 5740.654 5740.703 5746.100 5746.675 5981.811 5983.932 6005.893 6040.857 6335.904

P (μs) 7587.8889 7587.8334 7588.4100 7588.5810 7587.8836 7587.8552 7589.1029 7589.1350 7589.1358

a (m s-2) - 0.92 ± 0.08 - 0.24 ± 0.08 - 1.68 ± 0.04 + 1.67 ± 0.06 + 0.72 ± 0.06 - 0.44 ± 0.08 + 0.52 ± 0.08 + 0.00 ± 0.04 + 0.00 ± 0.02

By plotting 𝑎(𝜙) as a function of 𝑃(𝜙), we can obtain a parametric curve. As evident from the relations above, this curve in the period-acceleration plane is an ellipse. In this problem, we estimate the intrinsic spin period, 𝑃0 , the orbital period, 𝑃B , and the orbital radius, 𝑟, by an analysis of this data set, assuming a circular orbit. (D1.1) Plot the data, including error bars, in the period-acceleration plane (mark your graph as “D1.1”).

7

(D1.2) Draw an ellipse that appears to be a best fit to the data (on the same graph “D1.1”).

2

(D1.3) From the plot, estimate 𝑃0 , 𝑃t and 𝑎t , including error margins.

7

(D1.4) Write expressions for 𝑃B and 𝑟 in terms of 𝑃0 , 𝑃t , 𝑎t .

4

(D1.5) Calculate approximate value of 𝑃B and 𝑟 based on your estimations made in (D1.3), including error margins.

6

(D1.6) Calculate orbital phase, 𝜙, corresponding to the epochs of the following five observations in the above table: data rows 1, 4, 6, 8, 9.

4

(D1.7) Refine the estimate of the orbital period, 𝑃B , using the results in part (D1.6) in the followin g way: (D1.7a) First determine the initial epoch, 𝑇0 , which corresponds to the nearest epoch of zero orbital phase before the first observation.

2

(D1.7b) The expected time, 𝑇calc , of the estimated orbital phase angle of each observation is given by,

7

𝑇calc = 𝑇0 + (𝑛 +

𝜙 360 ∘

) 𝑃B ,

Data Analysis Examination Page 2 of 4 where n is the number of full cycle of orbital phases that may have elapsed between 𝑇0 and 𝑇 (or 𝑇calc ). Estimate n and 𝑇calc for each of the five observations in part (D1.6). Note down difference 𝑇O−C between observed 𝑇 and 𝑇calc . Enter these calculations in the table given in the Summary Answersheet. (D1.7c) Plot 𝑇O−C against 𝑛 (mark your graph as “D1.7”).

4

(D1.7d) Determine the refined values of the initial epoch, 𝑇0 ,r, and the orbital period, 𝑃B,r.

7

(D2) Distance to the Moon Geocentric ephemerides of the Moon for September 2015 are given in the form of a table. Each reading was taken at 00:00 UT. Date Sep 01 Sep 02 Sep 03 Sep 04 Sep 05 Sep 06 Sep 07 Sep 08 Sep 09 Sep 10 Sep 11 Sep 12 Sep 13 Sep 14 Sep 15 Sep 16 Sep 17 Sep 18 Sep 19 Sep 20 Sep 21 Sep 22 Sep 23 Sep 24 Sep 25 Sep 26 Sep 27 Sep 28 Sep 29 Sep 30

h 0 1 2 3 4 5 6 7 7 8 9 10 11 11 12 13 14 14 15 16 17 18 19 20 21 22 23 0 1 2

R.A. m 36 33 30 27 23 19 14 7 59 49 37 23 9 54 39 25 11 58 47 38 31 26 22 19 16 14 12 10 9 7

(α) s 46.02 51.34 45.03 28.48 52.28 37.25 19.23 35.58 11.04 0.93 11.42 57.77 41.86 49.80 50.01 11.64 23.13 50.47 54.94 50.31 40.04 15.63 17.51 19.45 55.43 46.33 43.63 48.32 5.89 39.02

Dec. (δ) ' '' 3 6 16.8 7 32 26.1 11 25 31.1 14 32 4.3 16 43 18.2 17 55 4.4 18 7 26.6 17 23 55.6 15 50 33.0 13 34 55.6 10 45 27.7 7 30 47.7 3 59 28.8 0 19 50.2 -3 20 3.7 -6 52 18.8 -10 9 4.4 -13 2 24.7 -15 24 14.6 -17 6 22.8 -18 0 52.3 -18 0 41.7 -17 0 50.6 -14 59 38.0 -11 59 59.6 -8 10 18.3 -3 44 28.7 0 58 58.2 5 38 54.3 9 54 16.1 ∘

Angular Size (θ) Phase (ϕ) '' 1991.2 0.927 1974.0 0.852 1950.7 0.759 1923.9 0.655 1896.3 0.546 1869.8 0.438 1845.5 0.336 1824.3 0.243 1806.5 0.163 1792.0 0.097 1780.6 0.047 1772.2 0.015 1766.5 0.001 1763.7 0.005 1763.8 0.026 1767.0 0.065 1773.8 0.120 1784.6 0.189 1799.6 0.270 1819.1 0.363 1843.0 0.463 1870.6 0.567 1900.9 0.672 1931.9 0.772 1961.1 0.861 1985.5 0.933 2002.0 0.981 2008.3 1.000 2003.6 0.988 1988.4 0.947

Elongation Of Moon 148.6 ∘ W 134.7 ∘ W 121.1 ∘ W 107.9 ∘ W 95.2 ∘ W 82.8 ∘ W 70.7 ∘ W 59.0 ∘ W 47.5 ∘ W 36.2 ∘ W 25.1 ∘ W 14.1 ∘ W 3.3 ∘ W 7.8 ∘ E 18.6 ∘ E 29.5 ∘ E 40.4 ∘ E 51.4 ∘ E 62.5 ∘ E 73.9 ∘ E 85.6 ∘ E 97.6 ∘ E 110.0 ∘ E 122.8 ∘ E 136.2 ∘ E 150.0 ∘ E 164.0 ∘ E 178.3 ∘ E 167.4 ∘ W 153.2 ∘ W

The composite graphic1 below shows multiple snapshots of the Moon taken at different times during the total lunar eclipse, which occurred in this month. For each shot, the centre of frame was coinciding with the central north-south line of umbra. For this problem, assume that the observer is at the centre of the Earth and angular size refers to angular diameter of the object / shadow.

1

Credit: NASA’s Scientific Visualization Studio

Data Analysis Examination Page 3 of 4

(D2.1) In September 2015, apogee of the lunar orbit is closest to New Moon / First Quarter / Full Moon / Third Quarter. Tick the correct answer in the Summary Answersheet. No justification for your answer is necessary.

3

(D2.2) In September 2015, the ascending node of lunar orbit with respect to the ecliptic is closest to New Moon / First Quarter / Full Moon / Third Quarter. Tick the correct answer in the Summary Answersheet. No justification for your answer is necessary.

4

(D2.3) Estimate the eccentricity, 𝑒, of the lunar orbit from the given data.

4

(D2.4) Estimate the angular size of the umbra, 𝜃umbra , in terms of the angular size of the Moon, 𝜃Moon . Show your working on the image given on the backside of the Summary Answersheet.

8

(D2.5) The angle subtended by the Sun at Earth on the day of the lunar eclipse is known to be 𝜃Sun = 1915.0′′. In the figure below, 𝑆1 𝑅1 and 𝑆2 𝑅2 are rays coming from diametrically opposite ends of the solar disk. The figure is not to scale.

9

Calculate the angular size of the penumbra, 𝜃penumbra , in terms of 𝜃Moon . Assume the observer to be at the centre of the Earth. (D2.6) Let 𝜃Earth be angular size of the Earth as seen from the centre of the Moon. Calculate the angular size of the Moon, 𝜃Moon , as would be seen from the centre of the Earth on the eclipse day in terms of 𝜃Earth .

5

(D2.7) Estimate the radius of the Moon, 𝑅Moon , in km from the results above.

3

(D2.8) Estimate the shortest distance, 𝑟perigee , and the farthest distance, 𝑟apogee , to the Moon.

4

(D2.9) Use appropriate data from September 10 to estimate the distance, 𝑑Sun , to the Sun from the Earth. (D3) Type IA Supernovae Supernovae of type Ia are considered very important for the measurements of large extragalactic distances. The brightening and subsequent dimming of these explosions follow a characteristic light curve, which helps in identifying these as supernovae of type Ia. Light curves of all type Ia supernovae can be fit to the same model light curve, when they are scaled appropriately. In order to achieve this, we first have to express the light curves in the reference frame of the host galaxy by taking care of the cosmological stretching/dilation of all observed time intervals, Δ𝑡obs , by a factor of (1 + 𝑧). The time interval in the rest frame of the host galaxy is denoted by Δ𝑡gal .

10

Data Analysis Examination Page 4 of 4 The rest frame light curve of a supernova changes by two magnitudes compared to the peak in a time interval Δ𝑡0 after the peak. If we further scale the time intervals by a factor of s (i.e. Δ𝑡𝑠 = 𝑠Δ𝑡gal ) such that the scaled value of Δ𝑡0 is the same for all supernovae, the light curves turn out to have the same shape. It also turns out that 𝑠 is related linearly to the absolute magnitude, 𝑀peak , at the peak luminosity for the supernova. That is, we can write 𝑠 = 𝑎 + 𝑏𝑀peak , where 𝑎 and 𝑏 are constants. Knowing the scaling factor, one can determine absolute magnitudes of supernovae at unknown distances from the above linear equation. The table below contains data for three supernovae, including their distance moduli, 𝜇 (for the first two), their recession speed, 𝑐𝑧, and their apparent magnitudes, 𝑚obs , at different times. The time Δ𝑡obs ≡ 𝑡 − 𝑡peak shows number of days from the date at which the respective supernova reached peak brightness. The observed magnitudes have already been corrected for interstellar as well as atmospheric extinction. Name μ (mag) cz (km s-1) Δtobs (days) -15.00 -10.00 -5.00 0.00 5.00 10.00 15.00 20.00 25.00 30.00

SN2006TD 34.27 4515 mobs (mag) 19.41 17.48 16.12 15.74 16.06 16.72 17.53 18.08 18.43 18.64

SN2006IS 35.64 9426 mobs (mag) 18.35 17.26 16.42 16.17 16.41 16.82 17.37 17.91 18.39 18.73

SN2005LZ 12060 mobs (mag) 20.18 18.79 17.85 17.58 17.72 18.24 18.98 19.62 20.16 20.48

(D3.1) Compute Δ𝑡gal values for all three supernovae, and fill them in the given blank boxes in the data tables on the BACK side of the Summary Answersheet. On a graph paper, plot the points and draw the three light curves in the rest frame (mark your graph as “D3.1”).

15

(D3.2) Take the scaling factor, 𝑠2 , for the supernova SN2006IS to be 1.00. Calculate the scaling factors, 𝑠1 and 𝑠3 , for the other two supernovae SN2006TD and SN2005LZ, respectively, by calculating Δ𝑡0 for them.

5

(D3.3) Compute the scaled time differences, Δ𝑡s , for all three supernovae. Write the values for Δ𝑡s in the same data tables on the Summary Answersheet. On another graph paper, plot the points and draw the 3 light curves to verify that they now have an identical profile (mark your graph as “D3.3”).

14

(D3.4) Calculate the absolute magnitudes at peak brightness, 𝑀peak ,1, for SN2006TD and 𝑀peak ,2 , for SN2006IS. Use these values to calculate 𝑎 and 𝑏.

6

(D3.5) Calculate the absolute magnitude at peak brightness, 𝑀peak,3 , and distance modulus, 𝜇 3, for SN2005LZ.

4

(D3.6) Use the distance modulus 𝜇 3 to estimate the value of Hubble's constant, 𝐻0 . Further, estimate the characteristic age of the universe, 𝑇H .

6

Data Analysis Examination Page 1 of 20 (D1) Binary Pulsar Through systematic searches during the past decades, astronomers have found a large number of millisecond pulsars (spin period < 10 ms). Majority of these pulsars are found in binaries, with nearly circular orbits. For a pulsar in a binary orbit, the measured pulsar spin period (P ) and the measured line-of-sight acceleration (a) both vary systematically due to orbital motion. For circular orbits, this variation can be described mathematically in terms of orbital phase φ (0 ≤ φ ≤ 2π) as, 2πP0 r P (φ) = P0 + Pt cos φ where Pt = cPB 4π 2 r a(φ) = −at sin φ where at = PB2 where PB is the orbital period of the binary, P0 is the intrinsic spin period of the pulsar and r is the radius of the orbit. The following table gives one such set of measurements of P and a at different heliocentric epochs, T , expressed in truncated Modified Julian Days (tMJD), i.e. number of days since MJD = 2,440,000.

No. 1 2 3 4 5 6 7 8 9

T (tMJD) 5740.654 5740.703 5746.100 5746.675 5981.811 5983.932 6005.893 6040.857 6335.904

P (µs) 7587.8889 7587.8334 7588.4100 7588.5810 7587.8836 7587.8552 7589.1029 7589.1350 7589.1358

a (m s−2 ) −0.92 ± 0.08 −0.24 ± 0.08 −1.68 ± 0.04 +1.67 ± 0.06 +0.72 ± 0.06 −0.44 ± 0.08 +0.52 ± 0.08 +0.00 ± 0.04 +0.00 ± 0.02

By plotting a(φ) as a function of P (φ), we can obtain a parametric curve. As evident from the relations above, this curve in the period-acceleration plane is an ellipse. In this problem, we estimate the intrinsic spin period, P0 , the orbital period, PB , and the orbital radius, r, by an analysis of this data set, assuming a circular orbit. (D1.1) Plot the data, including error bars, in the period-acceleration plane (mark your graph as “D1.1”). Solution: Graph Number : D1.1

7

Data Analysis Examination Page 2 of 20

• Plot uses more than 50% of graph paper: 0.5 • Axes labels (P and a): 0.5 • Dimensions of axes: 0.5 • Ticks and values on axes (or scale written explicitly): 0.5 • Points correctly plotted: Points plotted 9 8 Marks given 4.0 3.5

7 3.0

6 2.0

5 1.0

<5 0

Correctness of points: deduction of 0.5 for each wrong point. • Errorbars on points (at least 5): 1.0 (D1.2) Draw an ellipse that appears to be a best fit to the data (on the same graph “D1.1”). Solution: See above • Elliptical curve with visual best fit: 1.0 • Curve symmetric about a = 0 line: 0.5

2

Data Analysis Examination Page 3 of 20

• Curve symmetric about some value of P (P ≈ 7588.48): 0.5 (D1.3) From the plot, estimate P0 , Pt and at , including error margins.

7

Solution: Values are determined from lengths of axes of ellipse and mid-point of P -axis. Error margins may be determined by estimating extreme ellipses covering the points with errorbars. Any reasonable method of estimating error margins will be accepted. 2Pt = (1.34 ± 0.04) µs

[(13.4 ± 0.4) cm on graph]

Pt = (0.67 ± 0.02) µs

2.0

P0 = (7588.48 ± 0.02) µs

2.0

2at = (3.42 ± 0.12) m s−2

[(17.1 ± 0.6) cm on graph]

3.0

at = (1.71 ± 0.06) m s−2 • Marking table: Parameter Half credit Minimum Pt (µs) 0.59 δPt (µs) 0.01 P0 (µs) 7588.38 δP0 (µs) 0.01 at (m s−2 ) 1.61 −2 δat (m s ) 0.04

Full credit Minimum Maximum 0.63 0.71 0.02 0.04 7588.43 7588.53 0.02 0.04 1.65 1.77 0.05 0.07

Half credit Maximum 0.75 0.05 7588.58 0.05 1.81 0.08

• Wrong values due to wrong/poor plot/fit in (D1.1) and (D1.2) WILL BE penalised. • Error estimation is based on graph drawing. Quoted values correspond to the envelope of possible ellipses drawn to include all points with errorbars. Any reasonable method to estimate error to be given credit. (D1.4) Write expressions for PB and r in terms of P0 , Pt , at .

4

Solution: We can easily recover the orbital period (PB ) and the radius of the orbit (r) in a circular orbit: 4π 2 at = 2 r PB PB2 at 4π 2 2πP0 r Pt = × PB c 2πP0 PB2 at P0 PB at = × = 2 PB c 4π 2πc

∴r=

1.0

Data Analysis Examination Page 4 of 20

Pt 2πc P0 at   Pt 2πc 2 at r= 2 4π P0 at  2 2 c Pt ∴ r= P0 at

∴ PB =

1.0 1.0 1.0

Alternative algebraic routes accepted. Each of PB and r carry 2.0 marks. (D1.5) Calculate approximate value of PB and r based on your estimations made in (D1.3), including error margins.

6

Solution: Pt 2πc P0 at 0.67 2π × 2.998 × 108 = × s 7588.48 1.71 = 96 260 s = 1.125 70 d s      ∆Pt 2 ∆P0 2 ∆at 2 ∆PB = PB + + Pt P0 at s   2   0.02 2 0.02 0.06 2 = 1.12570 × + + d 0.67 7588.48 1.71 PB =

1.0 1.0

= 1.12570 × 0.0461 ' 0.052 d ∴ PB = (1.13 ± 0.05) d

c2 at 2

Pt P0



0.67 7588.48

r= =

2



2.998 × 108 × 1.71

1.0


2 m

∴ r = 4.097 39 × 108 m = 2.738 90 × 10−3 AU s      2∆P0 2 ∆at 2 2∆Pt 2 ∆r = r + + Pt P0 at s 2     2 × 0.02 2 × 0.02 2 0.06 2 −3 = 2.738 90 × 10 × + + AU 0.67 7588.48 1.71

1.0 1.0

= 2.738 90 × 10−3 × 0.069 25 AU ' 0.19 × 10−3 AU r = (2.74 ± 0.19) × 10−3 AU

1.0

Errors in PB and r can be also estimated as maximum possible (worst case) error. In such case, errors would be about 1.5 times the standard error calculated above (δPB = 0.07, δr = 0.25). (D1.6) Calculate orbital phase, φ, corresponding to the epochs of the following five observations

4

Data Analysis Examination Page 5 of 20 in the above table: data rows 1, 4, 6, 8, 9. Solution: Using these newly determined orbital parameters, we can calculate the angular orbital phase for each data point, i.e., for each pair of acceleration and period measured (P , a).   a Pt −1 − φ = tan at P − P0 Care has to be taken to choose the value of the phase from among φ, π ± φ, 2π − φ, depending on the sign of cos φ and sin φ. Sr. T P a φ no. (tMJD) (µs) (m s−2 ) 1 5740.654 7587.8889 −0.92 148.62◦ 4 5746.675 7588.5810 +1.67 278.77◦ 6 5983.932 7587.8552 −0.44 164.57◦ 8 6040.857 7589.1350 +0.00 0.00◦ 9 6335.904 7589.1358 +0.00 0.00◦ • Credit for each correct value: 1.0 for first three, 0.5 for last two. • Credit for π ± φ or 2π − φ is 0.5 per value. • All values wrong due to wrong expression for φ gets a maximum of 1.0 mark. • Values in radians accepted. (D1.7) Refine the estimate of the orbital period, PB , using the results in part (D1.6) in the following way: (D1.7a) First determine the initial epoch, T0 , which corresponds to the nearest epoch of zero phase before the first observation. Solution: T1 − T0 φ1 φ1 = ⇒ T0 = T1 − PB PB 2π 2π 148.62◦ × 1.12570 tMJD 360◦ T0 = 5740.189 tMJD

2

1.0

T0 = 5740.654 −

1.0

Tolerance: ±0.002 tMJD. Using P0 instead of PB gets zero. (D1.7b) The expected time, Tcalc , of the estimated phase of each observation is given by   φ Tcalc = T0 + n + PB , 360◦ where n is the number of full cycle of orbital phases elapsed between T0 and Tcalc . Estimate n and Tcalc for each of the five observations in part (D1.6). Note down difference TO–C between observed T and Tcalc . Enter these calculations in the table given in the Summary Answersheet. Solution:

7

Data Analysis Examination Page 6 of 20

 Tcalc = T0 + n +

φ 360◦

 PB

where n = Integer part of [(T − T0 )/PB ]. Sr. T φ n Tcalc TO–C No. (tMJD) (MJD) (days) 1 5740.654 148.62◦ 0 5740.654 0.000 ◦ 4 5746.675 278.77 5 5746.689 −0.014 6 5983.932 164.57◦ 216 5983.855 0.077 ◦ 8 6040.857 0.00 267 6040.751 0.106 9 6335.904 0.00◦ 529 6335.684 0.220 Deduction for each wrong/missing value of n, Tcalc and TO–C : 0.5 No double penalty in one row. (D1.7c) Plot TO–C against n (mark your graph as “D1.7”).

4

Solution: Graph Number : D1.7

• Plot uses more than 50% of graph paper: 0.5 • Axes labels (TO–C and n) including dimensions: 0.5 • Ticks and values on axes (or scale written explicitly): 0.5 • Points correctly plotted: 0.5 for each point • Goodness of linear fit credited in next part (D1.7d) Determine the refined values of the initial epoch, T0,r , and the orbital period, PB,r .

7

Data Analysis Examination Page 7 of 20

Solution: A linear fit to the plot of TO–C vs n gives the offset of period per cycle (slope) and the shift in the zero-phase point (intercept). This concept, which may be evident in the subsequent calculation, gains the credit, explicit statement is not necessary. From a linear fit, Slope = 0.000 43 d/n

Intercept = −0.010 d

2.0

3.0

• Credit for good visual linear fit: 1.0 • Correct values of slope and intercept: 1.0 each • Tolerance: ±0.00002 in slope and ±0.002 in intercept.

T0,r = 5740.189 − 0.010 = 5740.179 tMJD

1.0

PB,r = (1.12570 + 0.00043) d = 1.126 13 d PB = 1.1261 d Incorrect sign of correction applied carries penalty of 0.5 for each quantity.

1.0

Data Analysis Examination Page 8 of 20 (D2) Distance to the Moon Geocentric ephemerides of the Moon for September 2015 are given in the form of a table. Each reading was taken at 00:00 UT. Date Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep Sep

01 02 03 04 05 06 07 08 09 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30

h 0 1 2 3 4 5 6 7 7 8 9 10 11 11 12 13 14 14 15 16 17 18 19 20 21 22 23 0 1 2

R.A. m 36 33 30 27 23 19 14 7 59 49 37 23 9 54 39 25 11 58 47 38 31 26 22 19 16 14 12 10 9 7

(α) s 46.02 51.34 45.03 28.48 52.28 37.25 19.23 35.58 11.04 0.93 11.42 57.77 41.86 49.80 50.01 11.64 23.13 50.47 54.94 50.31 40.04 15.63 17.51 19.45 55.43 46.33 43.63 48.32 5.89 39.02

Dec. (δ) ◦

0

3 7 11 14 16 17 18 17 15 13 10 7 3 0 -3 -6 -10 -13 -15 -17 -18 -18 -17 -14 -11 -8 -3 0 5 9

6 32 25 32 43 55 7 23 50 34 45 30 59 19 20 52 9 2 24 6 0 0 0 59 59 10 44 58 38 54

Angular Size (θ) 00

00

16.8 26.1 31.1 4.3 18.2 4.4 26.6 55.6 33.0 55.6 27.7 47.7 28.8 50.2 3.7 18.8 4.4 24.7 14.6 22.8 52.3 41.7 50.6 38.0 59.6 18.3 28.7 58.2 54.3 16.1

1991.2 1974.0 1950.7 1923.9 1896.3 1869.8 1845.5 1824.3 1806.5 1792.0 1780.6 1772.2 1766.5 1763.7 1763.8 1767.0 1773.8 1784.6 1799.6 1819.1 1843.0 1870.6 1900.9 1931.9 1961.1 1985.5 2002.0 2008.3 2003.6 1988.4

Phase (φ) 0.927 0.852 0.759 0.655 0.546 0.438 0.336 0.243 0.163 0.097 0.047 0.015 0.001 0.005 0.026 0.065 0.120 0.189 0.270 0.363 0.463 0.567 0.672 0.772 0.861 0.933 0.981 1.000 0.988 0.947

Elongation of Moon 148.6◦ W 134.7◦ W 121.1◦ W 107.9◦ W 95.2◦ W 82.8◦ W 70.7◦ W 59.0◦ W 47.5◦ W 36.2◦ W 25.1◦ W 14.1◦ W 3.3◦ W 7.8◦ E 18.6◦ E 29.5◦ E 40.4◦ E 51.4◦ E 62.5◦ E 73.9◦ E 85.6◦ E 97.6◦ E 110.0◦ E 122.8◦ E 136.2◦ E 150.0◦ E 164.0◦ E 178.3◦ E 167.4◦ W 153.2◦ W

Data Analysis Examination Page 9 of 20 The composite graphic1 below shows multiple snapshots of the Moon taken at different times during the total lunar eclipse, which occurred in this month. For each shot, the centre of frame was coinciding with the central north-south line of umbra. For this problem, assume that the observer is at the centre of the Earth and angular size refers to angular diameter of the relevant object / shadow.

(D2.1) In September 2015, apogee of the lunar orbit is closest to New Moon / First Quarter / Full Moon / Third Quarter. Tick the correct answer in the Summary Answersheet. No justification for your answer is necessary. Solution: From the table we see that the angular size of Moon is smallest close to the New Moon day. Thus, the answer is New Moon . Justification is NOT necessary for full credit.

3

3.0

(D2.2) In September 2015, the ascending node of lunar orbit with respect to the ecliptic is closest to New Moon / First Quarter / Full Moon / Third Quarter. Tick the correct answer in the Summary Answersheet. No justification for your answer is necessary. Solution: As there is an eclipse happening in this month, the lunar nodes are close to Full Moon day and New Moon day. Next we notice that lowest declination of Moon is just 18◦ . This means that after the New Moon day, the orbit of Moon is above the ecliptic. In other words, the ascending node is near the New Moon . Justification is NOT necessary for full credit.

4

4.0

(D2.3) Estiamte the eccentricity, e, of the lunar orbit from the given data. Solution: The largest angular size of the Moon in the ephemerides is 2008.300 and the smallest angular size is 1763.700 . The distance is inversely proportional to the angular size. Hence ratio of distance at perigee to the distance at apogee is: rperigee a 1−e 0 Ratio = =  × rapogee 1+e a 0  1

Credit: NASA’s Scientific Visualization Studio

4

2.0

Data Analysis Examination Page 10 of 20

∴

1−e 1763.7 = = 0.87821 1+e 2008.3 1 − 0.87821 ∴e= = 0.064846 1 + 0.87821 e ' 0.065

2.0

Rounding off is done to account for the fact that our data are not continuous, hence exact angular sizes at perigee and apogee are not known. A non-rounded answer will also receive full credit. (D2.4) Estimate the angular size of the umbra, θumbra , in terms of the angular size of the Moon, θMoon . Show your working on the image given on the backside of the Summary Answersheet. Solution: The following construction is shown.

Only two chords are necessary to determine the centre. The credit is divided in two parts for the drawing: • Realisation that centre of umbra circle needs to be determined to find θumbra : 1.5 • Accurate determination of centre of umbra circle by geometric construction: 2.5 Determination of centre of hand-drawn circle: maximum 1.5

8

5.0

Data Analysis Examination Page 11 of 20

• Measuring diameters of umbra and Moon: 0.5 each By estimating approximate centre of the shadow in the image, we find out, θumbra dumbra = θMoon dMoon 9.1 = = 2.76 3.3 ∴ θumbra ' 2.76θMoon

2.0

1.0

Acceptable range: ±0.10. (D2.5) The angle subtended by the Sun at Earth on the day of the lunar eclipse is known to be θSun = 1915.000 . In the figure below, S1 R1 and S2 R2 are rays coming from diametrically opposite ends of the solar disk. The figure is not to scale. S1

R2 θSun R1

S2 Earth

Moon

Calculate the angular size of the penumbra, θpenumbra , in terms of θMoon . Assume the observer to be at the centre of the Earth. Solution: The following diagram needs to be drawn. A

P

θ1

Q

Moon’s path Angular size of umbra is θumbra = 2]BOQ Angular size of penumbra is θpenumbra = 2]AOQ We have QA = QP + P A = OE + P E tan θ1

θSun

θ2

B J

S1

E

S2 R⊕ O

9

Data Analysis Examination Page 12 of 20

≈ R⊕ + dMoon θ1 θSun ≈ R⊕ + dMoon 2

since P A  P E since θ1 ≈ θ2 ≈ θSun /2

2.0

and QB = QP − P B = OE − P E tan θ2 ≈ R⊕ − dMoon θ2 θSun ≈ R⊕ − dMoon 2

2.0

∴ θpenumbra = 2]AOQ = 2 tan R⊕ + dMoon =2

dMoon

−1



QA OQ

 ≈2

QA OQ

since QA  OQ

θSun 2 = 2R⊕ + θ Sun dMoon

1.0

and θumbra = 2]BOQ = 2 tan R⊕ − dMoon =2

dMoon

−1



QB OQ

 ≈2

QB OQ

θSun 2 = 2R⊕ − θ Sun dMoon

1.0

Subtracting, θpenumbra − θumbra = 2θSun ⇒ θpenumbra = θumbra + 2θSun

1.0

We have, θumbra = 2.76θMoon

and θSun = 1915.000

From the given data, θMoon = 2008.300 . Therefore, 1915.0 θMoon θpenumbra = 2.76θMoon + 2 2008.3 θpenumbra = 4.67θMoon Acceptable range: 4.57θMoon to 4.77θMoon .

Alternative solution:

In the figure below, rays HEA and IF C are coming from one edge of solar disk and rays HF D and GEB are coming from the opposite edge. The observer (O) is assumed to be at the centre of the Earth. The Moon travels along the path ABCD during the course of eclipse.

1.0

1.0

Data Analysis Examination Page 13 of 20

G

A θSun

B J

E

θSun

O

θSun

C

θSun

F

H θSun

D

I

Moon’s path A θSun

E

B J

O

θSun

C θSun

F

D Moon’s path From figure, ]AEB = ]GEH = ]HF I = ]DF C = ]EJF = θSun θumbra = ]BOC = 2.76θMoon θpenumbra = ]AOD

3.0 1.0 1.0

]AOD = ]AOB + ]BOC + ]COD ]AOB = ]AEB

1.0

]COD = ]CF D

1.0

θpenumbra ' ]AEB + θumbra + ]CF D = 2θSun + 2.76θMoon = 2 × 1915.000 + 2.76 × 2008.300 θpenumbra = 9372.900 = 4.67θMoon

2.0

(D2.6) Let θEarth be angular size of the Earth as seen from the centre of the Moon. Calculate the angular size of the Moon, θMoon , as would be seen from the centre of the Earth on the eclipse day in terms of θEarth .

5

Solution: From the Moon, 2R⊕ θEarth = dMoon

1.0

Data Analysis Examination Page 14 of 20

From part (D2.5), θumbra + θpenumbra = 2θEarth ∴ θEarth =

2.0

θumbra + θpenumbra 2.76 + 4.67 = θMoon = 3.72θMoon 2 2

θMoon = = 0.269θEarth

2.0

Alternative solution: Let us say that the Moon is at position of B. Thus, angular size of Earth as seen from this position will be, (see figure in the previous part) θEarth = ]EBF = ]BF D

2.0

= ]BF C + ]CF D ' θumbra + θSun

1.0

The angular size of the Full Moon on 28 September as seen in the table is 2008.300 . θEarth = 2.76 × 2008.300 + 1915.000 = 7453.000 θMoon = 0.269θEarth =

θEarth 3.72

2.0

(D2.7) Estimate the radius of the Moon, RMoon , in km from the results above. Solution: Thus, the radius of Moon will be, R⊕ RMoon = 3.72 6371 RMoon = 3.72 RMoon ' 1713 km

3

1.0 2.0

Acceptable range: ±20 km. (D2.8) Estimate the shortest distance, rperigee , and the farthest distance, rapogee , to the Moon.

4

Solution: The shortest and longest distances will be, 2 × 1713 × 206265 rperigee = 2008.3 rperigee = 3.52 × 105 km rapogee =

2.0

2 × 1713 × 206265 1763.7

rapogee = 4.01 × 105 km

(D2.9) Use appropriate data from September 10 to estimate the distance, dSun , to the Sun from the Earth.

2.0

10

Data Analysis Examination Page 15 of 20

Solution: Moon

M

φM

S Sun

φS

E Earth

φE

Moon’s orbit On September 10, phase of Moon is 0.097 and elongation of Moon is 36.2◦ . Angular size of the Moon on this day is 1792.000 . Therefore, distance to Moon (from Earth) on September 10 is 2 × 1713 × 206265 dMoon,10 = 1792.0 = 3.94 × 105 km

2.0

Let ]EM S = φM ]ESM = φS ]SEM = φE ∴ φE = 36.2◦ 1 + cos φM phase = 2 −1 ∴ φM = cos (2 × phase − 1)

1.0 2.0

= cos−1 (2 × 0.097 − 1) = cos−1 (−0.806) = 143.71◦

2.0

◦

φS = 180 − φE − φM = 180◦ − 36.2◦ − 143.71◦ = 0.09◦ Now using sine rule, dSun sin φM = dMoon,10 sin φS 3.94 × 108 × sin 143.71◦ ∴ dSun = sin 0.09◦ dSun = 1.48 × 1011 m

1.0

2.0

1.0

Data Analysis Examination Page 16 of 20 (D3) Type IA Supernovae Supernovae of type Ia are considered very important for the measurements of large extragalactic distances. The brightening and subsequent dimming of these explosions follow a characteristic light curve, which helps in identifying these as supernovae of type Ia. Light curves of all type Ia supernovae can be fit to the same model light curve, when they are scaled appropriately. In order to achieve this, we first have to express the light curves in the reference frame of the host galaxy by taking care of the cosmological stretching/dilation of all observed time intervals, ∆tobs , by a factor of (1 + z). The time interval in the rest frame of the host galaxy is denoted by ∆tgal . The rest frame light curve of a supernova changes by two magnitudes compared to the peak in a time interval ∆t0 after the peak. If we further scale the time intervals by a factor of s (i.e. ∆ts = s∆tgal ) such that the scaled value of ∆t0 is the same for all supernovae, the light curves turn out to have the same shape. It also turns out that s is related linearly to the absolute magnitude, Mpeak , at the peak luminosity for the supernova. That is, we can write s = a + bMpeak , where a and b are constants. Knowing the scaling factor, one can determine absolute magnitudes of supernovae at unknown distances from the above linear equation. The table below contains data for three supernovae, including their distance moduli, µ (for the first two), their recession speed, cz, and their apparent magnitudes, mobs , at different times. The time ∆tobs ≡ t − tpeak shows number of days from the date at which the respective supernova reached peak brightness. The observed magnitudes have already been corrected for interstellar as well as atmospheric extinction. Name SN2006TD SN2006IS SN2005LZ µ (mag) 34.27 35.64 −1 cz (km s ) 4515 9426 12060 ∆tobs (days) mobs (mag) mobs (mag) mobs (mag) −15.00 19.41 18.35 20.18 −10.00 17.48 17.26 18.79 −5.00 16.12 16.42 17.85 0.00 15.74 16.17 17.58 5.00 16.06 16.41 17.72 10.00 16.72 16.82 18.24 15.00 17.53 17.37 18.98 20.00 18.08 17.91 19.62 25.00 18.43 18.39 20.16 30.00 18.64 18.73 20.48 (D3.1) Compute ∆tgal values for all three supernovae, and fill them in the given blank boxes in the data tables on the BACK side of the Summary Answersheet. On a graph paper, plot the points and draw the three light curves in the rest frame (mark your graph as “D3.1”).

15

Solution: Redshifts for the three supernovae are z1 = 0.0151, z2 = 0.0314 and z3 = 0.0402.

1.5

Filling in the three tables (∆tgal , third column)

3.5

Data Analysis Examination Page 17 of 20

SN2006TD

SN2006IS

SN2005LZ

∆tobs

mobs

∆tgal

∆ts

∆tobs

mobs

∆tgal

∆ts

∆tobs

mobs

∆tgal

∆ts

(d)

(mag)

(d)

(d)

(d)

(mag)

(d)

(d)

(d)

(mag)

(d)

(d)

−15.00

19.41

−14.78

−20.00

−15.00

18.35

−14.54

−14.54

−15.00

20.18

−14.42

−17.03

−10.00

17.48

−9.85

−13.34

−10.00

17.26

−9.70

−9.70

−10.00

18.79

−9.61

−11.35

−5.00

16.12

−4.93

−6.67

−5.00

16.42

−4.85

−4.85

−5.00

17.85

−4.81

−5.68

0.00

15.74

0.00

0.00

0.00

16.17

0.00

0.00

0.00

17.58

0.00

0.00

5.00

16.06

4.93

6.67

5.00

16.41

4.85

4.85

5.00

17.72

4.81

5.68

10.00

16.72

9.85

13.34

10.00

16.82

9.70

9.70

10.00

18.24

9.61

11.35

15.00

17.53

14.78

20.00

15.00

17.37

14.54

14.54

15.00

18.98

14.42

17.03

20.00

18.08

19.70

26.67

20.00

17.91

19.39

19.39

20.00

19.62

19.23

22.70

25.00

18.43

24.63

33.34

25.00

18.39

24.24

24.24

25.00

20.16

24.03

28.38

30.00

18.64

29.56

40.01

30.00

18.73

29.09

29.09

30.00

20.48

28.84

34.06

Full marks of 3.5 for all correct values. Penalty for incorrect values (3×7 independent values): Incorrect 1-3 Deduction 0.5

4-6 1.0

7-9 1.5

10-12 2.0

13-15 2.5

16-18 3.0

19-21 3.5

The light curves in galaxy frame would appear as follows Graph Number: D3.1

• Plot uses more than 50% of graph paper: 0.5

10.0

Data Analysis Examination Page 18 of 20

• Both axes labels (∆tgal and mobs ) present: 0.5 • Both dimensions of axes (days and mag) present: 0.5 • Ticks and values on axes (or scale written explicitly): 0.5 • Points correctly plotted: All points correctly plotted: 5.0 Penalty for incorrect or missing points: Incorrect 1 Deduction 0

2-4 0.5

5-7 1.0

8-10 1.5

11-13 2.0

14-16 2.5

17-19 3.0

20-22 3.5

23-25 4.0

26-30 5.0

• Smooth curve through points: 1.0 per curve

(D3.2) Take the scaling factor, s2 , for the supernova SN2006IS to be 1.00. Calculate the scaling factors, s1 and s3 , for the other two supernovae SN2006TD and SN 2005LZ, respectively, by calculating ∆t0 for them. Solution: From the graph D3.1, SN2006IS took 22.0 d to fade by 2 magnitudes. That is, ∆t0 (SN2006IS) = 22.0 d. Similarly, ∆t0 (SN2006TD) = 16.4 d. And ∆t0 (SN2005LZ) = 18.8 d. Acceptable range: ±1.0 days Thus, stretching factors for these two supernovae are s1 =

22.2 = 1.354 16.4

s3 =

22.2 = 1.181 18.8

5

3.0

2.0

(D3.3) Compute the scaled time differences, ∆ts , for all three supernovae. Write the values for ∆ts in the same data tables on the Summary Answersheet. On another graph paper, plot the points and draw 3 light curves to verify that they now have an identical profile (mark your graph as “D3.3”).

14

Solution: Filling the scaled values in the fourth column of the table (∆ts in table above) Full marks of 3.5 for all correct values. Penalty for incorrect values (3×7 independent values): Incorrect Deduction

1-3 0.5

4-6 1.0

7-9 1.5

10-12 2.0

13-15 2.5

16-18 3.0

The scaled light curves would appear as follows, Graph Number: D3.3

19-21 3.5

3.5

Data Analysis Examination Page 19 of 20

10.5 • Plot uses more than 50% of graph paper: 0.5 • Both axes labels (∆ts and mobs ) present: 0.5 • Both dimensions of axes (days and mag) present: 0.5 • Ticks and values on axes (or scale written explicitly): 0.5 • Points correctly plotted: All points correctly plotted: 5.0 Penalty for incorrect or missing points: Incorrect 1 Deduction 0

2-4 0.5

5-7 1.0

8-10 1.5

11-13 2.0

14-16 2.5

17-19 3.0

20-22 3.5

23-25 4.0

26-30 5.0

• Smooth curve through points: 1.0 per curve • The curves should show identical profiles.

0.5

(D3.4) Calculate the absolute magnitudes at peak brightness, Mpeak,1 , for SN2006TD and Mpeak,2 , for SN2006IS. Use these values to calculate a and b.

6

Solution: To get a and b, Mpeak,1 = mpeak,1 − µ1 = 15.74 − 34.27 mag = −18.53 mag Mpeak,2 = mpeak,2 − µ2 = 16.17 − 35.64 mag = −19.47 mag s1 − s2 1.354 − 1 0.354 ∴b= = mag−1 = mag−1 Mpeak,1 − Mpeak,2 −18.53 − (−19.47) 0.94

2.0

Data Analysis Examination Page 20 of 20

b = 0.3762 mag−1

2.0

a = s2 − bMpeak,2 = 1 − 0.3762 × (−19.47) = 1 + 7.325 a = 8.325

2.0

No penalty for missing mag−1 in b. (D3.5) Calculate the absolute magnitude at peak brightness, Mpeak,3 , and distance modulus, µ3 , for SN2005LZ.

4

Solution: s3 = a + bMpeak,3 s3 − a 1.181 − 8.325 −7.144 ∴ Mpeak,3 = = mag = mag b 0.3762 0.3762 Mpeak,3 = −18.99 mag

2.0

Distance modulus to SN2005LZ is µ3 = mpeak,3 − Mpeak,3 = 17.58 − (−18.99) mag µ3 = 36.57 mag

2.0

(D3.6) Use the distance modulus µ3 to estimate the value of Hubble’s constant, H0 . Further, estimate the characteristic age of the universe, TH .

6

Solution: Distance to SN2005LZ is µ3 µ3 d = 10( 5 +1) pc = 10( 5 +1−6) Mpc 3

36.57 = 10( 5 −5) Mpc = 102.314 Mpc

' 206 Mpc 12060 cz3 H0 = = km s−1 Mpc−1 d3 206 H0 = 58.5 km s−1 Mpc−1

4.0

1 3.086 × 1022 = yr H0 58.5 × 103 × 3.156 × 107 TH = 16.7 Gyr

2.0

TH =

Extra factor of 2/3 allowed in the value of TH .

Observational Examination (OM) Page 1 of 1

Instructions (1)

Total duration of the examination is 30 minutes.

(2)

Your answersheet consists of an A3 size sheet on which a rectangular sky map (Mercator Projection) is printed. In Mercator projection the circles of constant R.A. and circles of constant declination appear as straight lines. Please make sure that you write your contestant code on the map sheet in the box provided for the purpose.

(3)

Some inaccuracy in marking of a position will be tolerated, but the contestant will only get partial credit.

(4)

After the examination, fold your map, put it back inside the envelope and hand it over to the volunteers.

(5)

Markings with incorrect/missing codes will not get any marks.

Observational Examination (OM) Page 1 of 1 (OM1) Mark any 5 (five) of the following stars on the map by putting a circle (O) around the appropriate star and writing its code next to it. If you mark more than 5 stars, only the first 5 in serial order will be considered. Code S1 S2 S3 S4

Name Caph Asellus Australis Acrux Alphard

Bayer Name β Cas δ Cnc α Cru α Hya

Code S5 S6 S7 S8

Name Sheliak Albireo Rasalhague Kaus Australis

Bayer Name β Lyr β Cyg α Oph ϵ Sgr

(OM2) Mark location of any 3 (three) of the following galaxies on the map by putting a ‘+’ sign at appropriate place in the map and writing its code next to it. If you mark more than 3 galaxies, only the first 3 in serial order will be considered. Code G1 G2 G3

Name Triangulum Galaxy Whirlpool Galaxy Southern Pinwheel Galaxy

M number M 33 M 51 M 83

Code G4 G5

20

Name Virgo A Sombrero Galaxy

15

M number M 87 M 104

(OM3) Draw ecliptic on the map and label it as ‘E’.

5

(OM4) Show position of Autumnal Equinox (descending node of the ecliptic) on the map by a ‘+’ sign and label it as ‘A’.

5

(OM5) Draw local meridian for Bhubaneswar on Winter Solstice day (22nd December) at local midnight and label it as ‘M’.

5

Observational Examination (OP) Examination

Page 1 of 1

Instructions A.

Logistical (1) Total duration of the examination is 30 minutes. (2) After you take your designated seat inside planetarium dome, you will be given 5 minutes to read all the questions. Planetarium will be turned on only after this time. You will be given 5 minutes for familiarisation with the sky after it has been turned on. (3) At your seat you will find a writing board, a torch and a pen. Please use this pen to mark final answers on the maps provided (Map 1 and Map 2). Answers marked by other writing instruments will not be considered. (4) At any time during the examination, you may stand up at your place or turn around. However, you are not allowed to move out of your place. (5) You must maintain strict silence throughout the examination. (6) After your examination is over, keep writing board, torch and pen at the same place. Fold your map, put it back inside the envelope and hand it over to the volunteers.

B.

Academic (1) Your Summary Answersheet consists of an A4 size sheet on which sky maps are printed on both sides (Map 1 and Map 2). Please make sure that you write your contestant code on both sides of the sheet in the boxes provided for the purpose. (2) To mark the answers put a ‘+’ sign at appropriate place on the map and write code as given in the question next to it. (3) Some inaccuracy in marking of a position will be tolerated, with an appropriate penalty. (4) Start of a new part of each question will be announced. Simultaneously, a corresponding code will be displayed on the dome. Please pay attention.

Observational Examination (OP) Page 1 of 1 (OP1)

Eight well known historical supernovae will appear in the projected sky one at a time (not necessarily in chronological order). You have to identify the appropriate map (Map 1 / Map 2) where a particular supernova belongs and mark it in the corresponding map with ‘+’ sign and write codes ‘S1’ to ‘S8’ besides it. Each supernova code will be projected on dome for 10 seconds, followed by appearance of supernova for 60 seconds and then 20 seconds for you to mark the answers.

40

(OP1.1) For S1, S2, S3, S4 and S5, the projected sky corresponds to the sky as seen from Rio de Janeiro on the midnight of 21st May. (OP1.2) For S6, S7 and S8, the projected sky corresponds to the sky as seen from Beijing on the midnight of 20th November. There will be a gap of two minute after S5 for change over and adaptation to new sky. (OP2) We are now projecting sky of another planet. The sky will be slowly rotated for 5 minutes . Identify the visible celestial pole of this planet and mark it with a ‘+’ sign and label it as ‘P’ on the appropriate map (Map 1 / Map 2).

10

Observational Examination (OT) Page 1 of 1

Instructions (1) Total duration of the examination is 25 minutes. There will be a bell every five minutes. (2) The examiner at your station will give you a writing board, a torch and pen. Please use only the given pen to mark final answers on the map provided in the Summary Answersheet. Answers marked using any other writing instrument will not be considered. (3) After your examination is over, return the writing board, torch, pen and the envelope including your answersheet to the examiner.

Observational Examination (OT) Examination

Page 1 of 1

When you arrive at your observing station, DO NOT disturb the telescope before attempting the first question (OT1). (OT1) The telescope is already set to a deep sky object. Identify the object and tick the correct box in the Summary Answersheet. Note: You can use any technique to identify the object. However, if you disturb the telescope, you will NOT be helped to bring it back to the original position. (OT2)

10

(OT2.1) Point the telescope to M45. Show the object to the examiner. Note: 1. After 5 minutes, 1 mark will be deducted for a delay of every minute (or part thereof) in pointing the telescope. 2. You have a single chance to be evaluated. If your pointing is incorrect the examiner will change the pointing to M45 for the next part of the question.

5

(OT2.2) Your Summary Answersheet shows telescopic field of M45. In the image, seven (7) brightest stars of the cluster are replaced by ‘+’ sign. Compare the image with the field you see in the telescope and number the ‘+’ marks from 1 to 7 in the order of decreasing brightness (brightest is 1 and faintest is 7) of the corresponding stars.

15

(OT3) The examiner will give you a moon filter, an eyepiece with a cross-wire and a stopwatch. Point the telescope towards the Moon. Attach the filter to the telescope. On the surface of the Moon, you will see several “seas” (maria) which are nearly circular in shape. Estimate the diameter of Mare Serenitatis, 𝐷MSr , labelled as “1” in the figure below, as a fraction of the lunar diameter, 𝐷Moon , by measuring the telescope drift times, 𝑡Moon and 𝑡MSr , for the Moon and the mare, respectively.

20

Group Examination Page 1 of 2 (G1) A spacecraft of mass m and velocity 𝑣⃗ approaches a massive planet of mass M and orbital velocity 𝑢 ⃗⃗, as measured by an inertial observer. We consider a special case, where the incoming trajectory of the spacecraft is designed in a way such that velocity vector of the planet does not change direction due to the gravitational boost given to the spacecraft. In this case, the gravitational boost to the velocity the spacecraft can be estimated using conservation laws by measuring asymptotic velocity of the spacecraft before and after the interaction and angle of approach of the spacecraft.

(G1.1) What will be the final velocity (𝑣 ⃗⃗⃗⃗) ⃗⃗⃗⃗ and 𝑢 ⃗⃗⃗⃗ are exactly anti-parallel (see f of the spacecraft, if 𝑣 Figure 1).

3

(G1.2) Simplify the expression for the case where 𝑚 << 𝑀.

1

(G1.3) If angle between 𝑣 ⃗⃗⃗⃗ and −𝑢 ⃗⃗⃗⃗ is 𝜃 and 𝑚 << 𝑀 (see Figure 2), use results above to write expression for the magnitude of final velocity (𝑣f).

3

(G1.4) Table on the last page gives data of Voyager-2 spacecraft for a few months in the year 1979 as it passed close to Jupiter. Assume that the observer is located at the centre of the Sun. The distance from the observer is given in AU and 𝜆 is heliocentric ecliptic longitude in degrees. Assume all objects to be in the ecliptic plane. Assume that the orbit of the Earth to be circular. Plot appropriate column against the date of observation to find the date at which the spacecraft was closest to the Jupiter, and label the graph as G1.4.

8

(G1.5) Find the Earth-Jupiter distance, (𝑑E−J ) on the day of the encounter.

4

(G1.6) On the day of the encounter, around what standard time (𝑡std ) had the Jupiter transited the meridian in the sky of Bhubaneswar (20.27∘ N; 85.84∘ E; UT + 05: 30)?

6

(G1.7) Speed of the spacecraft (in km s −1) as measured by the same observer on some dates before the encounter and some dates after the encounter are given below. Here day n is the date of encounter. Use these data to find the orbital speed of Jupiter (𝑢) on the date of encounter and angle 𝜃.

12

date vtot date vtot

n-45 10.1408 n+5 21.8636

n-35 10.0187 n+15 21.7022

n-25 9.9078 n+25 21.5580

n-15 9.8389 n+35 21.3812

n-5 10.2516 n+45 21.2365

n 25.5150

(G1.8) Find eccentricity, 𝑒J , of Jupiter's orbit.

8

(G1.9) Find heliocentric ecliptic longitude, 𝜆p , of Jupiter's perihelion point.

5

Group Examination Page 2 of 2 Month Date June June June June June June June June June June June June June June June June June June June June June June June June June June June June June June July July July July July July July July July July July July July July July July

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

λ (∘) 135.8870 135.9339 135.9806 136.0272 136.0736 136.1200 136.1662 136.2122 136.2582 136.3040 136.3496 136.3951 136.4405 136.4857 136.5307 136.5756 136.6202 136.6647 136.7090 136.7532 136.7970 136.8407 136.8841 136.9273 136.9702 137.0127 137.0550 137.0969 137.1384 137.1795 137.2200 137.2600 137.2993 137.3378 137.3754 137.4118 137.4467 137.4798 137.5116 137.5628 137.6898 137.8266 137.9599 138.0903 138.2186 138.3453

Distance (AU) 5.1589731906 5.1629499712 5.1669246607 5.1708975373 5.1748689006 5.1788390741 5.1828084082 5.1867772826 5.1907461105 5.1947153428 5.1986854723 5.2026570402 5.2066306418 5.2106069354 5.2145866506 5.2185705999 5.2225596924 5.2265549493 5.2305575243 5.2345687280 5.2385900582 5.2426232385 5.2466702671 5.2507334797 5.2548156324 5.2589200110 5.2630505798 5.2672121872 5.2714108557 5.2756542053 5.2799520895 5.2843175880 5.2887686308 5.2933308160 5.2980426654 5.3029664212 5.3082133835 5.3140161793 5.3210070441 5.3312091210 5.3405592121 5.3466522674 5.3516661563 5.3561848203 5.3604205657 5.3644742164

Month Date July July July July July July July July July July July July July July July August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

λ (∘) 138.4707 138.5949 138.7183 138.8409 138.9628 139.0841 139.2048 139.3250 139.4448 139.5641 139.6831 139.8016 139.9198 140.0377 140.1553 140.2725 140.3895 140.5062 140.6225 140.7387 140.8546 140.9702 141.0856 141.2007 141.3157 141.4303 141.5448 141.6591 141.7731 141.8869 142.0006 142.1140 142.2272 142.3402 142.4530 142.5657 142.6781 142.7904 142.9024 143.0143 143.1260 143.2375 143.3488 143.4599 143.5709 143.6817

Distance (AU) 5.3684017790 5.3722377051 5.3760047603 5.3797188059 5.3833913528 5.3870310297 5.3906444770 5.3942369174 5.3978125344 5.4013747321 5.4049263181 5.4084696349 5.4120066575 5.4155390662 5.4190683021 5.4225956100 5.4261220723 5.4296486357 5.4331761326 5.4367052982 5.4402367851 5.4437711745 5.4473089863 5.4508506867 5.4543966955 5.4579473912 5.4615031166 5.4650641822 5.4686308707 5.4722034391 5.4757821220 5.4793671340 5.4829586711 5.4865569133 5.4901620256 5.4937741595 5.4973934544 5.5010200385 5.5046540300 5.5082955377 5.5119446617 5.5156014948 5.5192661222 5.5229386226 5.5266190687 5.5303075275

Group Examination Page 1 of 8 (G1) A spacecraft of mass m and velocity ~v approaches a massive planet of mass M and orbital velocity ~u, as measured by an inertial observer. We consider a special case, where the incoming trajectory of the spacecraft is designed in a way such that velocity vector of the planet does not change direction due to the gravitational boost given to the spacecraft. In this case, the amount of gravitational boost to the velocity the spacecraft can be roughly estimated using conservation laws by measuring asymptotic velocity of the spacecraft before and after the interaction and angle of approach of the spacecraft.

Planet

Planet

u

u

v spacecraft θ

v spacecraft

Figure 1

Figure 2

Figure 3

(G1.1) What will be the final velocity (v~f ) of the spacecraft, if ~v and ~u are exactly anti-parallel (see Figure 1).

3

Solution: Let v~f and u~f be the final velocity of the spacecraft and the planet respectively. As the planet For anti-parallel case, using conservation of linear momentum, M~u + m~v = M u~f + mv~f ∴ M u − mv = M uf + mvf m uf = u − (vf + v) M Now, using conservation of energy, M u2 + mv 2 = M u2f + mvf2 2 m m 2  m u2 + v = u− (vf + v) + v2 M M M f m 2 m m m m 2 u2 +  v = u2 + · (vf + v)2 − 2u  (vf + v) +  vf M M M M M     m 0= (vf + v)2 − 2u(vf + v) + (vf2 − v 2 ) M m   (v  0= (vf + v) (v (v f + v) − 2u f + v) +  f + v)(vf − v) M m 0= (vf + v) − 2u + vf − v M    m m ∴ vf 1 + = 2u + 1 − v M M
m 2u + 1 − M v
vf = m 1+ M Alternative solution in COM frame (G1.2) Simplify the expression for the case where m  M .

1

Group Examination Page 2 of 8

Solution: If m  M , vf ≈ 2u + v (G1.3) If angle between ~v and −~u is θ and m  M (see Figure 2), use results above to write expression for the magnitude of final velocity (vf ).

3

Solution: As velocity vector of the planet is not changing direction, there is no momentum transfer in direction perpendicular to ~u. We will resolve ~v and v~f into components parallel and perpendicular to ~u. vx = −v cos θ vfx = 2u + v cos θ

vy = v sin θ vfy = v sin θ

vf2 = vf2x + vf2y = (2u + v cos θ)2 + (v sin θ)2 = 4u2 + 4uv cos θ + v 2 cos2 θ + v 2 sin2 θ = 4u2 + 4uv cos θ + v 2 p ∴ vf = 4u2 + v 2 + 4uv cos θ (G1.4) Table on the last page gives data of Voyager-2 spacecraft for a few months in the year 1979 as it passed close to Jupiter. Assume that the observer is located at the centre of the Sun. The distance from the observer is given in AU and λ is heliocentric ecliptic longitude in degrees. Assume all objects to be in the ecliptic plane. Assume the orbit of the Earth to be circular. Plot appropriate column against the date of observation to find the date at which the spacecraft was closest to the Jupiter, and label the graph as G1.4.

8

Solution:

From the graph, it can be inferred that the encounter with Jupiter occured on 10th July (day 191) and its distance from the Sun on that day is 5.33121 AU (G1.5) Find the Earth-Jupiter distance, (dE−J ) on the day of the encounter.

4

Group Examination Page 3 of 8

Solution: The day number of Vernal Equinox is 80. Thus, ecliptic longitude of the Sun as seen from the Earth on the day of encounter will be, λ = (191 − 80) ∗ 360◦ /365.25 = 109.4045◦ Thus, the ecliptic longitude of the Earth as seen from the Sun on the day of encounter will be, λ⊕ = 180◦ + 109.4045◦ = 289.4045◦ Applying cosine rule, q d⊕−J = d2⊕ + d2J − 2d⊕ dJ cos ∆λ p = 12 + 5.33122 − 2 × 1 × 5.3312 × cos(289.4045◦ − 137.5628◦ ) = 6.2308 AU i.e. the Earth is 6.2308 AU from Jupiter on that day.

(G1.6) On the day of the encounter, around what standard time (tstd ) had the Jupiter transited the meridian in the sky of Bhubaneswar (20.27◦ N; 85.84◦ E; UT + 05:30)?

Solution: Thus, the angle of eastern elongation for Jupiter (]SEJ) on that day would be,   ◦ ◦ −1 5.3312 × sin(289.4045 − 137.5628 ) ξ = sin 6.2308 ◦ = 23.8146 It would rise 95 minutes after the Sun rise, i.e. around 7:35am. It would transit the meridian after around 6 hours i.e. around 13:35 local time or 13:22 IST . For more precise answer, R.A. of Jupiter on the day of encounter is approximately, λJgeocentric = 109.4045◦ + 23.8146◦ = 133.2191◦ tan αJ = tan λJgeocentric cos  = tan 133.2191◦ cos 23◦ 260 ∴ αJ = 135.68◦ = 9h 3m Thus, it will culminate at that sidereal time. On that day, sidereal time at noon is 07:24 (111 days from V.E. times 4 minutes). Thus, it will culminate 1 hour 39 minutes after the local noon i.e. at 13:39 local time or at about 13:26 IST .

6

Group Examination Page 4 of 8

To V.E.

S

δλ

S

E

A

φ1

D u

J

J

B Figure 4

φ2

Figure 5

C

(G1.7) Speed of the spacecraft (in km s−1 ) as measured by the same observer on some dates before the encounter and some dates after the encounter are given below. Here day n is the date of encounter. Use these data to find the orbital speed of Jupiter (u) on the date of encounter and angle θ. date vtot date vtot

n − 45 10.1408 n+5 21.8636

n − 35 10.0187 n + 15 21.7022

n − 25 9.9078 n + 25 21.5580

n − 15 9.8389 n + 35 21.3812

n−5 10.2516 n + 45 21.2365

n 25.5150

Solution: In Figure 5, path of Voyager-2 is shown as A-B-C. The Sun is shown as S and the Jupiter is shown as J. From the data we note that r is increasing continuously. The same should be reflected in the diagram. For practical purpose, J and B are the same points. The direction of velocity vector of Jupiter is given by JD. In the figure, ]ASB = δλ1

]ASB = δλ2

]ASC = δλ

]ABD = θ

]ABC = θ1

]DBC = ]ABC − ]ABD = θ1 − θ

]SAB = φ1

]SCB = φ2

Now the lines originating from the Sun indicate radial direction on the respective dates. Let us take speed of the spacecraft sufficiently far from the day 190, to avoid any influence of Jupiter in initial and final velocity estimation. We can choose dates 35 days on either side of July 10 i.e. June 5 and August 14. δλ1 = 137.5628◦ − 136.0736◦ = 1.4892◦ p l(AB) = l(SA)2 + l(SB)2 − 2 × l(SA) × l(SB) × cos δλ1 p = 5.174872 + 5.331212 − 2 × 5.17487 × 5.33121 × cos 1.4892◦ = 0.207 55 au     l(SB) sin δλ1 5.33121 × sin 1.4892◦ = sin−1 φ1 = sin−1 l(AB) 0.20755

12

Group Examination Page 5 of 8

= sin−1 (0.66755) φ1 = 41.8783◦ or 138.1217◦ δλ2 = 141.2007◦ − 137.5628◦ = 3.6379◦ p l(BC) = l(SC)2 + l(SB)2 − 2 × l(SC) × l(SB) × cos δλ2 p = 5.450852 + 5.331212 − 2 × 5.45085 × 5.33121 × cos 3.6379◦ = 0.362 53 au     ◦ −1 5.33121 × sin 3.6379 −1 l(SB) sin δλ2 = sin φ2 = sin l(BC) 0.36253 = sin−1 (0.66755) φ2 = 68.9199◦ or 111.0801◦ from the figure, φ1 should be obtuse and φ2 may be acute. In SABC δλ = λ2 − λ1 = 141.2007◦ − 136.0736◦ = 5.1271◦ ∴ θ1 = 360◦ − δλ − φ1 − φ2 = 360◦ − 5.1271◦ − 138.1217◦ − 68.9199◦ θ1 = 147.8313◦ In 4SBC, we notice ]SBC = 180◦ − φ2 − δλ2 = 180◦ − 68.9199◦ − 3.6379◦ = 107.4422◦ vy v sin θ tan ]DBC = f = v xf v cos θ + 2u sin θ tan(θ1 − θ) = cos θ + 2 uv 2u sin θ ∴ = − cos θ v tan(θ1 − θ) We use this expression to find |~u|. vf2 = 4u2 + v 2 + 4uv cos θ vf2

4u2 4u +1+ cos θ 2 2 v v v      v 2 2u 2 2u f = +1+2 cos θ v v v   2  sin θ sin θ = − cos θ + 1 + 2 − cos θ cos θ tan(θ1 − θ) tan(θ1 − θ)  2   sin θ 2 sin θ cos θ 2 sin θ cos θ 2 = − + cos θ + 1 + − 2 cos2 θ     tan(θ1 − θ) tan(θ − θ) tan(θ − θ) 1 1   sin2 θ = + 1 − cos2 θ tan2 (θ1 − θ)
sin2 θ = + sin2 θ = sin2 θ cot2 (θ1 − θ) + 1 2 tan (θ1 − θ) =

Group Examination Page 6 of 8

sin2 θ sin2 (θ1 − θ) vf sin θ sin θ ∴ = = v sin(θ1 − θ) sin θ1 cos θ − cos θ1 sin θ v ∴ = sin θ1 cot θ − cos θ1 vf sin θ1 tan θ = v vf + cos θ1 =

=

sin 147.8313◦ = −1.4088 10.0187 ◦ 21.3812 + cos 147.8313

∴ θ = 180◦ − 54.6328◦ = 125.3672◦ vf2 = 4u2 + v 2 + 4uv cos θ 21.38122 = 4u2 + 10.01872 + 4u × 10.0187 cos 125.3672◦ (457.1557 − 100.3743) 0 = u2 − 5.7990u − 4 2 0 = u − 5.7990u − 89.1953 √ 5.7990 + 5.79902 + 4 × 89.1953 ∴u= 2 = 12.7789 km s−1 Jupiter’s orbital velocity on the day of encounter is 12.779 km s−1 and the angle between the initial velocity of the spacecraft and Jupiter’s velocity vectors is 125◦ 220 . (G1.8) Find eccentricity, eJ , of Jupiter’s orbit. Solution: The angle between ~r and ~u on the day of encounter will be, ψ = ]SBC − (θ1 − θ) = 107.4422◦ − 147.8313◦ + 125.3672◦ = 84.9781◦ Now we use angular momentum conservation to estimate eccentricity. If up and rp represent perihelion velocity and perihelion distance of Jupiter, s   GM 1 + e rp up = aJ (1 − e) aJ 1−e p = GM aJ (1 − e2 ) rp up = ru sin ψ ∴ 1 − e2 =

r2 u2 sin2 ψ GM aJ


2 5.331212 × 1.496 × 1011 × 12.7789 × 103 sin2 84.9781◦ = 6.6741 × 10−11 × 1.9891 × 1030 × 5.202 60 = 0.99761 √ ∴ e = 1 − 0.99761 = 0.0489

8

Group Examination Page 7 of 8

The eccentricity of Jupiter’s orbit is 0.0489 . (G1.9) Find heliocentric ecliptic longitude, λp , of Jupiter’s perihelion point. Solution: To estimate longitude of perihelion, one should estimate true anomaly of Jupiter on that day. a(1 − e2 ) 1 + e cos Θ a(1 − e2 ) 5.20260 × 0.99761 ∴ 0.0489 cos Θ = −1= −1 r 5.33121 = −0.02646 r=

Θ = 122.754◦ Thus, the longitude of perihelion of Jupiter is, λp = λJ − Θ = 137.5628◦ − 122.754◦ λp = 14.809◦

5

Group Examination Page 8 of 8 Month

Date

λ (o )

June June June June June June June June June June June June June June June June June June June June June June June June June June June June June June July July July July July July July July July July July July July July July July

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16

135.8870 135.9339 135.9806 136.0272 136.0736 136.1200 136.1662 136.2122 136.2582 136.3040 136.3496 136.3951 136.4405 136.4857 136.5307 136.5756 136.6202 136.6647 136.7090 136.7532 136.7970 136.8407 136.8841 136.9273 136.9702 137.0127 137.0550 137.0969 137.1384 137.1795 137.2200 137.2600 137.2993 137.3378 137.3754 137.4118 137.4467 137.4798 137.5116 137.5628 137.6898 137.8266 137.9599 138.0903 138.2186 138.3453

Distance (AU) 5.1589731906 5.1629499712 5.1669246607 5.1708975373 5.1748689006 5.1788390741 5.1828084082 5.1867772826 5.1907461105 5.1947153428 5.1986854723 5.2026570402 5.2066306418 5.2106069354 5.2145866506 5.2185705999 5.2225596924 5.2265549493 5.2305575243 5.2345687280 5.2385900582 5.2426232385 5.2466702671 5.2507334797 5.2548156324 5.2589200110 5.2630505798 5.2672121872 5.2714108557 5.2756542053 5.2799520895 5.2843175880 5.2887686308 5.2933308160 5.2980426654 5.3029664212 5.3082133835 5.3140161793 5.3210070441 5.3312091210 5.3405592121 5.3466522674 5.3516661563 5.3561848203 5.3604205657 5.3644742164

Month

Date

λ (o )

July July July July July July July July July July July July July July July August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August August

17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31

138.4707 138.5949 138.7183 138.8409 138.9628 139.0841 139.2048 139.3250 139.4448 139.5641 139.6831 139.8016 139.9198 140.0377 140.1553 140.2725 140.3895 140.5062 140.6225 140.7387 140.8546 140.9702 141.0856 141.2007 141.3157 141.4303 141.5448 141.6591 141.7731 141.8869 142.0006 142.1140 142.2272 142.3402 142.4530 142.5657 142.6781 142.7904 142.9024 143.0143 143.1260 143.2375 143.3488 143.4599 143.5709 143.6817

Distance (AU) 5.3684017790 5.3722377051 5.3760047603 5.3797188059 5.3833913528 5.3870310297 5.390644477 5.3942369174 5.3978125344 5.4013747321 5.4049263181 5.4084696349 5.4120066575 5.4155390662 5.4190683021 5.4225956100 5.4261220723 5.4296486357 5.4331761326 5.4367052982 5.4402367851 5.4437711745 5.4473089863 5.4508506867 5.4543966955 5.4579473912 5.4615031166 5.4650641822 5.4686308707 5.4722034391 5.4757821220 5.4793671340 5.4829586711 5.4865569133 5.4901620256 5.4937741595 5.4973934544 5.5010200385 5.5046540300 5.5082955377 5.5119446617 5.5156014948 5.5192661222 5.5229386226 5.5266190687 5.5303075275