# Chapter Five Class 9th

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CHP # 5

GRAVITATION

Q.1 A.

What is meant by gravitational force? The force with which earth attract every other object towards itself is called gravitational force. It is denoted by “ Fg ”.

Q.2 A.

State and explain Newton’s law of universal gravitation? Statement This law states that “Every object in the universe attracts every other object with a force. This force of attraction is directly proportional to the product of their masses and inversely proportional to the square of the distance between their centers.” Explaination Consider two spherical objects of masses “m1” and “m2” are placed at a distance of “r” from each other.

r

m1

m2

Mathematical expression

Fg  m1 m2    (1) Fg 

1      (2) r2

Combining equation (1) and (2) we get

Fg 

m1m2 r2

Fg  (Constt) Fg  G

m1m2 r2

m1m2 r2

Where “G” is constant of proportionality and it is called universal gravitation constant, its value is 6.67  10 11 Nm2/Kg2.

Q.3 A.

Using the law of universal gravitation determined mass of the earth. Consider a body of mass “m” is placed at the surface of the earth. The mass of the earth and radius of the earth are “Me” and “Re” respectively. m

Re Me

By law of universal gravitation, we can write that;

Fg  G

M em 2 Re

We know that Fg = W = mg So

mg  G g G

mM e 2 Re

Me 2 Re 2

Me Me Me Me Me

R g  e G (6  4  106 ) 2 9  8  6  67  1011 40  96  1012  9  8   1011 6  67 401 408  101211  6  67  60  18  1023

M e  6  0  10231 M e  6  1024 Kg Q.4 A.

How is the value of “g” changing by going to higher altitude? Write the relevant formula. The value of “g” is decreasing by going to higher altitude. Suppose an object of mass “m” is placed at a height “h” from the surface of the earth. If mass of the earth is “Me” and radius of the earth is “Re”. So the distance between mass “m” and center of the earth is (Re +h). Then by law of universal gravitation;

Fg  G

M em ( Re  h) 2

We know that Fg = W = mg

mg  G

M em ( Re  h ) 2

g (h)  G

Me ( Re  h) 2

At the surface of the earth the value of “g” is:

g G

Me ( Re ) 2

It is clear from the above equation that the value of “g” decreasing by going to higher altitude. The value of “g” at the equator is greater than the poles. The value of “g” at sea level is greater than that of the hilly areas. Q.5 A.

What is satellite? How they are moving around the planets. Satellites These are the objects revolving around the planets in stable orbits. Moon is a natural satellite of the earth. Scientists send artificial satellites which carries instruments or passengers in space revolving around the earth. During the circular motion satellite get the required centripetal force from the earth gravitation force. Consider a satellite of mass “m” is revolving around the earth in an orbit of radius “r” with uniform speed “v”. Then the required centripetal force can be determined as;

mv 2 Fc  r

--------- (A)

As the gravitation force provide the necessary centripetal force, then

Fg  G

M em -------- (B) r2

Comparing equation (A) and (B)

mV 2 M em G r r2 Me r Me V  G r If satellite is at a distance r = (Re + h), Where h = 22 106 m V2 G

V  6.67 10

11

V  3753m / s

6 1024 6.4 106  22 106

CONCEPTUAL QUESTIONS 1. A.

What will be the weight of a body if it is raised above the earth equal to its radius? We know that w = mg, as the value of g decreases by going to higher altitude. At a distance equal to the radius of the earth the value of “g” decrease by one fourth so the weight of the body also decrease by one fourth.

2. A.

Moon is attracted by the earth, why it does not fall on earth? There exist a gravitational force between the earth and the moon. This gravitational force provides the necessary centripetal force to the moon due to which it revolves around the earth and does not fall on earth.

3.

Why is water does not fall out of a bucket when it is whirled in a vertical circle? When the bucket of water is whirled in vertical circle, a centripetal force is providing to water. Due to which the velocity of water is along the tangent to circular path and does not fall.

A.

4. A.

Why it is not easy to whirl a hammer by a longer chain? To whirl the hammer with longer chain greater torque is required. The athlete could not provide the required torque. Therefore it is not easy to whirl a hammer by a longer chain.

5.

Explain why, if a stone held in our hand is released, it falls towards the earth? A gravitational force exists between the earth and the stone. The mass of the earth is greater than the mass of the stone. Therefore the stone is attracted by the earth and falls to the earth.

A.

6. A.

What is the value of gravitation constant “G” on the moon? The value of gravitational constant “G” is same everywhere, therefore the value of “G” on the moon will be 6.67  1011 Nm / Kg 2 .

7.

If the distance between two objects is tripled, what is the decrease in the gravitational force? The gravitational force between the two objects is given as;

A.

F G

m1 m2 r2

Let for the distance r / = 3r the gravitation force be F /

F/ G

m1 m 2 (r / ) 2

F/ G

m1 m 2 (3r ) 2

F/ G

m1 m2 9r 2

1 mm F /  (G 1 2 2 ) 9 r F/ 

1 F 9

It shows that the gravitational force will decrease by one nine.

8. A.

What is the difference between force of gravity and force of gravitation? Force of gravity The force with which earth attract a body towards its centre is called the force of gravity. Force of gravitation The force of attraction which exists between any two objects is called the force of gravitation.

9. A.

If a mass in earth gravitational field is doubled what will happen to the force exerted by the field upon mass. The gravitational force between the earth and mass “m” is

M em R 2e M e ( 2m) F/  G 2 Re F G

F /  2G

M em 2 Re

F /  2F If mass of the body is double the gravitational force will be doubled. 10. A.

What provide the force that produce centripetal acceleration in orbit? The gravitational force between the earth and satellite provide the necessary centripetal force which produce the centripetal acceleration in orbit.

11.

A.

A satellite is moving around the earth. On which of the following does the speed depends. a) Mass of satellite b) distance of satellite c) mass of earth The speed of satellite in the orbit is given by

V 

GM e Re  h

The speed of satellite depends on the a) Mass of satellite b) distance of satellite

NUMERICAL QUESTIONS 1.

Calculate the force of gravitation due to the earth on a child weighing 10Kg standing on the ground. Given data

m  10Kg

g  10m / s 2 We know that

Fg  W  mg Fg  1010 Fg  100N 2.

Calculate the gravitation force of attraction between a stone weighing 1Kg and earth. What will be the acceleration produce in the stone? Given data

m  1Kg

g  9.8m / s 2 We know that

Fg  W  mg

F  1 9.8 F  9.8N

a  g  9.8m / s 2

3.

Find the gravitational force of attraction between lead spheres each of mass 1000Kg place with their centers 1m apart? Given data

m1  1000Kg

m2  1000Kg

r  1m G  6.67  1011 Nm / Kg 2

Fg  ? Fg  G

m1 m2 r2

Fg  6.67  1011

1000 1000 (1) 2

Fg  6.67  1011  106 Fg  6.67  105 N 4.

A body of mass 25Kg is placed on the surface of the earth calculate the gravitational force. If the body is raised to a distance equal to the radius of the earth how will the weight of the body change? Given data

m1  25Kg

M e  61024 Kg

Re  6.4  106 m G  6.67  1011 Nm / Kg 2

Fg  ? Fg  G

m1 M e ( Re ) 2

Fg  6.67  1011

Fg 

25  6  1024 (6.4  106 ) 2

1000.5  10 11  10 24 40.96  1012

Fg  24.43  10241112

Fg  24.43  10

Fg  244.3N We know that

W  Fg  244.3N At a distance equal to the radius of earth the weight become one fourth so; W 

1 (244.3) 4

W  61.75N 5.

Two spherical objects of masses 10Kg and 100Kg are 90cm apart. Find the gravitational force between them? Given data

m1  10Kg m2  100Kg

r  90cm  0.9m G  6.67  1011 Nm / Kg 2

Fg  ? Fg  G

m1 m2 r2

Fg  6.67  1011

10  100 (0.9) 2

Fg 

6670  10 11 0.81

Fg  8234.57  1011 Fg  8.23  108 N 6.

The mass of earth is 6  1024 Kg . Using the law of universal gravitational, find the radius of earth? Given data

M e  61024 Kg We know that

W  Fg mg  G

mM e ( Re ) 2

( Re ) 2  G

Re  G

Me g

Me g

Re  6.67  1011 Re  4.08  1013 Re  40.8  1012

Re  6.39 106 m

6  1024 9.8

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