This set of Class 11 Physics Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Gravitation – Earth Satellite – 2”.

1. The time period of a satellite depends on _____

a) the mass of the satellite

b) radius of its orbit

c) both mass of satellite and radius of the orbit

d) neither mass of satellite nor radius of the orbit

View Answer

Explanation: The time period of a satellite is given by;

T = 2 x pi x [(R + h)

^{3}/ (G X M)]

^{1/2}

Where;

R = Radius of planet

h = Height above the planet

M = Mass of the planet

(R+h) = Radius of orbit

2. Consider two satellites A and B. Both move around the earth in the same orbit but the mass of B is twice that of the mass of A.

a) Orbital speeds of A and B are equal

b) The orbital speed of A is twice that of B

c) The orbital speed of B is twice that of A

d) The kinetic energy of both A and B are equal

View Answer

Explanation: Since orbital velocities are independent of the mass of satellites and only depends on the radius of orbits, they are equal for both A and B.

The kinetic energy of a satellite depends on the mass of the satellite. Hence, it differs for satellite A and B.

3. Two satellites of masses 50 kg and 100 kg revolve around the earth in circular orbits of radii 9R and 16R. What is the ratio of speeds of the two satellites?

a) 3:4

b) 4:3

c) 9:16

d) 16:9

View Answer

Explanation: Orbital velocity is inversely proportional to the square root of the orbital radius.

Hence;

v

_{1}: v

_{2}= 16

^{1/2}: 9

^{1/2}

= 4 : 3

4. A satellite is seen after every 8 hours. If it rotates opposite to that of the earth, what is its angular velocity about the centre of the earth? Assume the earth is perfectly spherical and the satellite is in the equatorial plane.

a) pi/2

b) pi/4

c) pi/6

d) pi/8

View Answer

Explanation: Angular velocity of the satellite if the earth were stationary; w’ = (2 x pi) / 8

= pi / 4

Angular velocity of earth; w’’ = (2 x pi) / 24

= pi / 12

Therefore, the angular velocity about the centre (w) = w’ – w’’

= pi/4 – pi/12

= pi/6

5. If a satellite is orbiting as close to the earth’s surface as possible _____

a) its speed is maximum

b) its speed is minimum

c) it’s kinetic energy will be minimum

d) it is not possible to quantify its kinetic energy

View Answer

Explanation: The orbital velocity is inversely proportional to the square root of the radius of the orbit. This implies that a minimum orbital radius would generate maximum orbital velocity.

Minimum orbital radius is acquired closer to the earth’s surface. Hence, If a satellite is orbiting as close to the earth’s surface as possible its speed is maximum.

6. The radius of a planet is “R” and mass is “M”. A satellite revolves around it in an orbital radius of “r” and an orbital velocity “v”. How would you express the acceleration due to gravity on the surface of the planet?

a) (r^{3}*v/R)

b) (r^{2}*v^{3}/R)

c) (r^{3}*v^{2}/R)

d) (r*v^{2}/R^{2})

View Answer

Explanation: Orbital velocity (v) = [(G x M) / r]

^{1/2}

(G x M) = v

^{2}x r

Acceleration due to gravity (g) = (G x M) / R

^{2}

Therefore; g = (v

^{2}x r) / R

^{2}.

7. A satellite revolves around a planet of radius “Q”. If the angular velocity of the satellite is “w” and the radius of the orbit is “4nQ”, where “n” is an integer, what is the acceleration due to gravity on the surface of the planet?

a) (w^{2} x (nQ) ^{3}/Q^{2})

b) 16 (w^{2} x (nQ) ^{3}/Q^{2})

c) 64(w^{2} x (nQ) ^{3}/Q^{2})

d) 32(w^{2} x (nQ) ^{3}/Q^{2}1)

View Answer

Explanation: Orbital velocity (v) = [(G x M) / (4nQ)]

^{1/2}

(G x M) = v

^{2}x (4nQ)

Acceleration due to gravity (g) = (G x M) / Q

^{2}

Therefore; g = (v

^{2}x (4nQ)) / Q

^{2}

Angular velocity (w) = v/(4nQ)

v = w x (4nQ)

Therefore; g = (w

^{2}x (4nQ)

^{3}/Q

^{2})

= 64 x (w

^{2}x (nQ)

^{3}/Q

^{2}).

8. The orbital velocity of a satellite orbiting the earth is half the escape velocity of the earth. What is the height above the surface of the earth at which it is orbiting? (Let the radius of the earth (R) = 6400 km).

a) 6400 km

b) 3200 km

c) 9600 km

d) 4800 km

View Answer

Explanation: Orbital velocity (v) = [(G x M) / (R+h)]

^{1/2}

Escape velocity (v’) = [(2 x G x M) / R]

^{1/2}

We know; v = v’/2

v

^{2}= v’

^{2}/4

[(G x M) / (R+h)] = [(2 x G x M) / R]/4

1 / (R+h) = 1 / (2R)

2R = R + h

h = R

Therefore; h = 6400km.

9. The only natural satellite of India is the moon.

a) True

b) False

View Answer

Explanation: The entire planet Earth has only 1 natural satellite; the moon. All other satellites are called “artificial satellites”. India has launched approximately 118 artificial satellites since 1975.

**Sanfoundry Global Education & Learning Series – Physics – Class 11**.

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