# Class 11 Physics MCQ – Gravitation – Geostationary and Polar Satellites

This set of Class 11 Physics Chapter 8 Multiple Choice Questions & Answers (MCQs) focuses on “Gravitation – Geostationary and Polar Satellites”.

1. CARTOSAT is a group of geostationary satellites sent up by India.
a) True
b) False

Explanation: INSAT is a group of geostationary satellites sent up by India for telecommunications, broadcasting, meteorology, and search and rescue operations. Commissioned in 1983, INSAT is the largest domestic communication system in the Indo-Pacific Region.

2. Most waves used for communication purposes rely on geostationary satellites because _____
a) they cannot transmit data at long distances due to curvature of the earth
b) they are reflected by the atmosphere
c) they are very cheap
d) it does not occupy space on the earth’s surface

Explanation: Most communication waves cannot transmit data accurately over long distances due to the curvature of the earth. They can only do so in the line of sight. However, they are not reflected by the ionosphere like radio waves and can be relayed to a wide area using geostationary satellites.

3. A geostationary satellite seems to be fixed in the sky because it does not orbit the earth.
a) True
b) False

Explanation: A geostationary satellite appears to be fixed in the sky because its time period of revolution around the earth is equal to the time period of revolution of the earth around its own axis, i.e., 24 hours.

4. The radius of orbit of a geostationary satellite is given by _____ (M = Mass of the earth; R = Radius of the earth; T = Time period of the satellite)
a) [(T2*G*M)/(4*pi2)]3/2
b) [(T2*G*M)/(4*pi2)]2/3 – R
c) [(T2*G*M)/(4*pi2)]1/3 – R
d) [(T2*G*M)/(4*pi2)]1/3

Explanation: The time period of a satellite is given by;
T = 2 x pi x (R+h) 3/2/ (G x M)1/2
Solving for orbital radius “(R+h)”, we get;
(R+h) = [(T2 x G x M)/(4 x pi2)]1/3.

5. The height of the geostationary satellites above the earth’s surface is approximately 36,000 km.
a) True
b) False

Explanation: The orbital radius “r” for a geostationary satellite is as follows;
r = [(T2 x G x M)/(4 x pi2)] 1/3
= [(864002 x (6.67 x 10-11) x (6 x 1024))/(4 x pi2)] 1/3
= 4,22,00,000 m
= 42,200 km
r = R + h
= 6,400 km
h = height of satellite above the earth’s surface
42,200 = 6,400 + h
h = 35,800 km
= approx. 36,000 km.
Sanfoundry Certification Contest of the Month is Live. 100+ Subjects. Participate Now!

6. Polar satellites are high altitude satellites.
a) True
b) False

Explanation: Polar satellites orbit close to the earth at distances of about 500 – 800 km above the surface and have time periods of about 90 – 100 minutes.

7. Polar satellites are used for high-resolution imaging of the earth’s surface because _____
a) they have better cameras
b) they are very high above the surface of the earth and travel slowly to gather more information
c) they are closer to the surface of the earth and can cover vast areas very quickly
d) they can be launched by most countries in the world

Explanation: Since polar satellites orbit close to the surface of the earth, it is easier to click images of the planet’s surface with greater detail. Their high orbital velocity enables them to view most of the surface strip-by-stip very quickly with which high-resolution images can be constructed.

8. A geostationary satellite is orbiting the earth at a height of 4R above the earth’s surface. The time period of another satellite at a height of 2R is _____
a) 24 x (3/5)1/2 hr
b) 12 x (3/5)1/2 hr
c) 6 x (3/5)1/2 hr
d) 72 x (3/5)1/21 hr

Explanation: The time period of a satellite is given by;
T = 2 x pi x (R+h)3/2 / (G x M)1/2
For a geostationary satellite;
24 = (5R)3/2 x (constant); [As per giver information] Required;
t = (3R)2/2 x (constant)
t = Time period of the satellite at a height of 2.5R
t/24 = (3R/5R)3/2
= (3/5)3/2
t = 24 x (3/5)1/2 hr.

9. A polar orbit is ideal for a spy satellite.
a) True
b) False

Explanation: A polar orbit is ideal for a spy satellite because it is closer to the earth’s surface and can click high-resolution images of the surface and also pick up communication data that is being transmitted.

10. How much is a polar orbit’s inclination with the earth’s equatorial plane?
a) 0 degrees
b) 45 degrees
c) 90 degrees
d) 180 degrees

Explanation:

From the figure, we can see that the polar orbit has an inclination of 90 degrees from the equatorial plane.

11. What is the angular velocity of parking satellites?

Explanation: Parking satellites (or) geostationary satellites have a time period of 24 hours. This means that it takes 24 hours to complete 1 revolution, i.e., 2*pi radians.
Therefore; angular velocity = (2 x pi)/24 rad/hr

12. The PSLV rockets of India launch satellites into the high-earth orbit.
a) True
b) False

Explanation: PSLV stands for “Polar Satellite Launch Vehicle”. They launch polar satellites which are low-earth orbit satellites. They orbit the earth at about 500 – 800 km from the earth’s surface.

13. A geosynchronous satellite always appears fixed in the sky.
a) True
b) False

Explanation: A geosynchronous satellite is a satellite with a time period of 24 hours. It may or may not lie in the equatorial plane and hence it may not always appear fixed in the sky. However, it will return to the same position after every 24 hours.

14. Every geostationary satellite is a geosynchronous satellite.
a) True
b) False

Explanation: A geosynchronous satellite is a satellite with a time period of 24 hours. It may or may not lie in the equatorial plane and hence it may not always appear fixed in the sky. However, it will return to the same position after every 24 hours.
Therefore, a geosynchronous satellite orbiting the earth in the equatorial plane is a geostationary satellite.
Every geostationary satellite is a geosynchronous satellite but not vice versa.

15. A geostationary satellite cannot orbit in any plane other than the equatorial plane.
a) True
b) False

Explanation: A geostationary satellite cannot orbit in any plane other than the equatorial plane because then it would have to be at rest, which cannot happen because the gravitational pull of the earth would eventually crash the satellite on the surface.

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

To practice all chapters and topics of class 11 Physics, here is complete set of 1000+ Multiple Choice Questions and Answers.