A geostationary satellite is orbiting the earth at a height of 5R above that surface of the earth, R being the radius of the earth.The time period of another satellite in hours at a height of 2R from the surface of the earth is 

1. 5 hours                             

2. 10 hours

3. 6$\sqrt{2}$ hours                  

4. 6/$\sqrt{2}$ hours

A geostationary satellite is orbiting the earth at a height of \(5R\) above the surface of the earth, \(R\) being the radius of the earth. The time period of another satellite in hours at a height of \(2R\) from the surface of the earth is:
1. \(5\)
2. \(10\)
3. \(6\sqrt2\)
4. \(\frac{6}{\sqrt{2}}\)

A satellite is in an elliptical orbit around a planet \(P\). It is observed that the velocity of the satellite when it is farthest from the planet is \(6\) times less than that when it is closest to the planet. The ratio of distances between the satellite and the planet at closest and farthest points is:
1. \(1:3\)
2. \(1:2\)
3. \(3:4\)
4. \(1:6\)

A satellite of mass \(M\) is revolving around the Earth in a stationary orbit with a time period \(T.\) If \(10\%\) of the satellite's mass is detached, what will happen to its time period?
1. remain the same
2.  increase by \(10\%\)
3.  decrease by \(10\%\)
4. decrease by \(20\%\)

For a satellite moving in an orbit around the earth, the ratio of kinetic energy to potential energy is:
1. \(\frac{1}{\sqrt{2}}\)
2. \(2\)
3. \(\sqrt{2}\)
4. \(\frac{1}{2}\)

The distance of a planet from the sun is 5 times the distance between the earth and the sun. The time period of the planet is -

1. $5^{3/2}$ years                       

2. $5^{2/3}$ years

3. $5^{1/3}$ years                       

4. $5^{1/2}$ years

Consider a satellite orbiting the Earth in a circular orbit. Then,