For the moon to cease as the earth's satellite, its orbital velocity has to be increased by a factor of:

How many times is escape velocity, of orbital velocity for a satellite revolving near earth?

1.$\sqrt{2}$ times                   
2. 2 times
3. 3 times                      
4. 4 times

The value of escape velocity on a certain planet is 2 km/s. Then the value of orbital speed

for a satellite orbiting close to its surface is:

1. 12km/s                                                                 

2. 1km/s

3. $\sqrt{2<mspace\; width="1em"></mspace>}<mspace\; width="1em"></mspace>km/s$                                                             

4. $2\sqrt{2}<mspace\; width="1em"></mspace>km/s$

If $v_e$ and $v_o$ represent the escape velocity and orbital velocity of a satellite corresponding

to a circular orbit of radius R, then 

1. $v_e=v_o$                     

2. $\sqrt{2}{v}_o=v_e$

3. $v_e=v_o/\sqrt{2}$               

4. $v_e$ and $v_o$ are not related

An earth satellite of mass m revolves in a circular orbit at a height h from the surface of

the earth. R is the radius of the earth and g is acceleration due to gravity at the surface

of the earth. The velocity of the satellite in the orbit is given by:

1. $\frac{g{R}^2}{R+h}$                     

2. gR

3. $\frac{gR}{R+h}$                     

4. $\sqrt{\frac{g{R}^2}{R+h}}$