Discussion Forum



1)

A satellite is moving with a constant speed u in a circular orbit about the earth. An object of mass m is ejected from the satellite such that it just escapes from the gravitational pull of the earth. At the time of its ejection, the kinetic energy of the object is 


A) $\frac{1}{2}mv^{2}$

B) $mv^{2}$

C) $\frac{3}{2}mv^{2}$

D) $2mv^{2}$

Answer:

Option B

Explanation:

In circular orbit of a satellite, potential energy

 $=-2 \times $(kinetic energy)

 =$-2 \times \frac{1}{2} mv^{2}=-mv^{2}$

Just to escape from the gravitational pull, its total mechanical energy should
be zero. Therefore' its kinetic energy should be $+mv^{2}$

 $\therefore$  Correct answer is (b)