Question

A satellite is moving with a constant speed ${}^{\prime }{v}^{\prime }$ in a circular orbit about the earth. An object of mass ${}^{\prime }{m}^{\prime }$ 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. $\frac{1}{2}m{v}^2$
  
• B. $m{v}^2$
  
• C. $\frac{3}{2}m{v}^2$
• D. $2m{v}^2$

Answer 1

The correct option is B $m{v}^2$


At height $r$ from center of earth, orbital velocity of a satellite of mass $m$ around the earth of mass $M$,
$$v=\sqrt{\frac{GM}{r}}...\left(1\right)$$ Just to escape from the gravitational pull, its total mechanical energy should be zero.

$\text{K.E+P.E}=0$

$\Rightarrow \text{K.E}+\left(\frac{-GMm}{r}\right)=0$

Substituting the value of $GM$ from $\left(1\right)$, we get

$\Rightarrow \text{K.E}+\frac{-v^{2}rm}{r}=0$

$\therefore \text{K.E}=m{v}^2$

Hence, option (b) is correct answer.

$\text{Alternate Solution:}$

In circular orbit of a satellite,

$\text{Potential energy =}-2\times \text{kinetic energy}$

$=-2\times \frac{1}{2}m{v}^2$

$=-m{v}^2$

Just to escape from the gravitational pull, its total mechanical energy should be zero.i.e.

$\text{K.E+P.E}=0$

$\Rightarrow \text{K.E}=-\text{P.E}=+m{v}^2$

Therefore, its $\text{K.E}$ should be $+m{v}^2$