Q48 / 60 · JEE Advanced 2011 (Paper 2) · Physics

A satellite is moving with a constant speed $v$ 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
    $\frac{1}{2} m v^2$
  • Correct answer
    B
    $m v^2$
  • C
    $\frac{3}{2} m v^2$
  • D
    $2 m v^2$

Solution

In circular orbit of a satellite, potential energy $$ \begin{aligned} & =-2 \times(\text { kinetic energy) } \\ & =-2 \times \frac{1}{2} m v^2=-m v^2 \end{aligned} $$ Just to escape from the gravitational pull, its total mechanical energy should be zero. Therefore, its kinetic energy should be $+m v^2$. $\therefore$ Correct answer is (b). Analysis of Question (i) Question is moderately difficult. (ii) In circular orbit of satellite $U=-\frac{G M m}{r}$ and $K=\frac{G M m}{2 r}$ or $U=-2 K$ Here, $U$ is potential energy and $K$ is kinetic energy.

Hence, the correct answer is option: 'B'.

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