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JEE Advanced · Physics · 9. Gravitation

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

  1. A \(\frac{1}{2} m v^2\)
  2. B \(m v^2\)
  3. C \(\frac{3}{2} m v^2\)
  4. D \(2 m v^2\)
Verified Solution

Answer & Solution

Correct Answer

(B) \(m v^2\)

Step-by-step Solution

Detailed explanation

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.
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