Radius of orbit of satellite of earth is R. Its kinetic ene…

Physics Gravitation Gravitation Field, Force,Mass, Potential Energy and Escape Velocity Single Correct MCQ

Published by: CGP EDU Academic Team

Published on: September 12, 2026

Radius of orbit of satellite of earth is R. Its kinetic energy is proportional to

$\frac{1}{R}$

$\frac{1}{\sqrt{R}}$

$\frac{1}{R^{3/2}}$

Text Solution

Verified by Experts

The correct answer is:

Kinetic energy of the satellite

$\mathrm{KE} = \frac{1}{2} m v_0^2$

Now putting the value of $v_0$ is eq. (1), we get

$\mathrm{KE} = \frac{1}{2} m \left( \sqrt{\frac{\mathrm{GM}}{R}} \right)^2$

$= \frac{1 \text{ mGM}}{2 \text{ R}} Hence, KE \propto \frac{1}{R}$

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