Q65 / 120 · NEET 2010 (Mains) · Physics
The additional kinetic energy to be provided to a satellite of mass $m$ revolving around a planet of mass $M$, to transfer it from a circular orbit of radius $R_1$ to another of radius $R_2\left(R_2>R_1\right)$ is
- A$\mathrm{GmM}\left(\frac{1}{\mathrm{R}_1^2}-\frac{1}{\mathrm{R}_2^2}\right)$
- B$\mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
- C$2 \mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
- Correct answerD$\frac{1}{2} \mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$
Solution
The kinetic energy changing the orbit of satellite
$$
\begin{aligned}
& \mathrm{KE}=\left(-\frac{\mathrm{GMm}}{2 \mathrm{R}_2}\right)-\left(-\frac{\mathrm{GMm}}{2 \mathrm{R}_1}\right) \\
& \mathrm{KE}=\frac{\mathrm{GMm}}{2}\left[\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right]
\end{aligned}
$$
Hence, the correct answer is option: 'D'.