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 answer
    D
    $\frac{1}{2} \mathrm{GmM}\left(\frac{1}{\mathrm{R}_1}-\frac{1}{\mathrm{R}_2}\right)$

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'.

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