Q135 / 150 · MHT CET 2020 (16 Oct Shift 2) · Physics
The ratio of energy required to raise a satellite of mass 'm' to a height ' $\mathrm{h}$ ' above the earth's surface to that required to put it into the orbit at same height is $[\mathrm{R}=$ radius of the earth $]$
- A$\frac{\mathrm{h}}{\mathrm{R}}$
- B$\frac{4 \mathrm{~h}}{\mathrm{R}}$
- C$\frac{3 \mathrm{~h}}{\mathrm{R}}$
- Correct answerD$\frac{2 \mathrm{~h}}{\mathrm{R}}$
Solution
(D)
Energy required to raise the satellite to a height $h$ from surface of earth is given by
$\begin{aligned}
U &=-\frac{G M m}{(R+h)}-\left(-\frac{G M m}{R}\right) \\
&=\frac{G M m}{R}-\frac{G M m}{(R+h)}=G M m\left[\frac{1}{R}-\frac{1}{R+h}\right] \\
&=G M m\left[\frac{R+h-R}{R(R+h)}\right]=G M m \frac{h}{R(R+h)}
\end{aligned}$
KE of the satellite is given by
$\begin{aligned}
\mathrm{K} &=\frac{1}{2} \mathrm{mV}_{0}^{2}=\frac{1}{2} \mathrm{~m} \frac{\mathrm{GM}}{(\mathrm{R}+\mathrm{h})} \\
\therefore \quad \frac{\mathrm{U}}{\mathrm{K}} &=\frac{\mathrm{GMmh}}{\mathrm{R}(\mathrm{R}+\mathrm{h})} \times \frac{(\mathrm{R}+\mathrm{h}) 2}{\mathrm{mGM}}=\frac{2 \mathrm{~h}}{\mathrm{R}}
\end{aligned}$
Hence, the correct answer is option: 'D'.