MHT CET · Physics · Gravitation
The ratio of energy required to raise a satellite of mass ' \(\mathrm{m}\) ' to height ' \(h\) ' above the earth's surface to that required to put it into the orbit at same height is \([\)\(\mathrm{R}=\) radius of earth\(]\)
- A \(\frac{\mathrm{h}}{\mathrm{R}}\)
- B \(\frac{2 \mathrm{~h}}{\mathrm{R}^2}\)
- C \(\frac{3 \mathrm{~h}}{\mathrm{R}^2}\)
- D \(\frac{2 h}{R}\)
Answer & Solution
Correct Answer
(D) \(\frac{2 h}{R}\)
Step-by-step Solution
Detailed explanation
Energy required to raise to satellite of \(m\) to a height \(h\) is equal to change in its potential energy.
\(
\therefore \mathrm{W}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}+\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMmh}}{(\mathrm{R}+\mathrm{h}) \mathrm{R}}
\)
The energy of a satellite moving in a circular orbit is given by
\(
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})} \\
& \therefore \frac{\mathrm{W}}{\mathrm{E}}=\frac{2 \mathrm{~h}}{\mathrm{R}}
\end{aligned}
\)
\(
\therefore \mathrm{W}=-\frac{\mathrm{GMm}}{\mathrm{R}+\mathrm{h}}+\frac{\mathrm{GMm}}{\mathrm{R}}=\frac{\mathrm{GMmh}}{(\mathrm{R}+\mathrm{h}) \mathrm{R}}
\)
The energy of a satellite moving in a circular orbit is given by
\(
\begin{aligned}
& \mathrm{E}=\frac{\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})} \\
& \therefore \frac{\mathrm{W}}{\mathrm{E}}=\frac{2 \mathrm{~h}}{\mathrm{R}}
\end{aligned}
\)
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