MHT CET · Physics · Gravitation
The minimum energy required to launch a satellite of mass \(m\) from the surface of a planet of mass M and radius R in a circular orbit at an altitude of \(2 R\) is
- A \(\frac{5 \mathrm{GMm}}{6 \mathrm{R}}\)
- B \(\frac{2 \mathrm{GMm}}{3 \mathrm{R}}\)
- C \(\frac{\mathrm{GMm}}{2 \mathrm{R}}\)
- D \(\frac{\mathrm{GMm}}{3 \mathrm{R}}\)
Answer & Solution
Correct Answer
(A) \(\frac{5 \mathrm{GMm}}{6 \mathrm{R}}\)
Step-by-step Solution
Detailed explanation
Orbital energy, \(\mathrm{E}_0=\frac{-\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}\)
\(\therefore \quad \mathrm{E}_0=\frac{-\mathrm{GMm}}{2(\mathrm{R}+2 \mathrm{R})}=\frac{-\mathrm{GMm}}{6 \mathrm{R}} \quad \ldots(\because \mathrm{~h}=2 \mathrm{R})\)
Energy at surface \(E=\frac{-\mathrm{GMm}}{R}\)
\(\begin{aligned}
\therefore \quad \text { Min. energy required } & =\mathrm{E}_0-\mathrm{E} \\
& =\frac{-\mathrm{GMm}}{6 \mathrm{R}}-\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right) \\
& =\frac{5 \mathrm{GMm}}{6 \mathrm{R}}
\end{aligned}\)
\(\therefore \quad \mathrm{E}_0=\frac{-\mathrm{GMm}}{2(\mathrm{R}+2 \mathrm{R})}=\frac{-\mathrm{GMm}}{6 \mathrm{R}} \quad \ldots(\because \mathrm{~h}=2 \mathrm{R})\)
Energy at surface \(E=\frac{-\mathrm{GMm}}{R}\)
\(\begin{aligned}
\therefore \quad \text { Min. energy required } & =\mathrm{E}_0-\mathrm{E} \\
& =\frac{-\mathrm{GMm}}{6 \mathrm{R}}-\left(\frac{-\mathrm{GMm}}{\mathrm{R}}\right) \\
& =\frac{5 \mathrm{GMm}}{6 \mathrm{R}}
\end{aligned}\)
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