Question 289: Gravitation

Question

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

Accepted Answer

Orbital energy, $\mathrm{E}_0=\frac{-\mathrm{GMm}}{2(\mathrm{R}+\mathrm{h})}$
$$	herefore \quad \mathrm{E}_0=\frac{-\mathrm{GMm}}{2(\mathrm{R}+2 \mathrm{R})}=\frac{-\mathrm{GMm}}{6 \mathrm{R}} \quad \ldots(\because \mathrm{~h}=2 \mathrm{R})$$
Energy at surface $E=\frac{-G M m}{R}$
$$\begin{aligned}
	herefore \quad 	ext { Min. energy required } & =\mathrm{E}_0-\mathrm{E} \
& =\frac{-\mathrm{GMm}}{6 \mathrm{R}}-\left(\frac{-\mathrm{GMm}}{\mathrm{R}}ight) \
& =\frac{5 \mathrm{GMm}}{6 \mathrm{R}}
\end{aligned}$$

Answer

$\frac{5 \mathrm{GMm}}{6 \mathrm{R}}$

Answer

$\frac{2 \mathrm{GMm}}{3 \mathrm{R}}$

Answer

$\frac{\mathrm{GMm}}{2 \mathrm{R}}$

Answer

$\frac{\mathrm{GMm}}{3 \mathrm{R}}$

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

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