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A satellite orbits the earth at a height of $400 \mathrm{~km}$ above the surface. How much energy must be expended to rocket the satellite out of the gravitational influence of the earth? Mass of the satellite is $\mathbf{2 0 0} \mathrm{kg}$, mass of the earth $=6 \times 10^{24} \mathrm{~kg}$, radius of the earth $=6.4 \times 10^6 \mathrm{~m}, \mathrm{G}$ $=6.67 \times 10^{-11} \mathrm{Nm}^2 / \mathrm{kg}^2$
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Verified Answer
Total energy of the satellite at a height
$$
\begin{aligned}
&h=-\frac{G M m}{(R+h)}+\frac{1}{2} m v^2 \\
&=-\frac{G M m}{R+h}+\frac{1}{2} m \frac{G M}{R+h}=-\frac{G M m}{2(R+h)}
\end{aligned}
$$
Energy expended to rocket the satellite out of the earth's gravitational field $=-(\mathrm{T} . \mathrm{E}$. of the satellite $)=\frac{G M m}{2(R+h)}$ $=\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2\left(6.4 \times 10^6+4 \times 10^5\right)}=5.9 \times 10^9 \mathrm{~J}$.
$$
\begin{aligned}
&h=-\frac{G M m}{(R+h)}+\frac{1}{2} m v^2 \\
&=-\frac{G M m}{R+h}+\frac{1}{2} m \frac{G M}{R+h}=-\frac{G M m}{2(R+h)}
\end{aligned}
$$
Energy expended to rocket the satellite out of the earth's gravitational field $=-(\mathrm{T} . \mathrm{E}$. of the satellite $)=\frac{G M m}{2(R+h)}$ $=\frac{6.67 \times 10^{-11} \times 6 \times 10^{24} \times 200}{2\left(6.4 \times 10^6+4 \times 10^5\right)}=5.9 \times 10^9 \mathrm{~J}$.
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