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An artificial satellite moves in a circular orbit around the earth. Total energy of the satellite is given by $E$. The potential energy of the satellite is
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Verified Answer
The correct answer is:
$-2 E$
We know that
The potential energy of the satellite
$$
U=-\frac{G M_{e} m}{R_{0}}
$$
The kinetic energy of the satellite
$$
K=\frac{1}{2} \frac{G M_{e} m}{R_{e}}
$$
The total energy
$$
\begin{aligned}
E=U+K &=-\frac{G M_{e} m}{R}+\frac{1}{2} \frac{G M_{e} m}{R_{0}} \\
E &=-\frac{1}{2} \frac{G M_{e} m}{R} \\
2 E &=-\frac{G M_{e} m}{R_{0}} \Rightarrow-2 E=U
\end{aligned}
$$
So. $\mathrm{PE}=-2(\mathrm{TE})$
$$
\mathrm{PE}=-2 E
$$
The potential energy of the satellite
$$
U=-\frac{G M_{e} m}{R_{0}}
$$
The kinetic energy of the satellite
$$
K=\frac{1}{2} \frac{G M_{e} m}{R_{e}}
$$
The total energy
$$
\begin{aligned}
E=U+K &=-\frac{G M_{e} m}{R}+\frac{1}{2} \frac{G M_{e} m}{R_{0}} \\
E &=-\frac{1}{2} \frac{G M_{e} m}{R} \\
2 E &=-\frac{G M_{e} m}{R_{0}} \Rightarrow-2 E=U
\end{aligned}
$$
So. $\mathrm{PE}=-2(\mathrm{TE})$
$$
\mathrm{PE}=-2 E
$$
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