An astronaut takes a ball of mass $m$ from earth to space. He throws the ball into a circular orbit about earth at an altitude of $318.5 \mathrm{~km}$. From earth's surface to the orbit, the change in total mechanical energy of the ball is $x \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}$. The value of $x$ is (take $\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km})$ :
Solution
<p>At earth surface, $E_1=-\frac{G M_m}{R_e}$</p>
<p>in the orbit, $E_2=-\frac{G M_m}{2 r}$</p>
<p>$$\begin{aligned}
\Delta E & =G M_m\left[\frac{1}{R_e}-\frac{1}{2 r}\right] \\\\
& =G M_m\left[\frac{1}{R_e}-\frac{1}{2.1 R_e}\right] \\\\
& =\frac{11}{21} \frac{G M_m}{R_e} \\\\
\Rightarrow & x=11
\end{aligned}$$</p>
About this question
Subject: Physics · Chapter: Gravitation · Topic: Kepler's Laws
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