JEE MAIN - Physics (2024 - 9th April Morning Shift - No. 18)
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 $$\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km})$$ :
12
11
9
10
Explanation
At earth surface, $$E_1=-\frac{G M_m}{R_e}$$
in the orbit, $$E_2=-\frac{G M_m}{2 r}$$
$$\begin{aligned} \Delta E & =G M_m\left[\frac{1}{R_e}-\frac{1}{2 r}\right] \\\\ & =G M_m\left[\frac{1}{R_e}-\frac{1}{2.1 R_e}\right] \\\\ & =\frac{11}{21} \frac{G M_m}{R_e} \\\\ \Rightarrow & x=11 \end{aligned}$$
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