Verified Solution • 2025–26 Session moderate

QJEEM24Phe407: JEE Main 2024 Physics - Class 11 JEEM24Phe407 Solution: An astronaut takes a ball of mass from earth to space. He…

An astronaut takes a ball of mass mm from earth to space. He throws the ball into a circular orbit about earth at an altitude of 318.5 km318.5 \mathrm{~km} . From earth's surface to the orbit, the change in total mechanical energy of the ball is xGMem21Rex \frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}} . The value of xx is (take Re=6370 km)\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km}) :

Multiple Choice Options

  • A 12
  • B 11
  • C 9
  • D 10

Step-by-Step Verified Solution

At earth surface, E1=−GMmReE_1=-\frac{G M_m}{R_e}

in the orbit, E2=−GMm2rE_2=-\frac{G M_m}{2 r}

ΔEamp;=GMm[1Re−12r]amp;=GMm[1Re−12.1Re]amp;=1121GMmRe⇒amp;x=11\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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