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Physics · Gravitation

JEE Main 2024 — 9 April, Shift 1 — Question 42

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 km . From earth's surface to the orbit, the change in total mechanical energy of the ball is xGMem21Rex \frac{G M_{e} m}{21 R_{e}}. The value of xx is (take Re=6370 km\mathrm{R}_{\mathrm{e}}=6370 \mathrm{~km} ):

  1. Option A:

    11

    Correct
  2. Option B:

    9

  3. Option C:

    12

  4. Option D:

    10

Answer: A

Step-by-step solution

h=318.5≈(Re20)\mathrm{h}=318.5 \approx\left(\frac{\mathrm{R}_{\mathrm{e}}}{20}\right)

T⋅Ei=−GMemReT \cdot E_{i}=\frac{-\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\mathrm{R}_{\mathrm{e}}} T⋅Ef=−Gem2(Re+h)=−GMem2(Re+Re20)T \cdot E_{f}=\frac{-G_{e} m}{2\left(R_{e}+h\right)}=\frac{-G M_{e} m}{2\left(R_{e}+\frac{R_{e}}{20}\right)} ⇒T⋅Ef=−10GMem21Re\Rightarrow \mathrm{T} \cdot \mathrm{E}_{\mathrm{f}}=\frac{-10 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{R}_{\mathrm{e}}}

Change in total mechanical energy

=TEf−TEi=\mathrm{TE}_{\mathrm{f}}-\mathrm{TE}_{\mathrm{i}} =GMemRe⁡[1−1021]=11GMem21Re=\frac{\mathrm{GM}_{\mathrm{e}} \mathrm{m}}{\operatorname{Re}}\left[1-\frac{10}{21}\right]=\frac{11 \mathrm{GM}_{\mathrm{e}} \mathrm{m}}{21 \mathrm{Re}}

Answer key and solution verified before publishing.

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Exam
JEE Main 2024
Paper
9 April, Shift 1
Subject
Physics
Chapter
Gravitation
Topic
Motion of Satellites and Escape Speed