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Question: Answered & Verified by Expert
A planet in a distant solar system is 10 times more massive than the earth and its radius is 10 times smaller. Given that the escape velocity from the earth is $11 \mathrm{kms}^{-1}$, the escape velocity from the surface of the planet would be
PhysicsGravitationJEE MainJEE Main 2008
Options:
  • A
    $1.1 \mathrm{kms}^{-1}$
  • B
    $11 \mathrm{kms}^{-1}$
  • C
    $110 \mathrm{kms}^{-1}$
  • D
    $0.11 \mathrm{kms}^{-1}$
Solution:
1193 Upvotes Verified Answer
The correct answer is:
$110 \mathrm{kms}^{-1}$
$$
v_{\mathrm{esc}}=\sqrt{\frac{2 \mathrm{GM}}{R}}=\sqrt{\frac{2 \mathrm{G} \times 10 \mathrm{M}}{\mathrm{R} / 10}}=10 \times 11=110 \mathrm{~km} / \mathrm{s}
$$

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