Question:

The escape velocity for earth is v. A planet having 9 times mass that of earth and radius, 16 times that of earth, has the escape velocity of:

Updated On: Dec 9, 2024
  • \(\frac{v}{3}\)
  • \(\frac{2v}{3}\)
  • \(\frac{3v}{4}\)
  • \(\frac{9v}{4}\)
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The Correct Option is C

Solution and Explanation

Escape velocity ($v_e$) is given by $v_e = \sqrt{\frac{2GM}{R}}$, where G is the gravitational constant, M is the mass of the planet, and R is the radius.


For the new planet, $M' = 9M$ and $R' = 16R$.

$v_e' = \sqrt{\frac{2G(9M)}{16R}} = \frac{3}{4}\sqrt{\frac{2GM}{R}} = \frac{3}{4}v_e$

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