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Gravitation — JEE physics STD 12 Science — Question

CBSE BoardEnglish MediumSTD 12 ScienceJEE physicsGravitation1 Mark
MCQ
The escape velocity for a rocket from earth is $11.2 \mathrm{~km} / \mathrm{sec}$. Its value on a planet where acceleration due to gravity is double that on the earth and diameter of the planet is twice that of earth will be in $\mathrm{km} / \mathrm{sec}$
  • A
    $11.2$
  • B
    $5.6$
  • ✓
    $22.4$
  • D
    $53.6$
✓

Answer

Correct option: C.
$22.4$
$ \frac{v_p}{v_e}=\sqrt{\frac{g_p}{g_e} \times \frac{R_p}{R_e}}=\sqrt{2 \times 2}=2 $
$ \Rightarrow v_p=2 \times v_e=2 \times 11.2=22.4\ km / s$

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