1. Tardigrade
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  3. Physics
  4. The escape velocity of a body from earth is about 11.2 km / s. Assuming the mass and radius of the earth to be about 81 and 4 times the mass and radius of the moon, the escape velocity in km / s from the surface of the moon will be

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Q. The escape velocity of a body from earth is about $11.2 \, km / s$. Assuming the mass and radius of the earth to be about $81$ and $4$ times the mass and radius of the moon, the escape velocity in $km / s$ from the surface of the moon will be

Gravitation

Solution:

$V_{\text {escape }}=\sqrt{\frac{G M}{R}}$
$\frac{V_{\text {escape Earth }}}{V_{\text {escape moon }}}=\sqrt{\frac{M_{e}}{R_{e}} \times \frac{R_{m}}{M_{m}}}$
$=\sqrt{\frac{81}{4}}=\left(\frac{9}{2}\right)$
$\Rightarrow V_{\text {moon }}=\frac{2}{9} \times 11.2=2.48\, km / s$