The escape velocity of a body from the Earth | vedclass

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Question

(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.Given that the mass of the Earth $(M_e)$ is $81$ times the mass of the Moon $(M_m)

Vedclass · Accepted Answer

$2.48$

Vedclass · Answer

(B) The formula for escape velocity is $V_e = \sqrt{\frac{2GM}{R}}$.Given that the mass of the Earth $(M_e)$ is $81$ times the mass of the Moon $(M_m)$,so $M_e = 81 M_m$.The radius of the Earth $(R_e)$ is $4$ times the radius of the Moon $(R_m)$,so $R_e = 4 R_m$.The ratio of escape velocities is given by:$\frac{V_{e,e}}{V_{e,m}} = \sqrt{\frac{M_e}{M_m} 	imes \frac{R_m}{R_e}} = \sqrt{81 	imes \frac{1}{4}} = \sqrt{\frac{81}{4}} = \frac{9}{2} = 4.5$.Therefore,the escape velocity from the Moon is $V_{e,m} = \frac{V_{e,e}}{4.5} = \frac{11.2}{4.5} \approx 2.48 \, km/s$.

The escape velocity of a body from the 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,respectively,the escape velocity in $km/s$ from the surface of the Moon will be .......

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