The escape velocity from the moon's surface i | vedclass

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Question

(B) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet/moo

Vedclass · Accepted Answer

The radius of the moon is less than that of the earth.

Vedclass · Answer

(B) The formula for escape velocity is given by $v_e = \sqrt{\frac{2GM}{R}}$,where $G$ is the gravitational constant,$M$ is the mass of the planet/moon,and $R$ is its radius.Since the mass of the moon $(M_m \approx 7.35 	imes 10^{22} \ kg)$ is much smaller than the mass of the earth $(M_e \approx 5.97 	imes 10^{24} \ kg)$ and its radius $(R_m \approx 1.74 	imes 10^6 \ m)$ is also smaller than the earth's radius $(R_e \approx 6.37 	imes 10^6 \ m)$,the ratio $\frac{M}{R}$ for the moon is significantly smaller than that for the earth.Consequently,the escape velocity from the moon's surface is approximately $2.38 \ km/s$,which is much less than the earth's escape velocity of approximately $11.2 \ km/s$.Therefore,the correct reason is that the moon's mass and radius result in a smaller gravitational potential at its surface compared to the earth.

The escape velocity from the moon's surface is less than that on the earth's surface,because:

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