The escape velocity of an object from the Ear | vedclass

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Question

(A) The escape velocity $v$ is given by the energy conservation principle: $\frac{-GMm}{R} + \frac{1}{2}mv^2 = 0$.Solving for $v$,we get $v = \sqrt{\f

Vedclass · Accepted Answer

$v = R\sqrt{\frac{8\pi}{3}G\rho}$

Vedclass · Answer

(A) The escape velocity $v$ is given by the energy conservation principle: $\frac{-GMm}{R} + \frac{1}{2}mv^2 = 0$.Solving for $v$,we get $v = \sqrt{\frac{2GM}{R}}$.The mass of the Earth $M$ in terms of its mean density $ho$ and radius $R$ is $M = ho 	imes \frac{4}{3}\pi R^3$.Substituting this value of $M$ into the escape velocity formula:$v = \sqrt{\frac{2G}{R} 	imes ho 	imes \frac{4}{3}\pi R^3}$.Simplifying the expression:$v = \sqrt{\frac{8}{3}G\pi R^2 ho} = R\sqrt{\frac{8\pi}{3}Gho}$.

The escape velocity of an object from the Earth depends upon the mass of the Earth $(M)$,its mean density $(\rho)$,its radius $(R)$,and the gravitational constant $(G)$. Thus,the formula for escape velocity is:

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