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(B) The escape velocity $v$ from the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}}$.Let the escape velocity from a height $h = R$ be $v'$

Vedclass · Accepted Answer

$\frac{1}{\sqrt{2}}$

Vedclass · Answer

(B) The escape velocity $v$ from the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}}$.Let the escape velocity from a height $h = R$ be $v'$. According to the principle of conservation of energy,the total energy at the platform must be zero for the body to escape to infinity.$-\frac{GMm}{R+R} + \frac{1}{2}m(v')^2 = 0$$-\frac{GMm}{2R} + \frac{1}{2}m(v')^2 = 0$$(v')^2 = \frac{GM}{R} \Rightarrow v' = \sqrt{\frac{GM}{R}}$.Given that $v' = fv$,we have $f \sqrt{\frac{2GM}{R}} = \sqrt{\frac{GM}{R}}$.$f = \frac{\sqrt{GM/R}}{\sqrt{2GM/R}} = \frac{1}{\sqrt{2}}$.

The Earth is assumed to be a sphere of radius $R$. $A$ platform is arranged at a height $R$ from the surface of the Earth. The escape velocity of a body from this platform is $fv$,where $v$ is the escape velocity from the surface of the Earth. The value of $f$ is

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