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(C) The escape velocity from the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.For a body at a height $h = R$ from the surf

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

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(C) The escape velocity from the surface of the Earth is given by $v = \sqrt{\frac{2GM}{R}} = \sqrt{2gR}$.For a body at a height $h = R$ from the surface,the distance from the center of the Earth is $r = R + h = 2R$.The total energy required to escape from this distance must be zero:$-\frac{GMm}{2R} + \frac{1}{2}m(fv)^2 = 0$.Solving for the escape velocity at height $R$ $(v')$:$\frac{1}{2}m(v')^2 = \frac{GMm}{2R} \Rightarrow v' = \sqrt{\frac{GM}{R}}$.Since $v = \sqrt{\frac{2GM}{R}}$,we can write $v' = \frac{1}{\sqrt{2}} \sqrt{\frac{2GM}{R}} = \frac{v}{\sqrt{2}}$.Given that the escape velocity from the platform is $fv$,we have $fv = \frac{v}{\sqrt{2}}$.Therefore,$f = \frac{1}{\sqrt{2}}$.

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 its escape velocity from the surface of the Earth. The value of $f$ is

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