MCQ
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
  • A
    $\;\frac{1}{2}$
  • B
    $\;\sqrt 2 $
  • $\;\frac{1}{{\sqrt 2 }}$
  • D
    $\;\frac{1}{3}$

Answer

Correct option: C.
$\;\frac{1}{{\sqrt 2 }}$
c
           Escape velocity of the body from the surface of earth is $v = \sqrt {2gR} $

For escape velocity of the body from the platform $ potential\,\, energy + kinetic\,\, energy = 0$

$ - \frac{{GMm}}{{2R}} + \frac{1}{2}m{v^2} = 0$

$ \Rightarrow f{v_{escape}} = \sqrt {\frac{{GM}}{{{R^2}}} \cdot R}  = \sqrt {gR}  = fv$

From the surface of the earth, ${v_{escape}} = \sqrt {2gR} $

$\therefore \,\,f{v_{escape}} = \frac{{{v_{escape}}}}{{\sqrt 2 }}.\,\,\,\,\,\therefore \,\,\,f = \frac{I}{{\sqrt 2 }}.$

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