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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
$\sqrt{2}$
B
$\dfrac{1}{\sqrt{2}}$
C
$\dfrac{1}{3}$
D
$\dfrac{1}{2}$
Detailed Solution
The platform is at a distance $R + R = 2R$ from the centre of the Earth.
By conservation of mechanical energy, for the body to just escape (zero energy at infinity): $-\dfrac{GMm}{R + R} + \dfrac{1}{2}m(fv)^2 = 0 + 0$
$\dfrac{1}{2}(fv)^2 = \dfrac{GM}{2R}$
$\Rightarrow fv = \sqrt{\dfrac{GM}{R}}$
But the escape velocity from the surface is $v = \sqrt{\dfrac{2GM}{R}}$
Therefore $f\sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{GM}{R}}$
$\Rightarrow f = \dfrac{1}{\sqrt{2}}$
By conservation of mechanical energy, for the body to just escape (zero energy at infinity): $-\dfrac{GMm}{R + R} + \dfrac{1}{2}m(fv)^2 = 0 + 0$
$\dfrac{1}{2}(fv)^2 = \dfrac{GM}{2R}$
$\Rightarrow fv = \sqrt{\dfrac{GM}{R}}$
But the escape velocity from the surface is $v = \sqrt{\dfrac{2GM}{R}}$
Therefore $f\sqrt{\dfrac{2GM}{R}} = \sqrt{\dfrac{GM}{R}}$
$\Rightarrow f = \dfrac{1}{\sqrt{2}}$