Question
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 $$f{v_e},$$ where $${v_e}$$ is its escape velocity from the surface of the earth. The value of $$f$$ is
A.
$$\sqrt 2 $$
B.
$$\frac{1}{{\sqrt 2 }}$$
C.
$$\frac{1}{3}$$
D.
$$\frac{1}{2}$$
Answer :
$$\frac{1}{{\sqrt 2 }}$$
Solution :
At a platform at a height $$h,$$
Escape energy = binding energy of sphere
$$\eqalign{
& {\text{or}}\,\,\frac{1}{2}m{\left( {f{v_e}} \right)^2} = \frac{{GMm}}{{R + h}} \cr
& {\text{or}}\,\,f{v_e} = \sqrt {\frac{{2GM}}{{R + h}}} = \sqrt {\frac{{2GM}}{2R}} \,.......\left( {\text{i}} \right)\left( {\because h = R} \right) \cr} $$
But at surface of the earth, $${v_e} = \sqrt {\frac{{2GM}}{R}} \,......\left( {{\text{ii}}} \right)$$
Dividing Eq. (ii) by Eq. (i).
Hence, $$\frac{{f{v_e}}}{{{v_e}}} = \frac{{\sqrt {\frac{{GM}}{R}} }}{{\sqrt {\frac{{2GM}}{R}} }} \Rightarrow f = \frac{1}{{\sqrt 2 }}$$