A) \[{{v}_{e}}\]
B) \[\frac{{{v}_{e}}}{\sqrt{2}}\]
C) \[\frac{{{v}_{e}}}{2}\]
D) zero
Correct Answer: B
Solution :
[b]: Escape velocity,\[{{v}_{e}}=\sqrt{2Rg}\]where R is the radius of the planet. Potential energy of the body on the surface of the planet, \[{{U}_{S}}=\frac{GMm}{R}\](where m is the mass of the body) Potential energy of the body at the centre of the planet,\[{{U}_{C}}=\frac{3}{2}\frac{GMm}{R}\] If v is the velocity acquired by the body while at the centre of the planet, then \[\frac{1}{2}m{{v}^{2}}={{U}_{S}}-{{U}_{C}}=-\frac{GMm}{R}-\left( -\frac{3GMm}{2R} \right)\] or\[\frac{1}{2}m{{v}^{2}}=-\frac{GMm}{R}+\frac{3}{2}-\frac{GMm}{R}\] or\[{{v}^{2}}=2\frac{GM}{R}\left( \frac{3}{2}-1 \right)=\frac{GM}{R}=Rg=\frac{v_{e}^{2}}{2}\]. \[\left( \because g=\frac{GM}{{{R}^{2}}} \right)\]or\[v=\frac{{{v}_{e}}}{\sqrt{2}}\]
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