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7.Gravitation
normal
The escape velocity for a planet is $v_e.$ A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be
A
$v_e$
B
$\frac{{{v_e}}}{{\sqrt 2 }}$
C
$\frac{{{v_e}}}{{ 2 }}$
D
zero
Solution
From energy conservation
Potential inside sphere $V=-\frac{G M}{2 R^{3}}\left(3 R^{2}-r^{2}\right)$
$V_{S}=-\frac{G M}{R}, V_{0}=-\frac{3 G M}{2 R}$
Loss of potential energy of particle = Gain of kinetic energy
$m\left(V_{S}-V_{0}\right)=\frac{1}{2} m v^{2}$
$v^{2}=2\left[-\frac{G M}{R}-\left(-\frac{3 G M}{2 R}\right)\right] \Rightarrow v=\sqrt{\frac{G M}{R}}$
$v=\sqrt{\frac{2 G M}{R}} \Rightarrow v=\frac{V_{e}}{\sqrt{2}}$
Standard 11
Physics
