Question

The escape velocity from a planet is ${}^{\prime }{V}_e^{\prime }.$ A tunnel is dug along along a diameter of the planet and a small body is dropped into it. The speed of the body at the centre of the planet will be

• A. $V_e$
• B. $\frac{V_e}{2}$
• C. $2V_e$
• D. $\frac{V_e}{\sqrt{2}}$

Answer 1

Answer: (D)

$V_e=\sqrt{2gR}$
According to the law of conservation of energy
$U_e+\frac{1}{2}m{V}^2=U_s\Rightarrow \frac{1}{2}m{V}^2=U_s-U_c=m(V_s-V_c)$
$\Rightarrow \frac{1}{2}m{V}^2=m\left[-\frac{GM}{R}-\left(-\frac{3GM}{2R}\right)\right]$
$\Rightarrow \frac{1}{2}m{V}^2=m\left(\frac{GM}{2R}\right)\Rightarrow V=\sqrt{\frac{GM}{R}}=\frac{V}{\sqrt{2}}$
[$\therefore$V =$\sqrt{gR}$ ]