Question

The escape velocity from a planet is $'V_e'.$ 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. $\frac{V_e}{\sqrt2}$

Answer 1

Answer: (D)

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