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 center of the planet will be

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

Answer 1

The correct option is D $\frac{V_e}{\sqrt{2}}$

$V_e=\sqrt{2gR}$

According to the law of conservation of energy

$U_c+\frac{1}{2}m{V}^2=U_s=\frac{1}{2}m{V}^2=U_s-U_c=m(V_s-V_c)$

$\frac{1}{2}m{V}^2=m[-\frac{GM}{R}-(-\frac{3GM}{2R}\left)\right]$

$\frac{1}{2}m{V}^2=m\left(\frac{GM}{2R}\right)$

$V=\sqrt{\frac{GM}{R}}=\frac{V_e}{\sqrt{2}}[\cdot .\cdot V=\sqrt{gR}]$