# A smooth tunnel is dug along the radius of earth that ends at center. A ball is released from the surface of earth along tunnel. What is the velocity when it strike the center?

$(a)\;v=\sqrt {\frac{GR}{2M}} \quad (b)\;v=\sqrt {\frac{GM}{2R}} \quad (c)\;v=\sqrt {\frac{2GM}{R}} \quad (d)\;v=\sqrt {\frac{GM}{R}}$

Gain in KE= change in potential energy
$\large\frac{1}{2}$$mv^2 =m [V_A-V_B]$
$\qquad= m\bigg[\large\frac{-GM}{2R} -\bigg(\frac{-GM}{R}\bigg)\bigg]$
$\qquad= \large\frac{GMm}{2R}$
$v= \sqrt {\large\frac{GM}{R}}$
Hence d is the correct answer.
edited Jul 3, 2014