Q: A tunnel is dug along the diameter of the earth. There is a particle of mass m at the centre of the tunnel. Find the minimum velocity given to the particle so that is just reaches to the surface of the earth. (R = radius of earth)
(a) $ \displaystyle \sqrt{\frac{GM}{R}} $
(b) $\displaystyle \sqrt{\frac{GM}{2R}} $
(c) $ \displaystyle \sqrt{\frac{2GM}{R}} $
(d) it will reach with the help of negligible velocity
Ans: (a)
Sol: Applying Energy conservation between centre & Surface
$ \displaystyle -\frac{G M m }{2 R^3}(3R^2 – 0) + \frac{1}{2}m v^2 = -\frac{G M m}{R} $
$ \displaystyle -\frac{3 G M m }{2 R} + \frac{1}{2}m v^2 = -\frac{G M m}{R} $
$ \displaystyle \frac{1}{2}m v^2 = \frac{3 G M m }{2 R} -\frac{G M m}{R} $
$ \displaystyle \frac{1}{2}m v^2 = \frac{ G M m }{2 R} $
$ \displaystyle v^2 = \frac{ G M }{R} $
$ \displaystyle v = \sqrt{\frac{ G M }{R}} $