Solution:
Using conservation of mechanical energy from infinity to the centre: \(0 = \frac{1}{2}mv^2 - \frac{3GmM}{2R}\). Since escape velocity is \(v_e = \sqrt{\frac{2GM}{R}}\), we get \(v^2 = \frac{3GM}{R} = 1.5 v_e^2⇒ v = \sqrt{1.5} v_e\).
The escape velocity for a planet is \(v_e\). A particle starts from rest at a large distance from the planet, reaches the planet only under gravitational attraction, and passes through a smooth tunnel through its centre. Its speed at the centre of the planet will be
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