Particle Speed at Centre of Planet – Rankers Physics

Gravitational Potential Energy: Practice Problem & Solution

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
\(\sqrt{1.5} v_e\)
\(\frac{v_e}{\sqrt{2}}\)
\(v_e\)
zero

Short Solution (Best for NEET):

To solve this problem, we apply the core principles of Gravitational Potential Energy. Understanding the underlying formula is key to arriving at the correct answer below:

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 \(v^2 = \frac{3GM}{R} = 1.5 v_e^2⇒ v = \sqrt{1.5} v_e\).

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Similar Practice Questions

Explore conceptually related problems to test your knowledge.

1. One goes from the centre of the earth to an altitude half the radius of the earth, where will the \(g\) be greatest ?
2. Two identical satellites are at height \(R\) and \(7R\) from earth surface, the ratio of their kinetic energies will be :
3. The escape velocity from the Earth's surface is \(v\). The escape velocity from the surface of another planet having a radius, four times that of Earth and same mass density is
4. An object is placed at a distance of \(R/2\) from the centre of earth. Knowing mass is distributed uniformly, acceleration of that object due to gravity at that point is: (\(g =\) acceleration due to gravity on the surface of earth and \(R\) is the radius of earth)
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