Question

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?
A: $\sqrt{1.5}{{v}_{e}}$
B: $\dfrac{{{v}_{e}}}{\sqrt{2}}$
C: ${{v}_{e}}$
D: zero

Answer 1

This problem can be solved easily by using the concept of gravitational potential. The gravitational potential can be defined at a point in the gravitational field as the work done per unit mass that would have to be done by some competent external force to bring a massive object to that point from some defined position of zero potential which is taken to be infinity.

Complete step by step answer:
We know that the value of gravitational potential at some distance from the earth’s surface is given by: $-GM[\dfrac{3{{R}^{2}}-{{r}^{2}}}{2{{R}^{2}}}]$
Where G is the universal gravitational constant, r is the distance from the centre of the earth to the point where gravitational potential is to be found, $\rho$is the density of the earth and R is the radius of the earth.
If we go at the centre of the earth then r=0 and the above expression reduce to
$\dfrac{-3GM}{2R}$
Now using energy conservation,
\dfrac{m{{v}^{2}}}{2}=m[0-(\dfrac{-3GM}{2R})] \\\ \Rightarrow {{v}^{2}}=\dfrac{3GM}{2R} \\\ \because GM=g{{R}^{2}} \\\ \Rightarrow {{v}^{2}}=\dfrac{3g{{R}^{2}}}{2R} \\\ \Rightarrow {{v}^{2}}=1.5gR \\\ v=\sqrt{1.5gR} \\\ \because {{v}_{e}}=\sqrt{gR} \\\ \therefore v=\sqrt{1.5}{{v}_{e}} \\\

So, the correct answer is “Option A”.

Note:
Gravitational potential has its origin in gravitational potential energy and the value of gravitational potential at a point is always negative and it assumes its maximum value only at infinity.
The gravitational potential is measured in the units of J/kg and its dimensional formula is $[{{M}^{0}}{{L}^{2}}{{T}^{-2}}]$. Here we have written the velocity in terms of escape velocity on the earth.