Question

The escape velocity for a planet is ${{\rm{v}}_{\rm{e}}}$. A tunnel is dug along the diameter of the planet and a small body is dropped into it at the surface. When the body reaches the center of the planet, its speed will be
(A). ${{\rm{v}}_{\rm{e}}}$
(B). $\dfrac{{{{\rm{v}}_{\rm{e}}}}}{{\sqrt 2 }}$
(C). $\dfrac{{{{\rm{v}}_{\rm{e}}}}}{2}$
(D). zero

Answer 1

To solve this problem, first we will write the formula for Gravitational potential at the surface of the earth, and then we will write the Gravitational potential energy at the center of the earth, and finally use the Law of conservation of energy, and escape velocity to obtain the required result.

Complete step by step answer:
The Gravitational potential at the surface of the earth is given by,
${v_s} = - \dfrac{{GM}}{R}$
Here, $G$ is the universal Gravitational constant, $M$ is the mass of the object, and $R$ is the radius.
The Gravitational potential energy at distance $r$ inside a solid sphere is given by,
$v = - \dfrac{{GM}}{{2{R^3}}}\left( {3{R^2} - {r^2}} \right)$
We need to check the potential energy at the center of the Earth, so, $r = 0$ then formula will become as follows.
$v = - \dfrac{{GM}}{{2{R^3}}}\left( {3{R^2} - {{\left( 0 \right)}^2}} \right)\\\ = - \dfrac{{GM}}{{2{R^3}}}\left( {3{R^2}} \right)\\\ = - \dfrac{{3GM}}{{2R}}$
Let the Gravitational potential energy at the center of the earth is ${V_0}$
${v_0} = - \dfrac{{3GM}}{{2R}}$
By the conservation of energy, the loss in potential energy will be equal to the gain in kinetic energy. So,
$m\left( {{v_s} - {v_0}} \right) = \dfrac{1}{2}m{v^2}\\\ {v^2} = 2\left( { - \dfrac{{GM}}{R} + \dfrac{{3GM}}{{2R}}} \right)\\\ = \dfrac{{2GM}}{R}\left( {\dfrac{3}{2} - 1} \right)\\\ = \dfrac{{GM}}{R}$
The escape velocity is that minimum velocity of an object which is needed for an object to escape from the Earth’s gravitational force. If the kinetic energy of an object is equal in magnitude to the potential energy, then in the absence of friction resistance it could escape from the Earth.
The formula of escape velocity is given by,
${v_{\rm{e}}} = \sqrt {\dfrac{{2GM}}{R}}$
By Squaring both sides of the equation ${v_{\rm{e}}} = \sqrt {\dfrac{{2GM}}{R}}$, we get
$v_{\rm{e}}^2 = \dfrac{{2GM}}{R}$
By substituting ${v^2}$ for $\dfrac{{GM}}{R}$ in the equation $v_e^2 = \dfrac{{2GM}}{R}$, we get
$v_{\rm{e}}^2 = 2{v^2}\\\ \implies {v^2} = \dfrac{{v_{\rm{e}}^2}}{2}\\\ \implies v = \dfrac{{{v_{\rm{e}}}}}{{\sqrt 2 }}$

Therefore, the option (B) is the correct answer.

Note:
To obtain the correct answer we should use the correct formulas for Gravitational potential at the surface of the earth, and Gravitational potential energy at the center of the earth.
If the kinetic energy of an object is equal in magnitude to the potential energy, then in the absence of friction resistance it could escape from the Earth.