Question

The escape velocity for a planet is \[{v_e}\]. A tunnel is dug along a diameter of the planet and a small body is dropped into it at the surface. When the body reaches the centre of the planet, its speed will be
A.\[{v_e}\]
B.\[\dfrac{{{v_e}}}{{\sqrt 2 }}\]
C.\[\dfrac{{{v_e}}}{2}\]
D.\[\;Zero\]

Answer 1

Hint: For a spherical, massive/huge body such as a star, or planet, the escape velocity i.e, \[{v_e}\] for that body, at a particular distance is calculated by the formula below,
\[{v_e} = \sqrt {\dfrac{{2GM}}{r}} \]
Where,
\[G\] is universal gravitational constant given as \[G \approx 6.67 \times {10^{ - 11}}{m^3}.k{g^{ - 1}}.{s^{ - 2}}\]
M the mass of the body to be escaped from, and r the distance from the center of mass of the body to the object.

Step by step answer: The escape velocity is the minimal velocity wanted for a free, non-propelled object to break out from the gravitational effect of a huge object body, that is, to acquire an countless distance from it. Escape velocity is a feature of the mass of the object body and distance to the center of mass of the object body.
An example of a rocket, constantly accelerated with the aid of using its exhaust, no longer attain ballistic escape velocity at any distance because it is provided with extra kinetic energy with the aid of using the expulsion of its reaction mass. It can acquire escape at any velocity, given in an appropriate mode of propulsion and enough propellant to offer the accelerating force at the object to escape from the gravitational force.
 \[{\left( {K + Ug} \right)_i} = {\left( {K + Ug} \right)_f}\]
Kƒ =0 because of final velocity is 0, and Ugƒ = 0 because of its final distance is infinity, so
\[ \Rightarrow \dfrac{1}{2}m{v_e}^2 + \dfrac{{ - GMm}}{r} = 0 + 0\]
\[ \Rightarrow {v_e} = \sqrt {\dfrac{{2GM}}{r}} = \sqrt {\dfrac{{2\mu }}{r}} \]
where μ is the std gravitational parameter= GM.
So, we have formula calculating escape velocity
\[{v_e} = \sqrt {\dfrac{{2GM}}{r}} \]
if the body is reached at the center from surface, change in energy is zero i.e, \[{E_f} - {E_i} = 0\]
\[{\left( {K + Ug} \right)_i} = {\left( {K + Ug} \right)_f}\]
\[\left( {0 + \left( {\dfrac{{ - GMm}}{R}} \right)} \right) - \left( {\dfrac{{m{v^2}}}{2} + \dfrac{{ - 3GMm}}{{2R}}} \right) = 0 \to v = \left( {\dfrac{{GM}}{R}} \right)\]
Substituting and taking ratio we get,
\[\dfrac{v}{{{v_e}}} = \sqrt {\dfrac{{GM}}{R}} \times \sqrt {\dfrac{R}{{2GM}}} \]
So, \[v = \dfrac{{{v_e}}}{{\sqrt 2 }}\]
Therefore answer is Option B. \[v = \dfrac{{{v_e}}}{{\sqrt 2 }}\].

Note: Kinetic energy K and gravitational potential energy Ug are the only two types of energy that we have to deal with, so by the conservation of energy, we get
\[{\left( {K + Ug} \right)_i} = {\left( {K + Ug} \right)_f}\]