Question

What is the gravitational potential energy of a particle of mass m kept at a distance x from the centre of a disc of mass M on its axis? The radius of the disc is R.
A. $ - \dfrac{{2GMm}}{{{R^2}}}\left( {\sqrt {{R^2} + {x^2}} - x} \right)$
B. $ - \dfrac{{4GMm}}{{{R^2}}}\left( {\sqrt {{R^2} + {x^2}} - x} \right)$
C. $ - \dfrac{{6GMm}}{{{R^2}}}\left( {\sqrt {{R^2} + {x^2}} - x} \right)$
D. $ - \dfrac{{9GMm}}{{{R^2}}}\left( {\sqrt {{R^2} + {x^2}} - x} \right)$

Answer 1

Hint: Potential energy is defined as the stored energy of position possessed by an object. While gravitational potential energy is defined as the energy stored in an object due to its vertical position or its height. In order to find the solution of the question, we will draw a rough diagram of the disc and apply the concept of the gravitational potential energy.
Formula used:
$F = G\dfrac{{Mm}}{{{R^2}}}$
$U = mV$

Complete answer:
Let us consider a uniform disc of radius ‘R’. We are required to find the gravitational potential energy of the particle of mass ‘m’ which is kept at a distance ‘x’ from the centre of a disc of mass ‘M’ on its axis.

We know that the acceleration due to force on a body is given by, $a = \dfrac{F}{m}$. Where ‘F’ is the gravitational force acting on the body of mass ‘m’, and ‘a’ is the acceleration of a free-falling body. The force is given by, $F = G\dfrac{{Mm}}{{{R^2}}}$ Here, ‘M’ is the mass of the earth, ‘G’ is the universal gravitational constant and ‘R’ is the radius of the earth. Now, acceleration due to gravity is given by, $a = \dfrac{F}{m} = \dfrac{{GM}}{{{R^2}}}$ From this relation, the acceleration produced n a body does not depend upon the mass of the body. Thus, the acceleration due to gravity is the same for all the bodies.

The gravitational potential energy of the differential ring is given as,
$dV = - G\dfrac{{dM}}{{{r^2}}}$ --(1)
Where $dM = \dfrac{M}{{\pi {R^2}}} \times 2\pi rdr$
$ \Rightarrow dM = \dfrac{{2Mrdr}}{{{R^2}}}$
Now, ${r^2} = \sqrt {{r^2} + {x^2}} $
Substituting the values in equation (1) we get,
$dV = - G\dfrac{{\dfrac{{2Mrdr}}{{{R^2}}}}}{{\sqrt {{r^2} + {x^2}} }}$
Now, the potential due to the entire disc at point P is given as,
$V = \int {dV = - \dfrac{{GM}}{{{R^2}}}} \int\limits_0^R {\dfrac{{2rdr}}{{\sqrt {{r^2} + {x^2}} }}} $
Let, ${r^2} + {x^2} = {t^2}$ $ \Rightarrow 2rdr = 2tdt$
$V = - \dfrac{{GM}}{{{R^2}}}\int {\dfrac{{2tdt}}{t}} $
$ \Rightarrow V = - \dfrac{{2GM}}{{{R^2}}}t$
$ \Rightarrow V = - \dfrac{{2GM}}{{{R^2}}}{\left[ {\sqrt {{r^2} + {x^2}} } \right]_0}^R$
$ \Rightarrow V = - \dfrac{{2GM}}{{{R^2}}}\left[ {\sqrt {{R^2} + {x^2}} - x} \right]$
The gravitational potential energy of a system is given by,
$U = mV$
$ \Rightarrow V = - \dfrac{{2GMm}}{{{R^2}}}\left[ {\sqrt {{R^2} + {x^2}} - x} \right]$

So, the correct answer is “Option A”.

Note:
 An electric potential is defined as the work needed to move a unit of electric charge from a time varying electric field and vice versa. The electrostatic potential between any two arbitrary charges ‘${q_1}$’ and ‘${q_2}$’ separated by a distance ‘r’ is stated by Coulomb’s law. Mathematically it is given as, $U = k \times \dfrac{{{q_1}{q_2}}}{{{r^2}}}$; where ‘U’ is denoted for the electrostatic potential energy and ‘${q_1}$’ and ‘${q_2}$’ are two charges.