Gravitational Potential

Gravitational field is a conservative force field. The work done by the gravitational field in taking a particle in a closed path is zero.

Equivalently, the work done by the gravitational force (or the external force, if the transportation is done very slowly) in transporting a particle from a point A to another point B is independent of the path and depends only on the initial and the final points (A & B).

**Work done by an external force against the gravitational forces in transporting a particle in a gravitational field is stored in the configuration of the field, and also known as potential energy of the field.**

When the particle is moved from A to B:

The stored potential energy,

$ \displaystyle U(B,A) = -\int_{A}^{B} \vec{F_{grav}}.\vec{dr}$

If we take the reference point A at ∞, we get,

Potential energy, $ \displaystyle U(P) = -\int_{\infty}^{P} \vec{F_{grav}}.\vec{dr} = -\int_{\infty}^{P} \vec{m E}.\vec{dr}$

The gravitational potential at a point P is the gravitational potential energy per unit mass placed at the point P.

Gravitational potential,

$ \displaystyle V(P)= \frac{U(P)}{m} =-\int_{\infty}^{P} \vec{E}.\vec{dr} $

The negative sign is due to the fact that the work done against the gravitational field is stored as Potential Energy.

Thus, $ \displaystyle dV = -\vec{E}.\vec{dr}$

$ \displaystyle E_x = -\frac{\partial V}{\partial x} $

$ \displaystyle E_y = -\frac{\partial V}{\partial y} $

$ \displaystyle E_z = -\frac{\partial V}{\partial z} $

Where $\vec{E}$ represents the intensity of the gravitational field.

The gravitational potential V at a point P at a distance r from a fixed point mass M is

$ \displaystyle V_P = -\int_{\infty}^{r} (-\frac{G M}{r^2}) dr $

$ \displaystyle V_P = -\frac{G M}{r} $

### Potential of a Uniform Solid Sphere :

Let a sphere of mass M and radius R

(i) at external point (r ≥ R )

$ \displaystyle V = – \frac{G M}{r}$

(ii) at internal point (r ≤ R )

$ \displaystyle V = – \frac{G M}{2 R^3}(3 R^2 – r^2) $

At the center , r = 0 ;

$ \displaystyle V = – \frac{3 G M}{2 R}$

At r = R

$ \displaystyle V = – \frac{G M}{R}$

### Potential due to a uniform thin Spherical Shell :

(i) At external Point (r ≥ R )

$ \displaystyle V = – \frac{G M}{r}$

(ii) at internal point (r ≤ R )

$ \displaystyle V = – \frac{G M}{R}$

### Potential due to a Uniform Ring at a Point on its axis:

$ \displaystyle V = – \frac{G M}{\sqrt{R^2 + x^2}}$

At x = 0 ,

$ \displaystyle V = – \frac{G M}{R}$

Illustration : Find the gravitational potential at a point where the gravitational field intensity is zero due to two particles of masses m_{1} = 1 kg and m_{2} = 4 kg separated through a distance l = 3m ?

Solution:

Let the gravitational field intensity due to the point masses m_{1} and m_{2} be zero at P.

∴ E_{P} = 0

$ \displaystyle | \vec{E_1} + \vec{E_2} | = 0 $

E_{1} = E_{2} (numerically)

$ \displaystyle \frac{G m_1}{x^2} = \frac{G m_2}{(l-x)^2} $

$ \displaystyle \frac{1}{x^2} = \frac{4}{(3-x)^2} $

$ \displaystyle \frac{3-x}{x} = \pm 2 $

3 – x = -2x ; or, 3 – x = + 2x

=> x = -3 (irrelevant) or x = 1

$ \displaystyle V_P = -G (\frac{m_1}{x} + \frac{m_2}{l-x}) $

Putting G = 6.67 × 10^{-11} N – m^{2}/kg^{2}

m_{1} = 1 kg, m_{2} = 4 kg, l = 3 m , x = 1 m, we obtain,

V_{p} = – 20.01 × 10^{-11} J/kg.

Illustration : What is the work done by the gravitational field in placing three identical particles each of mass m at the vertices of an equilateral triangle of side l from infinity ?

Solution :

We know that the gravitational field is conservative.

W = -(ΔPE)system ……(i)

ΔPE = (PE_{final} – PE_{initial})

Since initially the particles A, B and C were at infinity, their initial gravitational potential energy is zero.

∴ ΔPE = PE_{final}

The gravitational potential energy of the configuration of the particles = gravitational PE between A and B + gravitational. PE between B and C + gravitational PE between C and A

$ \displaystyle \Delta P.E = (-\frac{G m_A m_B}{l} ) + (-\frac{G m_B m_C}{l} ) + (-\frac{G m_C m_A}{l} )$

Putting m_{A} = m_{B} = m_{C} = m, we obtain,

$ \displaystyle \Delta P.E = -\frac{3 G m^2}{l} $ ….(ii)

Using (i) and (ii), we obtain,

$ \displaystyle W = \frac{3 G m^2}{l} $

Exercise : Three identical point objects, each of mass m, are placed at the vertices of an equilateral triangle of side l. What is the gravitational potential at the centre of the equilateral triangle due to the point masses?

Exercise : Find the gravitational potential of a spherical shell of mass M and radius R at a point r distant from centre for r < R and r > R.