Gravitational Potential Energy - Formulas, Derivations, Solved Problems

Gravitational potential energy is the energy possessed by an object because of its position in the gravitational field. One of the most common uses of gravitational potential energy is for an object near the surface of the Earth where the acceleration due to gravity can be assumed to be constant (9.8 m/s2).

Table of Content:

What is Gravitational Potential Energy?

When a body of mass (m) is moved from infinity to a point inside the gravitational influence of a source mass (M) without accelerating it, the amount of work done in displacing it into the source field is stored in the form of potential energy this is known as gravitational potential energy.

Explanation: We know that the potential energy of a body at a given position is defined as the energy stored in the body at that position. If the position of the body changes due to the application of external forces the change in potential energy is equal to the amount of work done on the body by the forces.

Under the action of gravitational force, the work done is independent of the path taken for a change in position so the force is a conservative force.

The gravitational influence on a body at infinity is zero, therefore, potential energy is zero, which is called a reference point.

Gravitational Potential Energy Formula

The equation for gravitational potential energy is:

⇒ GPE = mgh

Where,

  • m is the mass in kilograms,
  • g is the acceleration due to gravity (9.8 on Earth)
  • h is the height above the ground in meters

⇒ Also Read:

Derivation of Gravitational Potential Energy Equation

Consider a source mass ‘M’ is placed at a point along the x-axis, initially, a test mass ‘m’ is at infinity. A small amount of work done in bringing it without acceleration through a very small distance (dx) is given by

dw = Fdx

Here, F is an attractive force and the displacement is towards the negative x-axis direction so F and dx are in the same direction. Then,

dw = (GMm/x2)dx

Integrating on both sides

\(w=\underset{\infty }{\overset{r}{\mathop \int }}\,\frac{GMm}{{{x}^{2}}}dx\) \(w=-[\frac{GMm}{x}]_{\infty }^{r}\)

⇒ w = -GMm/r – (-GMm/∞)

⇒ w = -GMm/r

Since this work done is stored as its potential energy U, therefore gravitational potential energy at a point which is at a distance ‘r’ from the source mass is given by;

U = -GMm/r

If a test mass moves from a point inside the gravitational field to the other point inside the same gravitational field of source mass, then the change in potential energy of the test mass is given by;

ΔU = GMm (1/ri – 1/rf)

If ri > rf then ΔU is negative.

⇒ Check: Acceleration due to Gravity

Expression for Gravitational Potential Energy at Height (h) – Derive ΔU = mgh

If a body is taken from the surface of the earth to a point at a height ‘h’ above the surface of the earth, then r= R and r= R + h then,

ΔU = GMm [1/R – 1/(R+h)]

ΔU = GMmh/R(R + h)

When, h<<R, then, R + h = R and g = GM/R2.

On substituting this in the above equation we get,

Gravitational Potential Energy ΔU = mgh

⇒ Note: 

  • The weight of a body at the centre of the earth is zero due to the fact that the value of g at the centre of the earth is zero.
  • At a point in the gravitational field where the gravitational potential energy is zero, the gravitational field is zero.

What is Gravitational Potential?

The amount of work done in moving a unit test mass from infinity into the gravitational influence of source mass is known as gravitational potential.

Simply, it is the gravitational potential energy possessed by a unit test mass

⇒ V = U/m

⇒ V = -GM/r

⇒ Important Points:

  • The gravitational potential at a point is always negative, V is maximum at infinity.
  • The SI unit of gravitational potential is J/Kg.
  • The dimensional formula is M0L2T-2.

Dimensional Formulas of Physical Quantities

Relation between Gravitational Field Intensity and Gravitational Potential

Integral Form:

V = \(-\mathop{\int }\vec{E}.\overrightarrow{dr}\) ( If E is given and V has to be found using this formula)

Differential Form:

E = -dV/dr (If V is given and E has to be found using this formula)

\(\overrightarrow{~E}=\frac{\partial V}{\partial x}\hat{i}+\frac{\partial V}{\partial y}\hat{j}+\frac{\partial V}{\partial z}\hat{k}\) (components along x, y and z directions).

Gravitational Potential of a Point Mass

Consider a point mass M, the gravitational potential at a distance ‘r’ from it is given by;

V = – GM/r.

Gravitational Potential of a Spherical Shell

Consider a thin uniform spherical shell of the radius (R) and mass (M) situated in space. Now,

Case 1: If point ‘P’ lies Inside the spherical shell (r<R): 

As E = 0, V is a constant.

The value of gravitational potential is given by, V = -GM/R.

Case 2: If point ‘P’ lies on the surface of the spherical shell (r=R): 

On the surface of the earth, E = -GM/R2.

Using the relation \(V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\)over a limit of (0 to R) we get,

Gravitational Potential (V) = -GM/R.

Case 3: If point ‘P’ lies outside the spherical shell (r>R): 

Outside the spherical shell, E = -GM/r2.

Using the relation \(V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\) over a limit of (0 to r) we get,

V = -GM/r.

Gravitational Potential of a Uniform Solid Sphere

Consider a thin uniform solid sphere of the radius (R) and mass (M) situated in space. Now,

Case 1: If point ‘P’ lies Inside the uniform solid sphere (r < R): 

Inside the uniform solid sphere, E = -GMr/R3.

Using the relation \(V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\)over a limit of (0 to r).

The value of gravitational potential is given by,

V = -GM [(3R2 – r2)/2R2]

Case 2: If point ‘P’ lies On the surface of the uniform solid sphere ( r = R ):

On the surface of a uniform solid sphere, E = -GM/R2. Using the relation \(V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\) over a limit of (0 to R) we get,

V = -GM/R.

Case 3: If point ‘P’ lies Outside the uniform solid sphere ( r> R): 

Using the relation over a limit of (0 to r) we get, V = -GM/R.

Case 4: Gravitational potential at the centre of the solid sphere is given by V = -3/2 × (GM/R).

Gravitational Self Energy

The gravitational self-energy of a body is defined as the work done by an external agent in assembling the body from the infinitesimal elements that are initially at an infinite distance apart.

Gravitational Self Energy of a system of ‘n’ particles:

Let us consider n particle system in which particles interact with each other at an average distance ‘r’ due to their mutual gravitational attraction, there are n(n – 1)/2 such interactions and the potential energy of the system is equal to the sum of the potential energy of all pairs of particles i.e.,

\({{U}_{s}}=\frac{1}{2}Gn\left( n-1 \right)\frac{{{m}^{2}}}{{{r}^{2}}}\)

Solved Examples

Example 1. Calculate the gravitational potential energy of a body of mass 10Kg and is 25m above the ground.

Solution:

Given, Mass m = 10 Kg and Height h = 25 m

G.P.E is given as,

U = m × g × h = 10 Kg 9.8 m/s2 × 25 m = 2450 J.

Example 2. If the mass of the earth is 5.98 ×1024 Kg and the mass of the sun is 1.99 × 1030 Kg and earth is 160 million Kms away from the sun. Calculate the GPE of the earth.

Solution:

Given, the mass of the Earth (m) = 5.98 × 1024 Kg and mass of the Sun (M) = 1.99 × 1030 Kg

The gravitational potential energy is given by:

U = -GMm/r

U = 6.673 ∗ 10-11 ∗ 5.98 ∗ 1024 ∗1.99∗1030/(160∗109) = 8.29 × 108 J

Example 3. A basketball weighing 2.2 kg falls off a building to the ground 50 m below. Calculate the gravitational potential energy of the ball when it arrives below.

Solution:

GPE = (2.2 kg)(9.8 m/s2)(50 m) = 1078 J.

Example 4: A 2 kg body free falls from rest from a height of 12 m. Determine the work done by the force of gravity and the change in gravitational potential energy. Consider the acceleration due to gravity to be 10 m/s2.

Solution:

Since, W = mgh

Substituting the values in the above equation, we get

W = 2 × 12 × 10 = 240 N

The change in gravitational potential energy is equal to the work done by gravity.

Therefore, ΔEP = 100 Joule.

⇒ Also Read:


Practise This Question

A uniform ring of mass m is lying at a distance 1.73 a from the centre of a sphere of mass M just over the sphere where a is the small radius of the ring as well as that of the sphere. Then gravitational

force exerted is