Gravitational potential energy is the energy an object has due to its position in a gravitational field. In other words, it is the energy associated with gravity or the gravitational force.

The pencil held above the table will have greater gravitational potential energy than the pencil that is placed at the table. This is the most common example used to help understand the concept of gravitational potential energy.

We can understand from this that the position of a pencil or any other object in a gravitational field gives it the potential to do work. We will explore this concept in greater detail below.

Table of Contents:

• What is Gravitational Potential Energy?
• Gravitational Potential Energy Formula
• Derivation of Gravitational Potential Energy Equation
• What is Gravitational Potential?
• Gravitational Field Intensity and Gravitational Potential: A Comparison
• Gravitational Potential of a Point Mass
• Gravitational Potential of a Spherical Shell
• Gravitational Potential of a Uniform Solid Sphere
• Solved Examples

What is Gravitational Potential Energy?

Gravitational Potential Energy is the energy stored in an object due to its position in a gravitational field. It is equal to the work done by the gravitational force to move the object to its current position.

The gravitational potential energy, represented by the symbol Ug, is the amount of work done in displacing a body of mass (m) from infinity to a point inside the gravitational influence of a source mass (M) without accelerating it. This energy is stored in the form of potential energy.

Explanation: The potential energy of a body at a given position is the energy stored in the body at that position. If the position of the body is altered by external forces, the change in potential energy is equivalent 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, which makes it a conservative force. Additionally, all such forces have some potential energy associated with them.

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

Gravitational Potential Energy Formula

The equation for gravitational potential energy is:

GPE = m * g * h

Where

m is the mass in kilograms.

Gravity (g) is the acceleration due to gravity on Earth, which is 9.8 m/s²

H is the height above the ground in meters.

Also Read:

Gravitational Force

Kepler’s Laws

Gravitational Field Intensity

Derivation of the Gravitational Potential Energy Equation

Consider a source mass M 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

The net work done, W will be positive, i.e., W = F*dx > 0.

$dw=\frac{GMm}{x^2}dx$

Integrating on both sides

\(\int_{-\infty}^r \frac{GMm}{x^2} dx\)

\begin{array}{l}w = -\int_{\infty}^{r}\frac{GMm}{x},dx\end{array}

\begin{array}{l}w = \frac{GMm}{r} + \frac{GMm}{\infty}\end{array}

$$\frac{-GMm}{r} = w$$

Therefore, the gravitational potential energy at a point which is at a distance ‘r’ from the source mass is given by U, where U is the work done stored as its potential energy.

$$U = -\frac{GMm}{r}$$

If a test mass moves from one point to another point within the same gravitational field of a source mass, then the change in potential energy of the test mass is given by:

$$ΔU = GMm [\frac{1}{r_i} - \frac{1}{r_f}]$$

If $$r_i > r_f$$, then $$\Delta U$$ is negative.

⇒ Check: Acceleration due to Gravity

Derivation of Expression for Gravitational Potential Energy at Height (h):

$$\Delta U = mgh$$

Given:

• Mass (m)
• Gravitational acceleration (g)
• Height (h)

Derivation: $$\Delta U = mgh$$

$$\Delta U = m \times g \times h$$

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 $$r_i = R$$ and $$r_f = R + h$$.

$$\Delta U = GMm \left[ \frac{1}{R} - \frac{1}{R+h} \right]$$

$$\Delta U = \frac{GMmh}{R(R + h)}$$

When $$h \lt \lt R$$, then $$R + h = R$$ and $$g = \frac{GM}{R^2}$$.

Substituting this into the equation above yields:

ΔU = mgh
where:

• ΔU = Change in Gravitational Potential Energy
• m = mass
• g = acceleration due to gravity
• h = height

⇒ Note:

The weight of a body is zero at the centre of the earth due to the fact that the value of g at the centre of the earth is also zero.

At a point in the gravitational field where the gravitational potential energy is zero, the gravitational field strength is also zero.

What is Gravitational Potential?

Gravitational Potential is the amount of energy stored in a physical system due to its position in a gravitational field.

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

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

V = U/m

$$V = \frac{-GM}{r}$$

⇒ Important Points:

1. Make sure to review all important points
2. Pay close attention to the details
3. Take the time to double check your work

The gravitational potential at a point is always negative, and its maximum value is at infinity.

The SI unit of gravitational potential is J/Kg.

The dimensional formula is [M]⁰[L]²[T]^-2

⇒ Click Here to Learn About the Dimensional Formulas of Physical Quantities

The Relationship Between Gravitational Field Intensity and Gravitational Potential

Integral Form: $$\int f(x) dx$$

$$V = (\int \vec{E} \cdot d\vec{r}) (If (\vec{E})$$ is given, then V can be found using this formula)

Differential Form: $$\frac{dy}{dx} = f(x,y)$$

$$E = \frac{-dV}{dr}$$ (If V is given, E can be found using this formula)

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

Gravitational Potential of a Point Mass

The gravitational potential at a distance r from a point mass M is given by:

$$V = \frac{-GM}{r}$$

Gravitational Potential of a Spherical Shell

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

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

As E = 0, V remains constant.

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

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

$$E = -\frac{GM}{R^2}$$ on the surface of the Earth.

Using the relation $$V=-\int\vec{E}\cdot\overrightarrow{dr}$$ over a limit of $(0,R)$, 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 (\begin{array}{l}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array}) from 0 to r, we get,

V = \frac{-GM}{r}

Gravitational Potential of a Uniform Solid Sphere

Consider a thin uniform solid sphere of radius (R) and mass (M) situated in space. What would be the gravitational force between two such spheres?

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

Inside the uniform solid sphere, E = -\frac{GM}{r^3}

Using the relation (\displaystyle V=-\int_0^r \vec{E}\cdot\overrightarrow{dr} )

The gravitational potential is given by:

V = -GM $\frac{(3R^2 – r^2)}{2R^2}$

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

Using the relation \begin{array}{l}V=-\mathop{\int }\vec{E}.\overrightarrow{dr}\end{array} over a limit of (0 to R) with the surface electric field given by $$(E = -\frac{GM}{R^2})$$, we get,

$$V = \frac{-GM}{R}$$

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, $V = -\dfrac{GM}{R}$.

Case 4: The gravitational potential at the centre of a solid sphere is given by: $$V = \left(-\frac{3}{2}\right) \times \left(\frac{GM}{R}\right).$$

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 infinitesimal elements that are initially separated by an infinite distance.

Gravitational Self-Energy of a System of ’n’ Particles:

Let us consider an n particle system in which particles interact with each other at an average distance r due to their mutual gravitational attraction. There are $$\frac{n(n-1)}{2}$$ 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.,

$$U_{s} = \frac{1}{2}Gn(n-1)\frac{m^2}{r^2}$$

Solved Problems

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

Gravitational potential energy of the body = mgh

$$= 10 \times 9.8 \times 25$$

$$= 2450 \text{ J}$$

Solution:

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

G.P.E (Gravitational Potential Energy) is given as,

U = 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 the Earth is 160 million km away from the Sun, calculate the gravitational potential energy (GPE) of the Earth.

Solution:

The mass of the Earth (m) is 5.98 x 10²⁴ kg and the mass of the Sun (M) is 1.99 x 10³⁰ kg.

The gravitational potential energy is given by:

$$U = -\frac{GMm}{r}$$

$U=(6.673\times {10}^{-11}\times 5.98\times {10}^{24}\times 1.99\times {10}^{30})/(160\times {10}^9)=4.963\times {10}^{30}\text{J}$

Example 3. Calculate the gravitational potential energy of a basketball weighing 2.2 kg when it arrives on the ground 50 m below after falling off a building.

Solution:

GPE = $(2.2~kg)(9.8~m/s^2)(50~m) = 1078~J.

Answer:

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

Solution:

Since, $$W = mgh$$

Substituting the values in the equation above, we get.

W = 2 × 12 × 10 = 240 N

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

Therefore, Gravitational Potential Energy = 240 Joules.

Frequently Asked Questions on Gravitational Potential Energy

What is Gravitational Potential Energy?

Gravitational potential energy is the energy an object has due to its position in a gravitational field. It is equal to the work done by the gravitational force to move an object from a very far distance to its current position.

The gravitational potential at a point in the gravitational field of a body is denoted as V and is defined as the amount of work done in displacing a body of unit mass from infinity to that point in the field.

The expression for gravitational potential energy is U = mgh, where m is the mass of the object, g is the acceleration due to gravity, and h is the height of the object.

Gravitational Potential, V = -GMm/r

M is the source mass placed along the x-axis.

M is the test mass at infinity.

The gravitational potential energy is determined at a distance of r from the source mass.

What is the unit of gravitational potential?

The unit of gravitational potential is Joules per kilogram (J/kg).

The unit of gravitational potential is J/kg.

What is the dimensional formula of gravitational potential?

The dimensional formula of gravitational potential is $[M^0{L}^2{T}^{-2}]$.