Gravitational Potential Energy - Physics Class 11 - NEET PDF Download

Table of Contents
1. What is Gravitational Potential Energy?
2. Gravitational potential energy near Earth's surface - the mgh formula
3. Derivation of gravitational potential energy for point masses
4. Important remarks and clarifications
5. Gravitational potential
6. Relation between gravitational field intensity and potential
7. Gravitational potential of simple distributions
8. Gravitational self-energy
9. Solved Problems
View more

Gravitational potential energy is the energy possessed by an object because of its position in a gravitational field. In simple terms, an object at a greater height in a gravitational field has greater gravitational potential energy than the same object at a lower height. For example, a pencil held above a table has greater gravitational potential energy than the same pencil lying on the table. This energy represents the object's ability to do work as a result of its position in the gravitational field.

What is Gravitational Potential Energy?

Gravitational potential energy of a body of mass m is the work done in bringing it from a reference point (commonly from infinity) to its position in the gravitational field of a source mass M, without acceleration. The gravitational potential energy is denoted by U or U_g.

The gravitational force is a conservative force. Therefore, the work done by (or against) gravity in moving a body between two points is independent of the path taken; it depends only on the initial and final positions. By convention, gravitational potential energy is taken as zero at infinity for an isolated source mass.

Gravitational potential energy near Earth's surface - the mgh formula

The commonly used expression for gravitational potential energy when a body of mass m is raised through a vertical height h near the Earth's surface is

U = m g h

Where,

• m is the mass (kg)
• g is the acceleration due to gravity (≈ 9.8 m/s² near Earth's surface; often approximated as 10 m/s² in examples)
• h is the height above the chosen reference level (m)

Derivation of gravitational potential energy for point masses

Consider a source mass M placed at the origin. A test mass m is brought from infinity to a point at a distance r from the source mass, slowly (without acceleration). The magnitude of the gravitational force between them is

F(r) = G M m / r²

A small amount of work done in moving the mass over an infinitesimal radial displacement dr (directed towards the source) is

dW = F(r) dr

Gravitational Potential

Integrating the work done from infinity (r = ∞) to r = the final position gives the work stored as potential energy. Since the force is attractive and the displacement is towards decreasing r, the potential energy is negative with the chosen reference U(∞)=0.

Carrying out the integration yields

U(r) = - G M m / r

Thus the gravitational potential energy of mass m at distance r from source mass M is -GMm/r.

If the test mass moves from an initial radius r_i to a final radius r_f, the change in gravitational potential energy is

ΔU = U(f) - U(i) = -GMm (1/r_f - 1/r_i) = GMm (1/r_i - 1/r_f)

Sign convention: If r_f < r_i (mass moved closer to source), ΔU is negative (potential energy decreases) and work is done by gravity.

Derivation of ΔU = m g h for small heights above Earth

Let Earth have radius R and mass M. A body of mass m is moved from the surface (r = R) to a height h above the surface (r = R + h). From the general formula,

ΔU = GMm (1/R - 1/(R + h)).

When h ≪ R, expand or approximate 1/(R + h) ≈ 1/R - h/R², so

ΔU ≈ GMm (1/R - (1/R - h/R²)) = GMm (h/R²).

Using g = GM/R² near the Earth's surface, we obtain

ΔU = m g h

Important remarks and clarifications

• Weight at Earth's centre: The gravitational field intensity g inside Earth decreases with radius and becomes zero at the centre for a spherically symmetric mass distribution; hence the weight of a body at the centre is zero.
• Potential and field are different quantities: Gravitational potential (a scalar) can be zero at some points depending on reference choice (for example at infinity by convention). Zero potential at a chosen reference does not generally imply zero gravitational field. The gravitational field (a vector) is zero at locations where the net gravitational force vanishes (for example, the centre of a symmetric shell or certain points between masses), but potential there need not be zero.

Gravitational potential

Gravitational potential at a point is the gravitational potential energy per unit test mass placed at that point. If U is the potential energy of mass m then

V = U / m

For a point mass M at distance r,

V(r) = - G M / r

Important points:

• The SI unit of gravitational potential is J kg^-1 (or simply J/kg).
• Gravitational potential for an isolated point mass is negative for finite r and tends to zero from below as r → ∞. By the usual convention V(∞) = 0.
• The dimensional formula of gravitational potential is L² T^-2 (equivalently M⁰ L² T⁻²).

Relation between gravitational field intensity and potential

The gravitational field (intensity) E at a point is the negative gradient of the gravitational potential V.

In one dimension (radial),

E(r) = - dV / dr

In vector form,

→E = -∇V

Thus, if the potential is known, the field can be found by differentiating the potential; conversely, if the field is known, the potential difference between two points can be found by integrating the field along a path.

Gravitational potential of simple distributions

Point mass

For a point mass M, the potential at distance r is

V(r) = - GM / r

Thin uniform spherical shell (mass M, radius R)

• Case 1 - Inside the shell (r < R):

  The gravitational field inside a thin uniform spherical shell is zero; therefore the potential is constant throughout the interior and equal to its value on the surface.

  V(r) = - GM / R

• Case 2 - On the surface (r = R):

  The field just outside is E = GM/R² directed radially inward (magnitude GM/R²). The potential on the surface is

  V(R) = - GM / R

• Case 3 - Outside the shell (r > R):

  The shell behaves gravitationally as if all its mass were concentrated at its centre. For r > R,

  V(r) = - GM / r

Uniform solid sphere (mass M, radius R)

• Case 1 - Inside the sphere (r < R):

  The gravitational field inside a uniform solid sphere varies linearly with r:

  E(r) = G M r / R³ (directed radially inward in magnitude; sign convention depends on radial direction chosen).

  The gravitational potential inside (measured with V(∞)=0) is

  V(r) = - (G M / 2 R³) (3 R² - r²) = - G M (3 R² - r²) / (2 R³)

  At the centre (r = 0),

  V(0) = - 3 G M / (2 R)

• Case 2 - On the surface (r = R):

  The potential on the surface equals that of a point mass of same mass at distance R,

  V(R) = - G M / R

• Case 3 - Outside the sphere (r > R):

  As for the shell, the sphere acts as if all mass were concentrated at its centre:

  V(r) = - G M / r

Gravitational self-energy

Gravitational self-energy (sometimes called gravitational binding energy) of an extended body is the work required to assemble the body from infinitesimal pieces brought in from infinity. Equivalently, it is the energy released when the pieces are assembled under their mutual gravity.

For a system of n point masses, the gravitational potential energy is the sum over all distinct pairs:

For a continuous body one integrates over the mass distribution. For a uniform solid sphere of mass M and radius R, the gravitational self-energy is

U_self = - 3 G M² / (5 R)

This negative value indicates that work must be done against gravity to disperse the body to infinity; conversely, assembling the sphere releases this magnitude of energy.

Solved Problems

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.

Use U = m g h.

Substitute g = 9.8 m/s².

U = 10 × 9.8 × 25.

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 Kms away from the sun. Calculate the GPE of the earth.

Solution:

Given mass of Earth m = 5.98 × 10²⁴ kg.

Mass of Sun M = 1.99 × 10³⁰ kg.

Distance r = 160 million km = 160 × 10⁶ km = 160 × 10⁶ × 10³ m = 160 × 10⁹ m = 1.60 × 10¹¹ m.

Use U = - G M m / r.

Substitute G = 6.673 × 10^-11 N m² kg^-2, M = 1.99 × 10³⁰ kg, m = 5.98 × 10²⁴ kg, r = 1.60 × 10¹¹ m.

Compute G M m ≈ 6.673 × 10^-11 × 1.99 × 10³⁰ × 5.98 × 10²⁴ ≈ 7.943 × 10⁴⁴ (approximately).

Divide by r = 1.60 × 10¹¹ gives U ≈ - 4.96 × 10³³ J.

Therefore the gravitational potential energy of the Earth in the Sun's field (with the chosen sign convention) is approximately U ≈ - 4.96 × 10³³ 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:

Given mass m = 2.2 kg and height h = 50 m.

Use U = m g h with g = 9.8 m/s².

U = 2.2 × 9.8 × 50.

U = 1078 J.

The gravitational potential energy lost by the ball during the fall is 1078 J (which becomes kinetic energy just before impact, neglecting air resistance).

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:

Given mass m = 2 kg, height h = 12 m, and g = 10 m/s².

The work done by gravity equals the decrease in gravitational potential energy: W = m g h.

Substitute values: W = 2 × 10 × 12.

W = 240 J.

Therefore, the work done by gravity is 240 J, and the change in gravitational potential energy (final - initial) is -240 J (a decrease of 240 J).