Gravitational Potential Energy

Gravitational potential energy is the energy possessed by an object due to its position in a gravitational field. It arises because of the gravitational force acting on the object and depends on its location relative to a reference point. When the object moves to a different position, its potential energy changes as work is done by or against gravity.

• Gravitational Potential Energy is represented by the symbol U.
• Its SI unit is J.
• The dimensional formula of Gravitational Potential Energy is [M¹L²T^-2]

Formula

Mathematically, gravitational potential energy is given by the product of the mass of the object, acceleration due to gravity, and the height above the ground:

U = mgh

• m is the mass of the object
• g is the gravity of the Earth
• h is the height at which object is held

Gravitational Potential Energy is negative when mass from infinity approaches the gravitational field of a planet.

The work done on a body is represented by expression when it is elevated to a height from the earth's surface.

W = mgh

where 
m is the mass of the body

The work done on a body when it is in the ground label is zero since its height is zero. 

W = 0, at h = 0 

Because a body's height is 0 while it is in the ground label, the work done on it is zero.

When a force is applied to a body, it changes its position. The amount of work done on the body equals the change in the potential energy of the body. Consider a body that has been transported vertically from a point at height h₁ to a position at height h₂. The work done by the force put on it while raising it via the height between the two places is equal to the difference in gravitational potential energy at this moment.

W = mgh₂ - mgh₁

Derivation

Consider two particles of masses m₁ and m₂​ separated by a distance r. The gravitational force between them is:

When one particle is moved by an infinitesimal distance dr, the work done by the gravitational force is:

dW = −Fdr

Substituting the value of F:

The change in gravitational potential energy dU is defined as:

dU = -dW

To find the total change in potential energy when the distance changes from r₁​ to r₂, integrate:

Expression for Gravitational Potential Energy at Height (h)

When an object is taken off from the Earth's surface to a point 'h' above the Earth's surface and the mass of the Earth is taken to be M and the mass of the object is m, such that,

r₁ = R

r₂ = R +h

Then,

ΔU = GMm/R – GMm/(R+h)

ΔU = GMm [1/R – 1/(R+h)]

ΔU = GMmh/R(R + h)

For a very small value of h R + h >>> h such that, R +h = r and g = GM/R²

On Substituting:

ΔU = mgh

Gravitational Potential

Gravitational Potential is defined as the amount of work done in moving a unit test mass from infinity into the gravitational field of the source mass.

Gravitational Potential Energy possessed by the unit test mass is called Gravitational Potential. It is represented by the symbol V.

• V = U/m
• V = -GM/r

Gravitational potential at a point is always negative and is maximum at infinity.

Relation between Gravitational Field Intensity and Gravitational Potential

The integral form of the gravitational field is,

V = - 

• V is Gravitational Potential
• E is Gravitational Field Intensity

Differential form of Gravitational Field Intensity and Gravitational Potential is,

E = -dV/dr

Gravitational Potential of Point Mass

For a point mass of mass M, the gravitational potential at a distance ‘r’ from the point mass is given by the formula,

V = – GM/r

Gravitational Potential of a Spherical Shell

The Gravitational Potential of a Spherical Shell is calculated using the Gravitational Potential formula,

Now for calculating the Gravitational Potential of a Spherical Shell consider a thin uniform spherical shell of the radius (R) and mass (M), then

Case 1: At point ‘P’ which is inside the spherical shell such that r<R.

 V is a constant as E = 0 and the gravitational potential is given by, 

V = -GM/R

Case 2: At point ‘P’ which is at the surface of the spherical shell such that r = R.

E = -GM/R².

Now,

V = - 

solving the integral from (0 to R), Gravitational Potential as

V = -GM/R

Case 3: At point ‘P’ which is outside the spherical shell such that r>R

E = -GM/r².

Now,

V = - 

solving the integral from (0 to r), Gravitational Potential as

V = -GM/r

Gravitational Potential of a Uniform Solid Sphere

The Gravitational Potential of a uniform solid sphere can easily be calculated using the gravitational potential formula. Let's consider a thin uniform solid sphere of the radius (R) and mass (M) then,

Case 1: At point ‘P’ which is inside the solid sphere such that r<R, the gravitational potential is given by, 

E = -GMr/R³

Now,

V = - 

Integrating over the limit from (0 to r)

Now the Gravitational Potential is given by,

V = -GM [(3R² – r²)/2R²]

Case 2: At point ‘P’ at the surface of the solid sphere such that r = R

E = -GM/R²

V = - 

Integrating over the limit from (0 to r)

V = -GM/R

Case 3: At point ‘P’ outside the solid sphere such that r > R

V = -GM/R

Case 4: At the centre of the solid sphere gravitational potential is given by,

V =(-3/2) × (GM/R)

Gravitational Self Energy

Gravitational self energy is defined as the work done by an external agent in assembling a body by bringing its infinitesimal mass elements from infinity to their final positions against gravitational attraction.

Gravitational Self-Energy of a System of n Particles

Consider a system of n identical particles, each of mass m, separated by an average distance r. Each pair of particles interacts gravitationally.

Total number of interacting pairs:

Potential energy of one pair:

U =

Total gravitational self-energy of the system:

U_s =

Mass of the Earth

The mass of the Earth can be calculated using the relation for acceleration due to gravity at the surface:

...(1)

where 
G is the Universal Gravitational Constant
R is the radius of the earth
g is the acceleration due to gravity
M is the mass of earth

Rearranging the formula to find M:

Substituting the values:

The mass of the Earth is approximately 5.98 × 10²⁴ Kg

Solved Problems

Question 1: The Gravitational field intensity at a point 5 × 10⁴ km from the surface of the earth is 4 N/kg. Calculate the gravitational potential at that point.

Solution: Gravitational field intensity is given by, 

E = F / m

4 N/kg = (G × M × m) / (r² × m)

= (G × M) / r²,  where r is the distance between the centre of the earth and the body. 

i.e. r = R + 5×10⁴ km, R is the radius of the Earth, R = 6.4 × 10⁶ m.

r = 6.4 × 10⁶ + 5 × 10⁷ m 

= 5.64 × 10⁷ m

Therefore,

4 N/kg = (G × M) / (5.64 × 10⁷)²

4 N/kg × (5.64 × 10⁷)² = G × M

Gravitational Potential (V) = -(G × M) / r

= - (4 × (5.64 × 10⁷)² ) / (5.64 × 10⁷) J/kg

V  = - 2.256 × 10⁸ J/kg.

Question 2: Derive an expression for the amount of work done to move a body from the Earth's surface to infinity i.e. beyond Earth'sgravitational field.

Solution: Force × displacement.

this formula cannot use directly because gravitational force is not constant. This formula works when force remains constant throughout the motion.

Therefore use integration as,

dW = F × dr

dW = (G × M × m / r²) × dr

Integrating W from R to infinity because the body is being moved from the earth's surface to infinity.

W = _R∫^∞ ((G × M × m) / r² ) × dr

W = (GMm) / R 

Question 3: Discussthe variation of acceleration due to gravity with altitude and depth.

Solution: Acceleration due to gravity is maximum at the surface of the earth. It decreases with the increase in altitude and depth.

Variation of g with altitude:

g' = g - [(2 h g) / R]---(1)

Considering equation (1), it is clear that with the increase in height t(h) value of g' decreases because 2, g and R are constants.

g' is the acceleration due to gravity at a height h from the surface of the earth, g is the acceleration due to gravity at the surface of the earth, g = 9.8 m/s²,and R is the radius of the earth,

R = 6.38 × 10⁶ m

Variation of g with depth:

g' = g [1 -(d/R)]---(2)

Considering equation (2), it is clear that with the increase in depth(d) value of g' decreases because g and R are constants.

g' is the acceleration due to gravity at a depth d from the surface of the earth, g is a constant i.e. it is the acceleration due to gravity at the surface of the earth, g = 9.8 m/s² and R is the radius of the earth.

Hence the acceleration due to gravity is maximum at the surface of the earth.

Question 4: A 10 kg block free falls from rest from a height of 20 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: The work done by the force of gravity, W = mgh

where m is mass, g is the gravitational acceleration and h is the height.

Substituting the values in the above equation

W = 10 kg × 20 m × 10 m/s²

= 2000 J

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

Therefore, Gravitational Potential Energy is also equal to - 2000 Joule.

Unsolved Problems

Question 1: A body of mass 4 kg is lifted vertically to a height of 5 m above the ground. Calculate the gravitational potential energy of the body. ((g = 9.8 , m/s²))

Question 2: Find the gravitational potential energy of a body of mass 2 kg placed at a height of 10 m above the Earth’s surface. ((g = 10 , m/s²))

Question 3: A particle of mass m is moved from a distance R to 2R from the centre of the Earth. Find the change in gravitational potential energy in terms of ( G, M, R, ) and ( m ).

Question 4: Calculate the work done in moving a body of mass 3 kg from the surface of the Earth to infinity. (Given: radius of Earth = ( R ))

Question 5: The gravitational potential at a point is (-8 × 10⁷ , J/kg) and the distance from the Earth’s centre is (4 × 10⁷ , m). Find the gravitational field intensity at that point.

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