Solved Examples on Gravitational Potential Energy

Example 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.

Example 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’s gravitational field.

Solution:

We know Work done (W) = Force × displacement.

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

Therefore we will 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 

Example 3: Discuss the 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.

Example 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: 

We know that, 

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, we get

W = 10 kg × 20 m × 10 m/s²

    = 200 N

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

Therefore, Gravitational Potential Energy is also equal to 200 Joule.

The energy possessed by objects due to changes in their position in a gravitational field is called Gravitational Potential Energy. It is the energy of the object due to the gravitational forces. The work done per unit mass to bring the body from infinity to a location inside the gravitational field of any object is known as gravitational potential and the energy change here is called Gravitational Potential Energy. Let’s learn about Gravitational Potential Energy in detail in this article.