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Gravitational Potential Energy

  • Difficulty Level : Medium
  • Last Updated : 28 Sep, 2021

Anything, when released in free space, tends to move down. This is a common experience in our day to day life. Sir Isaac Newton was the first to analyze this tendency of everything to move towards the Earth when released in free space. He observed an apple falling from the tree and started his research to make an observation on this favoured downward motion. He arrived at a conclusion that it is not only the Earth that attracts everything towards itself, but everybody in this universe attracts every other body. This is a property of a body due to its mass.

The work that a body must do against gravity in order to get it into a certain location is referred to as gravitational potential energy. In other words, gravitational potential energy is the energy that an object has or gains as a result of a change in its gravitational field position. Because energy cannot be produced or destroyed, even at rest, a thing has some kind of energy (Which is converted to kinetic energy, when it starts to move). Potential energy is the name given to this type of energy. 

Only the amount of work done on the body by force alters the particle’s potential energy if its position changes as a result of its force. Conservative forces are those whose work is self-contained.  Thus, the Gravitational Potential Energy is the potential energy of a body that emerges from the strength of gravity, which is a conservative force.

  • The energy in a body’s potential energy is defined as the energy in that position. When external forces modify the position of the body, the energy shift is equal to the amount of work done by the forces on the body.
  • The work done by gravity is independent of the direction in which a change of position occurs, suggesting that it is a conserving force. 

  • Furthermore, each of these forces has the potential to be powerful. As a result, in an infinite body, the gravitational impact is 0 and the potential energy is zero, which is known as a reference point.
  • A gravitational acceleration of roughly 9.8 m/s2 for an object at the earth’s surface is the most common use of gravitational potential energy.

What is Gravitational Potential Energy?

Gravitational Potential Energy is described as the energy of a body that arises from the gravitational force, that is, when the gravitational force acts on a body and that force produces a type of potential energy known as gravitational potential energy. Now, if the position changes due to force, the amount of work done on the body by the force is the change in potential energy. 

Assume we have two items, one A and the other B. Now, suppose B’s location changes exclusively as a result of a certain force. Assume M’s location is changing owing to gravitational force, followed by a change in potential energy. The quantity of work done by the gravitational force is equal to the change in the location of the item.

As a result, the work done by the gravitational force will be equal to the change in potential energy going through from P to Q. Assume P is position 1 and Q is position 2. In this instance, the effort required to move the item from position 1 to position 2 is equal to the change in potential energy, which is equal to the potential energy at point 2 minus the potential energy at point 1.

  • The work done per unit mass to bring that body from infinity to that location is known as gravitational potential. 
  • U is the symbol for it. 
  • J/Kg is the SI unit for gravitational potential. 
  • It is the potential body that arises from the gravitational pull. If the position changes as a result of the force, the change in potential energy is the work done on the body by the force.

Formula for Gravitational Potential Energy

Mathematically, Gravitational Potential Energy is given by the product of the mass (m) of the object, acceleration due to gravity (g) and height (h) above the ground as 

GPE, V= mgh

A wide expression of the law of gravity is generated by gravitational potential power, which is comparable to the effort required to overcome gravity by transporting a mass to a certain location in space. Because the gravitational force is inversely proportional, it is worthwhile to choose the gravitational energy zero at an infinite distance. Gravity’s potential energy is thus negative when mass approaches a planet because gravity operates positively as mass approaches. This negative potential depicts a “bundled condition,” in which a mass is imprisoned when it is near a bigger body until it is provided with enough energy to escape.

  • 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 of Gravitational Potential Energy

When a particle is transported over an infinitely short distance, dr. The work done on the second particle by the gravitational force is represented by -Fdr.

dW = -Fdr                                                                                                                                                                                                                                      ……(1)

where F is the Gravitational force acting on the particles and is given by,

F =  G m1 m2 / r2

where G is the Gravitational constant, m1 and m2 are the masses of the two particles in contact respectively.

On substituting F in equation (1), we get,

dW = -(G m1 m2) dr / r2                                                                                                                                                                                                                 ……(2)

The negative sign in the equation comes from the fact that the displacement is in the opposite direction as the force. The change in gravitational potential energy of the two-particle system during this short movement is equal to the negative work done on the second particle by definition. When a particle is transported over an infinitely short distance, dr. 

The work done on the second particle by the gravitational force is denoted by – Fdr.

dU = -dW = (G m1 m2) dr / r2                                                                                                                                                                                                      ……(3)

As the second particle goes from B to C, the change in gravitational potential energy of the two-particle system is a function of distance r, and is represented by,

\begin{aligned}U(r_2)-U(r_1)&=\int^{r_2}_{r_1}{dU}\\&=\dfrac{G m_1m_2}{r^2} dr\\&=G m_1m_2\int^{r_2}_{r_1}\dfrac{1}{r^2}\\&=-Gm_1m_2\left[\dfrac{1}{r}\right]^{r_2}_{r_1}\\&=-Gm_1m_2\left[\dfrac{1}{r_2}-\dfrac{1}{r_1}\right]\\&=Gm_1m_2\left[\dfrac{1}{r_1}-\dfrac{1}{r_2}\right]\end{aligned}

Gravitational Potential Energy at a Height h: When an object is taken off from the earth’s surface to a point ‘h’ above the earth’s surface, then, r1 = r2

And r2 = R +h

So,

ΔU = Gm1 m2 [1/R – 1/(R+h)]

or

ΔU = Gm1 m2h/R(R + h)

For very very small value of h, g = GM/R2

On Substituting, we get-

ΔU = mgh

Sample Questions 

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.

Answer:

The 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.

So, r = 6.4 × 106 + 5 × 107

        = 5.64 × 107 m

Therefore,

4 N/kg = (G × M) / (5.64 × 107)2 

4 N/kg × (5.64 × 107)2 = G × M 

This implies, 

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’s gravitational field.

Answer:

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, dr is small change in r

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: What is the mass of the earth?

Answer:  

We know acceleration due to gravity at the surface of the earth is given by,

g = (G× M) / R2 

9.8 = (6.67 × 10-11 × M) / (6.38 × 106)2

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

M = (9.8 × (6.38× 106)2) / (6.67 ×10-11)

M = 5.98 × 1024 kg

Question 4: Discuss the variation of acceleration due to gravity with altitude and depth.

Answer: 

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 5: 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.

Answer: 

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.


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