Suppose you are driving a car and there is a hill, you slow down or stop because of the steep road. Where does the energy go? The answer to this that the energy turns into gravitational potential energy. Let us explore more about Gravitational Potential Energy.

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## Gravitational Potential Energy

Gravitational Potential is the work done per unit mass to bring that body from infinity to that point. It is represented by V. SI unit of gravitational potential is J/Kg. It is the potential body arising out of the force of gravity. If due to the force, if the position changes, then the change in the potential energy is the work done on the body by the force.

The most common use of gravitational potential energy is for an object near the surface of the Earth where the gravitational acceleration is a constant at about 9.8 m/s².

### Expression of Gravitational Potential Energy

#### Case 1. ‘g’ is Constant

Consider this image of the earth. Assume an object at point A and later it moves to point B. So, in this case, the work done is force × displacement.

W_{BA} = Force × displacement

Force is nothing but the gravitational force exerted by the earth. The height of point A from the surface of the earth is h_{2 }and that of point B is h_{1}

W_{BA }= mg ( h_{2}– h_{1}) = mg h_{2 }– mg h_{1}

the work done in moving the object is the difference in its potential energy between its final and initial position.

**Browse more Topics under Gravitation**

- Newton’s Universal Law of Gravitation
- Thrust, Pressure and Buoyancy
- Acceleration Due to Gravity
- Earth Satellites
- Escape Velocity
- Kepler’s Law
- Weightlessness

#### Case 2. ‘g’ is Not Constant

Let the position vector of the first object be r_{1 }and the position vector of the second object be r_{2 . }So here the work done in bringing the object from one position to another is:

W = \( \int_{r_1}^{r_2} F dr \)

where, F = \( \frac{GM_em}{r^2} \)

W = \( \int_{r_1}^{r_2}GM_em dr \)

= – GM_{e}m[ \( \frac{1}{r_2} \) – \( \frac{1}{r_1} \) ]

## Solved Examples For You

Q1. The gravitational potential difference between the surface of a planet and a point 20m above the surface is 2 joule/Kg. If the gravitational field is uniform then the work carrying a 5 Kg body to a height of 4m above the surface is:

- 2 Joule
- 20 Joule
- 40 Joule
- 10 Joule

Ans: A. Since the gravitational field is uniform,

Gradient of potential = \( \frac{Potential difference} {distance between the points} \) = \( \frac{2}{20} \) = 0.1 J/ Kgs

Total work is done = Mass ×distance × potential gradient = 5 × 4 × 0.1 = 2 Joule.

Q2. An object is taken from a point P to another point Q in a gravitational field

- assuming the earth to be spherical if both P and Q lie on the earth’s surface, the work done is zero.
- if P is on the earth’s surface and Q above it, the work done is minimum when it is taken along the straight line PQ.
- the work done depends only on the position of PO and Q and is independent of the path along which the particle is taken.
- there is no work if the object is taken from P to Q and then brought back to P along any path.

Ans: If P and Q both lie on the earth surface, this means both have same potential energy that implies same mechanical energy. Thus there is no work in moving an object from P to Q. As the gravitational field is the conservative field and thus work done by the gravitational force depends only on the position of P and Q and is independent of the path taken. Also, work done by a conservative force along a closed loop is zero.