The gravitational potential energy of an object is the potential energy per kilogram of an object placed at that specific point.

## For a Uniform Field

- △PE = mg△h,
- △PE / m = g△h

△v = g△h

## For a Non-Uniform Field

v = – GM / r

Where M = mass of the Earth (kg), r = distance from the center of the Earth (m) and G is the universal gravitational constant which is 6.67×10^-11 Nm²kg^-2. Therefore, the change in gravitational potential is:

△v = -GM/r2 – (-GM/r1)

Where r1 is the inital radius and r2 is the final radius.

## Example

**Calculate the △v for a satellite:****a) which has moved from the ground floor of a building of height 235m to the top floor.**

We will use both equations to see if there is a difference in the equations. However, if we should use one, it would be the uniform field equation because we are talking about an object close to the Earth’s surface.

#### Non-Uniform Field

- r1 = radius of the Earth = 6,378,000m
- r2 = radius of the Earth + height of building = 6,378,000m + 235m
- mass of Earth = 5.97×10^24 kg

**2300N**.

#### Uniform Field

**2303N**

**b) which has moved from the Earth’s surface to the height of a geostationary orbit.**

- r1 = radius of the Earth = 6,378,000m
- r2 = radius of the Earth + height of building = 6,378,000m + 4.23×10^7m

**54.3×10^6N**.

**Calculate the PE gained in both cases.**

△v x m = △PE

## Non-Uniform Gravitational Field

- – Gm1m2 / r² = m2g

g = -Gm1 / r²

- M = 6x^24kg.
- r = 6400km.
- g = – 6.67×10^-11 x 6×10^24 / (6400×10^3)²

**-9.77 N/kg**

- g = -GM1 / r²

## Summary

- The change in potential of a uniform field close to the Earth surface is g△h. By multiplying the △v with the object’s mass, you can then find out the change in potential energy.
- The change in potential energy for a non-uniform field is △v = -GM/r2 – (-GM/r1).
- The non-uniform equation will always be more accurate than the uniform equation because all gravitational fields are not uniform.

What would be the graph of gravitational potential against distance from the centre of the Earth to its surface?Have a look at A2 AQA physics page 65. Tell me the left graph situ.

thank you.