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

The equation below applies to when there is a uniform field such as when we are talking about being very close to the Earth's surface: the size of gravity does not change significantly when we are talking about close to the Earth's surface. Therefore, we can presume that the gravitational field is uniform.

So, if:

- △PE = mg△h,
- △PE / m = g△h

△PE / m is known as the change in gravitational potential being '△v'. Therefore,

△v = g△h

## For a Non-Uniform Field

This equation below applies to when the gravitational field is not uniform. This is when we are talking about objects far from the Earth's surface because as we move further away from the Earth, gravity attracting us to the Earth decreases more than when we are close to it.

v = - GM / rWhere M = mass of the Earth (kg), r = distance from the center of the Earth (m) and G is the universal gravitational constant which is 6.67x10^-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 = 5.97x10^24 kg

We now have all the information we need to put it into the non-uniform equation. We get an answer of

**2300N**.#### Uniform Field

△v = 9.8 x 235 =

**2303N**

Although the non-uniform equation will always be more accurate, the unifrom field is only 0.1% off the true value which is a of great accuracy. This is why we use the uniform for objects close to the surface of the Earth and non-uniform for objects far away from the Earth. If you don't know which one to use, use both.

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

We have already calculated the radius of a geostationary orbit and know this to be 4.23x10^7 metres. This is quite a distance from the surface of the Earth. Therefore, we will have to use the non-uniform field equation for this.

- r1 = radius of the Earth = 6,378,000m
- r2 = radius of the Earth + height of building = 6,378,000m + 4.23x10^7m

Therefore, from putting all the information into the non-uniform equation, we get an answer of

**54.3x10^6N**.**Calculate the PE gained in both cases.**

For this, all we have to do is multiply the gravitational potential by the mass of the object being the satellite because △PE / m = g△h. Therefore. because g△h is △v:

△v x m = △PE

## Non-Uniform Gravitational Field

To understand the concept of a non-uniform gravitational field, we will have a look at a few representations with the first being a field lines representation.

What this diagram displays is that when the field lines are spaced closer together, the gravitational field will be stronger. Therefore, the further away you are from the object, the weaker the gravitational force will be on the object. The closer to the center you are object, the stronger the gravitational force will be.

The second representation will be through the use of a vector representation:

In this diagram, the longer vectors represent the stronger gravitational field which means the small vectors represent the weaker gravitational field.

We already know that weight = mass of object multiplied by gravity or w = mg. We also know F = -Gm1m2 / r² (r = distance between centers of mass of objects 1 + 2, m1 = mass of the Earth, m1 = mass of object and G is the gravitational constant). However, the force (F) calculates the force of gravity on an object and so does the weight. Therefore, we can combine these two equations together.

- - Gm1m2 / r² = m2g

g = -Gm1 / r²

Where g = gravitational field strength, M = mass of planet (or object) and r = distance to the center of the planet (or object). From this, we can calculate the gravitational field strength at the Earth's surface.

- M = 6x^24kg.
- r = 6400km.
- g = - 6.67x10^-11 x 6x10^24 / (6400x10^3)²

which equals

**-9.77 N/kg**

We can create a graph for the non-uniform field strength of the Earth.

As you can see from the graph, the gravitational field strength is at it's highest on the surface of the Earth. It then decreases to zero which a constant gradient (presuming the density of Earth is universal). As we move further away from the surface of the Earth, the gravitational field strength decreases exponentially to an asymptote. This is where the line will never reach zero but will continuously keep getting closer and closer to zero. This can be supported by the equations from above:

- g = -GM1 / r²

When r equals infinity, g will equal zero. But, the graph will never reach zero: it will keep getting closer and closer to zero. This means that, in theory, there is a gravitational field strength between you right now and the most distant sun from us on the other side of the universe. However, the field strength will be infinitely small. However, there will still be a field strength greater than zero.

Here are some sheets of mine you might find useful looking at the graphs of uniform and non-uniform fields using different y and x axis.

## 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.

Be sure to check out my other articles on Physics A2.

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.

ReplyDeletethank you.