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# Gravitational Potential

Rather than talking about gravitational potential energy all the time, it is useful for a number of reasons to define a new quantity - **Gravitational Potential, Φ.**

It is a very simple idea. **Gravitational potential is the potential energy per kilogram at a point in a field.** So the units are Jkg^{-1}, joules per kilogram.

**The equation for potential is:**

**where**

**G** = the universal gravitational constant

**m** = the mass causing the field

**r** = the distance between the centre of the mass causing the field and the point you are considering.

**Note that:**

**1.** Just like potential energy, the biggest value of potential you can get is zero. All other values are less than zero - i.e. negative!!

**2.** Potential is not a vector even though it has a negative sign. It doesn't have a direction, only a magnitude.

**Example**

**If G = 6.67x10 ^{-11}Nm^{2}kg^{-2} and the mass of the Earth is 6.0x10^{24}kg, calculate the potential at the surface of the Earth if the radius of the Earth is 6.4x10^{6}m.**

**Answer**

The **potential** (Ep per kg) at the surface of the Earth is

- 63 MJ kg ^{-1}

Do 10 MJ kg^{-1} of work on raising an object from the Earth's surface and it will move up to a point where it's **potential** is - 53MJ kg^{-1}. That's 10 MJ kg^{-1} greater than on the surface because it is 10MJ kg^{-1} closer to zero.

Confusing! Look at the following diagram. It shows how potential drops as you move further from the surface of the Earth.

So we can define potential as "**the work done per kg by an object when it moves from infinity to a point in a field**".

Looking back at the example above, that means that if you let 1kg drop from infinity to the surface of the Earth it will lose 63MJ of potential energy. Ignore air resistance, and the object will have that much kinetic energy when it hits the surface of the planet.

The **potential Φ** at a point in a field is the **potential energy per kg.**

So, if you put a mass of 'm' kg at that point, its **potential energy** is:

**Ep = mΦ**

so, in a radial field,

To escape completely from the Earth's gravitational field you need to give an object 63MJ of kinetic energy per kg. (As it rises from the Earth it will lose kinetic energy and gain potential energy)

This allows you to calculate an object's escape velocity - how fast you would have to throw it to get it completely out of the Earth's gravity field.

If it needs 63MJkg^{-1} energy then it must start off with all this energy as Ek.

**So:**

where m = 1kg

**Rearrange to get:**

v = 11,200 ms^{-1}.

That's how fast you'd have to throw it.