A-level Physics (Advancing Physics)/Gravitational Potential Energy

If you throw a ball into the air, you give it kinetic energy. The ball then slows down because of the effect of the Earth's gravitational field on it. However, we know that energy cannot be created or destroyed. The kinetic energy you gave the ball is transformed into gravitational potential energy. The further away from the Earth you manage to throw the ball, the greater the potential there is for kinetic energy to be created on the way back down. However, for kinetic energy to be created, there must be an acceleration. If there is an acceleration, there must be a force.

You should already know that energy is the same as the work done to move something a distance &Delta;x:

$$\Delta E_{grav} = F\Delta x$$

Work done is given by the force applied multiplied by the distance moved in the direction of the force. To move an object against gravity, the force applied upwards must equal the downwards force gravity exerts on the object, mg. So, if I move an object against gravity a distance &Delta;x, the work done is given by:

$$\Delta E_{grav} = mg\Delta x$$

It is usual to call this x the height, so you will often see Egrav=mgh. The deltas are important. They mean that it doesn't matter which distance x I move the object across - I can decide the point at which gravitational potential energy is 0 in a way which makes the maths easy.

The difficulty with this simple formula is that g does not remain the same over large distances:

$$g = \frac{-GM}{r^2}$$

So, over a distance &Delta;r, x becomes r and so:

$$E_{grav} = \frac{-GMmr}{r^2} = \frac{-GMm}{r}$$

Despite the fact that this "derivation" is somewhat convincing, it's actually invalid. A proper derivation of this formula would use calculus (see below). The "distance r" is typically huge, and the gravitational force will not remain constant over such a distance. However, the above formla $$E_{grav} = mg \Delta x$$ does assume the gravitational force is constant, so it's invalid to use that formula in this context.

So, if you're dealing with gravitational potential energy over large distances, use this formula. If you're dealing with gravitational potential energy over short distances, such as with ramps on the Earth's surface, where g=9.81ms−2, use Egrav = mgh.

Graphs
We have just done something sneaky. You probably didn't notice. Let's see what happens when we integrate the gravitational force F with respect to r between r and ∞:

$$\int_{r}^{\infty}\frac{-GMm}{r^2}dr = \left [ \frac{GMm}{r}\right ]_{r}^{\infty} = \frac{GMm}{\infty} - \frac{GMm}{r}$$

Since dividing anything by infinity gets you practically 0:

$$\int_{r}^{\infty}F dr = - \frac{GMm}{r} = E_{grav}$$

And therefore:

$$\frac{dE_{grav}}{dr} = F$$

So, if you have a graph of gravitational potential energy against radius, the gradient of the graph is the gravitational force. If you have a graph of gravitational force against radius, the area under the graph between any point and the F-axis is the gravitational potential energy at this point. The area under the graph between any two points is the difference in gravitational potential energy between them.

Questions
1. A ball rolls down a 3m-high smooth ramp. What speed does it have at the bottom?

2. In an otherwise empty universe, two planets of mass 1025 kg are 1012 m apart. Both the planets have a radius of 106 m. What are their speeds when they collide?

3. What is the least work a 2000 kg car must do to drive up a 100m hill?

4. How does the speed of a planet in an elliptical orbit change as it nears its star?

/Worked Solutions/