Course 7.2Lesson 1 of 3

Gravitational field and potential energy

A mass sets up a gravitational field around it: the force per kilogram, $\vec{g} = \dfrac{GM}{r^2}$

Lesson 1 of 3Show all lessons

Try the code and the simulations beside the lesson.

A field around every mass

A mass changes the space around it: any other mass placed there feels a pull. We describe this with the gravitational field, the force on each kilogram:

\displaystyle g = \frac{F}{m} = \frac{GM}{r^2}

in $\text{N/kg}$ (the same as $\text{m/s}^2$ ), pointing toward the mass.

Potential energy far from the ground

Lifting a body takes work, which is stored as potential energy. Near the ground, $mgh$ works. But $g$ weakens with height, so for big heights we must add up the work with the real force. Taking $U = 0$ when the bodies are very far apart,

\displaystyle U = -\frac{GMm}{r}

It is negative, and rises toward zero as $r$ grows. The work to lift a body from $r_1$ to $r_2$ is

\displaystyle \Delta U = GMm\left(\frac{1}{r_1} - \frac{1}{r_2}\right)

A graph of the gravitational potential energy U of a body against the distance r from the centre of the Earth. The curve is negative everywhere. It starts at minus G M m over R at the surface, is half of this at 2 R, and rises toward zero far away.

Compare it with $mgh$ .

You should find the two agree near the ground, but $mgh$ is $10\%$ too big already at a tenth of the Earth's radius, and far too big beyond.

Watch out. The zero of $U$ is far away, so near the Earth $U$ is a large negative number. Only changes in $U$ matter for work.

A worked example

How much work lifts a $250\ \text{kg}$ satellite from the ground to $2000\ \text{km}$ up? ( $g = 9.8\ \text{m/s}^2$ , $R = 6400\ \text{km}$ .)

Exact. $\Delta U = mgR\dfrac{h}{R + h} = 250\ \text{kg} \times 9.8\ \text{m/s}^2 \times 6.4 \times 10^{6}\ \text{m} \times \dfrac{2000\ \text{km}}{8400\ \text{km}} = 3.73 \times 10^{9}\ \text{J}$ .
Shortcut. $mgh = 250\ \text{kg} \times 9.8\ \text{m/s}^2 \times 2.0 \times 10^{6}\ \text{m} = 4.90 \times 10^{9}\ \text{J}$ .
Compare. The shortcut is $31\%$ too big.

Think first. Is the height small next to $6400\ \text{km}$ ? If not, use $GM\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ .

Practise with new numbers

Try it in Python

The program compares the exact energy with $mgh$ for four heights.

Remember these

Field: $g = \dfrac{GM}{r^2}$ , in N/kg.
$U = -\dfrac{GMm}{r}$ , zero far away; $\Delta U = GMm\left(\dfrac{1}{r_1} - \dfrac{1}{r_2}\right)$ .
$mgh$ only for heights much smaller than $R$ .