Acceleration due to gravity and how it changes

Why $9.8\ \text{m/s}^2$?

A body of mass $m$ at the surface feels $\dfrac{GMm}{R^2}$, so its acceleration is

The body's own mass cancels: everything falls alike. This also lets us weigh the Earth: $M = \dfrac{gR^2}{G} = 6.0 \times 10^{24}\ \text{kg}$.

Going up

At a height $h$, the distance from the centre is $R + h$:

Going down

At a depth $d$ inside a uniform Earth, the outer shell pulls nothing (its pulls cancel), and only the ball of radius $R - d$ inside pulls:

So $g$ is largest at the surface and zero at the centre.

You should find $g$ rising in a straight line inside, peaking at the surface, and falling away outside. The shape stays the same for any planet; only the peak changes.

The Earth's spin and shape

The Earth bulges at the equator, and the spin needs a centripetal acceleration $\omega^2R$ there. Both make $g$ at the equator about $0.5\%$ smaller than at the poles.

Watch out. The shortcut $1 - \dfrac{2h}{R}$ is only for heights much smaller than $R$. For satellites, use the exact formula.

A worked example

What is $g$ at the top of Mount Everest, $8.85\ \text{km}$ high? ($R = 6400\ \text{km}$.)

1. Small height. $h \ll R$, so the shortcut works.
2. Fraction. $\dfrac{2h}{R} = \dfrac{17.7\ \text{km}}{6400\ \text{km}} = 0.0028$.
3. Value. $g' = 9.8\ \text{m/s}^2 \times (1 - 0.0028) = 9.77\ \text{m/s}^2$: a climber weighs about $0.3\%$ less.

Think first. Above the surface, use $\dfrac{1}{r^2}$; below it, use $r$. Never mix them.

Practise with new numbers

Try it in Python

The program works out $g$ at several heights and depths.

Remember these

• $g = \dfrac{GM}{R^2}$; $M = \dfrac{gR^2}{G}$.
• Height: $g\left(\dfrac{R}{R + h}\right)^2$, or $g\left(1 - \dfrac{2h}{R}\right)$ for small $h$.
• Depth: $g\left(1 - \dfrac{d}{R}\right)$, zero at the centre.

Check yourself

Try each one on your own first. If you are stuck, open a hint. Then check with Show the answer.

1. The Moon has about $\tfrac{1}{81}$ of the Earth's mass and $0.27$ of its radius. Estimate $g$ there.

2. At what height is $g$ half its surface value? ($R = 6400\ \text{km}$.)

3. Why would you weigh nothing at the Earth's centre?

4. Does a $5\ \text{kg}$ stone fall faster than a $1\ \text{kg}$ stone, ignoring air?