Acceleration due to gravity
Where does 9.8 come from?
Drop a cricket ball from a balcony. It speeds up by 9.8 metres per second every second. Let us see where this number comes from.
By the shell theorem, the whole Earth pulls the ball as if all its mass M were at the centre, a distance R away.
So the force on the ball is G M m over R squared. Newton’s second law says this force equals m times g.
The mass of the ball, m, cancels out. We get g equals G M over R squared. Every object falls with the same g, light or heavy, ignoring air.
Turn this around. We know g, R and G, so we can find the mass of the Earth: M equals g R squared over G.
Put in g = 9.8, R = 6.4 × 10⁶ metres and G = 6.67 × 10⁻¹¹. The answer is about 6 × 10²⁴ kilograms. Cavendish’s experiment let us weigh the Earth.
But g is not the same everywhere. It changes as you climb a mountain, fly in a plane, or go down a deep mine. Let us look at each case.
Going up, going down
First, go up. At a height h above the surface, the distance from the centre is R plus h. So g at height h is G M over R plus h, all squared.
Compare this with g on the surface. g(h) equals g divided by one plus h over R, squared. The higher you go, the smaller g gets.
When h is much smaller than R, a neat shortcut works. The binomial approximation gives g(h) ≈ g times one minus 2h over R.
On top of a 10 km high mountain, h over R is tiny, so g drops by only about 0.3 percent. Watch the graph: near the surface the curve looks almost like a straight line.
Now go down a mine shaft to a depth d. Treat the Earth as a uniform ball. The outer shell of thickness d is now around you, and a shell pulls nothing on a point inside it.
Only the inner ball of radius R minus d pulls you. Its mass is smaller in proportion to its volume, so g at depth d equals g times one minus d over R.
At the centre, d equals R, and g becomes zero. Every bit of the Earth pulls you equally in all directions.
So g is greatest at the surface. It falls in a straight line as you go down, and as one over r squared as you go up.
Try it: g from the centre outwards
- Drag the slider from the centre of the Earth out to 4R.
- Inside, g rises in a straight line. Outside, it falls as 1/r².
- Turn on g(1 − 2h/R) and see how far it stays close to the real curve.
- The [ and ] keys nudge the position.
Quick check
At what height above the Earth’s surface does g become one-quarter of its value on the surface? (R is the radius of the Earth.)
- h = R/4
- h = R/2
- h = R
- h = 3R
Answer
C. g(h) = g·R²/(R + h)². For g(h) = g/4 we need (R + h)² = 4R², so R + h = 2R and h = R. The shortcut g(1 − 2h/R) would wrongly give h = 3R/8, because it only works when h is much smaller than R.
Up or down?
g at a small height h above the surface equals g at a depth d below the surface. How are d and h related?
- d = h
- d = 2h
- d = h/2
- d = 4h
Answer
B. For small h, g(1 − 2h/R) = g(1 − d/R), so d = 2h. Going up lowers g twice as fast as going down.
Summary
- On the surface:
g = GM / R² - At height h:
g(h) = GM / (R + h)² ≈ g(1 − 2h/R) for h ≪ R - At depth d:
g(d) = g(1 − d/R) - Mass of the Earth:
M = gR² / G ≈ 6 × 10²⁴ kg
- g does not depend on the mass of the falling body, only on the Earth’s M and R.
- Going up, g falls as 1/r². The shortcut g(1 − 2h/R) works only when h is much less than R.
- Going down, only the inner part of the Earth pulls, so g falls in a straight line to zero at the centre.
- g is greatest at the surface.