The Acceleration Due to Gravity and How It Varies

g is the field at the surface

In the chapter on Motion in One Dimension, \(g\) was the acceleration of free fall, measured and used. Now we can calculate it. A body of mass \(m\) at the Earth's surface is at a distance \(R\) from the centre. By the shell theorems, the Earth pulls it as a point mass \(M\) at the centre: \(F = GMm/R^2\). This pull alone gives the body its acceleration, so \(g = F/m = GM/R^2\).

The mass \(m\) of the falling body has cancelled. That is why a heavy stone and a light one fall together in a vacuum. The acceleration due to gravity belongs to the planet: its mass and its radius. In this unit we take \(g = 9.8\,\text{m/s}^2\), \(G = 6.67 \times 10^{-11}\,\text{N m}^2/\text{kg}^2\) and \(R_E = 6.4 \times 10^6\,\text{m}\) for the Earth.

Turned round, the formula weighs the Earth. \(M = gR^2/G = 9.8 \times (6.4 \times 10^6)^2/(6.67 \times 10^{-11}) = 6.0 \times 10^{24}\,\text{kg}\). Divide by the Earth's volume, \(\tfrac43\pi R^3 = 1.1 \times 10^{21}\,\text{m}^3\), and the mean density is about \(5500\,\text{kg/m}^3\). Surface rocks are only about half as dense, so the inside of the Earth must be much denser.

Climbing above the surface

At a height \(h\) above the surface, the distance from the centre is \(R + h\). The Earth still acts from its centre, so the field there is \(g_h = GM/(R + h)^2\). Divide by \(g = GM/R^2\) to remove \(G\) and \(M\): \(g_h = g\,R^2/(R + h)^2\). At a height of one radius, \(h = R\), the field is \(g/4 = 2.45\,\text{m/s}^2\).

For heights much smaller than \(R\), a short approximation helps. Write \(g_h = g(1 + h/R)^{-2}\). The binomial approximation \((1 + x)^{-2} \approx 1 - 2x\), for small \(x\), gives \(g_h \approx g(1 - 2h/R)\). The fractional fall in \(g\) is \(2h/R\). It reaches \(1\,\%\) at \(h = 0.005R = 32\,\text{km}\).

On top of Mount Everest, \(8.85\,\text{km}\) high, \(2h/R = 0.0028\). So \(g\) there is about \(9.77\,\text{m/s}^2\), smaller by less than a third of one percent. At the height of the International Space Station, \(400\,\text{km}\), the exact formula gives \(8.68\,\text{m/s}^2\). The pull there is still \(89\,\%\) of its value on the ground. Why astronauts float in spite of it is a question for the chapter on Gravitational Energy and Orbits.

Going down a tunnel

Now go down instead of up. Imagine a narrow tunnel from the surface to the centre, and ask where along the whole line, inside and out, \(g\) is greatest. Predict first, then watch.

At a depth \(d\) the marker is \(r = R - d\) from the centre. By the shell theorems, the shell of rock above it pulls nothing. Only the sphere of radius \(R - d\) below it counts. For a uniform Earth, the field there is \(GMr/R^3 = g\,r/R\). So \(g_d = g(1 - d/R)\). It falls in a straight line, to zero at the centre.

Compare a small height and a small depth. At a height \(h\), \(g\) falls by the fraction \(2h/R\). At a depth \(d\), it falls by \(d/R\). So a given height lowers \(g\) twice as much as the same depth. At the bottom of the Kolar Gold Fields in Karnataka, about \(3.2\,\text{km}\) deep, the uniform model gives \(g\) smaller by only \(0.005\,\text{m/s}^2\).

The spinning Earth

The Earth turns once a day, with angular speed \(\omega = 2\pi/(24 \times 3600\,\text{s}) = 7.27 \times 10^{-5}\,\text{rad/s}\). A point at latitude \(\lambda\) goes round a circle of radius \(R\cos\lambda\) about the Earth's axis. In the chapter on Dynamics of Circular Motion you met the rotating frame. In the Earth's frame, a body at rest there feels an outward centrifugal force \(m\omega^2 R\cos\lambda\), directed away from the axis.

The part of the centrifugal term along the radius is \(\omega^2 R\cos\lambda \times \cos\lambda\). It points outward, against gravity. So the effective g, the acceleration a body at rest there actually shows, is \(g' = g - \omega^2 R\cos^2\lambda\). At the equator \(\omega^2 R = (7.27 \times 10^{-5})^2 \times 6.4 \times 10^6 = 0.034\,\text{m/s}^2\), so \(g' = 9.766\,\text{m/s}^2\). At the poles \(\cos\lambda = 0\), and the spin changes nothing.

The Earth's shape adds a second effect. The spin has made the Earth bulge slightly at the equator. The equatorial bulge makes the equatorial radius about \(21\,\text{km}\) larger than the polar radius. A point on the equator is therefore further from the centre, and \(g\) there is smaller for that reason too. Measured values are about \(9.780\,\text{m/s}^2\) at the equator and \(9.832\,\text{m/s}^2\) at the poles. The spin accounts for \(0.034\,\text{m/s}^2\) of this difference, and the shape for most of the rest.

Now return to the spring balance. On Everest it reads less because the top is \(8.85\,\text{km}\) further from the Earth's centre. At the equator it reads less than at the poles for two reasons. The spin takes away part of the pull, and the bulge puts the equator further from the centre.