Physics · Ch 6 — Gravitation

Variation of g with Altitude, Depth and Latitude

6.3.1

Variation of g with Altitude, Depth and Latitude

The value $g = 9.8\ \text{m s}^{-2}$ is only exact right at the Earth's surface near the equator. Moving up, down, or sideways to a different latitude all change the measured value slightly.

Variation with altitude. For an object at height $h$ above the surface, distance from Earth's centre becomes $R_e+h$ , so

g'=\frac{GM_e}{(R_e+h)^2}=\frac{GM_e}{R_e^2}\left(1+\frac{h}{R_e}\right)^{-2}. \qquad (6.45,\,6.46)

When $h\ll R_e$ , a first-order binomial expansion gives the simpler and very useful approximation

g'\approx g\left(1-\frac{2h}{R_e}\right).

Since $g'<g$ always, $g$ decreases as altitude increases. For example, a mango falling from just 15 m has a completely negligible change ( $g'\approx g$ to six decimal places), while a satellite 1600 km up experiences roughly $g'\approx g/1.5$ , a genuinely significant reduction. Note that this approximate formula assumed $h\ll R_e$ , so it should not be used for $h=R_e$ ; the exact formula (6.45) must be used instead in that case.

Variation with depth. For a particle at depth $d$ inside the Earth (e.g. in a mine), only the inner sphere of radius $(R_e-d)$ contributes to the pull felt there -- the spherical shell of material above the particle contributes zero net force, by the hollow-sphere cancellation result from Section 6.1.2. Assuming the Earth has uniform density $\rho = M/V$ , the enclosed mass scales as $(R_e-d)^3/R_e^3$ , and working through the algebra gives

g'=g\left(1-\frac{d}{R_e}\right). \qquad (6.47\text{-}6.50)

So $g'<g$ here too: $g$ decreases as depth increases, going all the way to zero at the Earth's centre ( $d=R_e$ ). Combined with the altitude result, this means $g$ is at its maximum right at the Earth's surface, and falls off whether you go up or down from there.

Variation with latitude. Because the Earth spins about its own axis, it is not a perfectly inertial frame, and an object resting at latitude $\lambda$ on the surface experiences an outward centrifugal effect from that spin, of magnitude $\omega^2 R'$ where $R'=R\cos\lambda$ is the perpendicular distance from the rotation axis. Only the component of this centrifugal acceleration directed opposite to true gravity actually reduces the measured weight, giving …

Figure 6.17(a)A mass at height h from Earth's centre

What this figure shows. The Earth of radius R_e is drawn with its centre O, and a mass m is placed at height h above the surface, so its total distance from the centre is R_e + h; this labelled distance is what goes into the denominator of g' = GM_e/(R_e+h)^2, the formula used to derive how g falls off with increasing altitude. …

Figure 6.17(b)A particle inside a mine at depth d

What this figure shows. A cutaway of the Earth showing a particle of mass m sitting inside a mine at depth d below the surface, at a distance R_e - d from the centre O. The diagram highlights that only the inner sphere of radius (R_e - d) contributes to the gravitational pull felt at that depth -- the spherical shell of Earth material above the particle contributes nothing, by the same hollow-sphere cancellation result used ea …

Figure 6.18Centrifugal effect due to Earth's spin at latitude lambda

What this figure shows. A cross-section of the spinning Earth showing a point P on the surface at latitude lambda, at a perpendicular distance R' = R cos(lambda) from the Earth's rotation axis. Arrows show the true weight mg pulling straight toward the centre and a separate outward centrifugal force F_c along R'; the diagram is used to show why the two forces only fully oppose each other at the equator, making measured gravity weakest there and strongest at the poles, where the spin contributes no outw …

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