7.2 Acceleration due to Gravity and Its Variation

A precision spring balance carried from a beach in Kerala to a hut high in the Himalaya shows a slightly smaller reading for the same bag of rice. And $400\ \text{km}$ up, where the International Space Station flies, the Earth's pull on the bag is still about $89\%$ of its value at the ground — not zero, as many people think. Your mass is the same everywhere — it is the pull of the Earth, and so $g$, that changes. The changes are small, a fraction of a percent at any height a person can reach, but sensitive instruments measure them easily: geologists hunt for ore bodies and oil traps by mapping tiny variations of $g$, and satellites track the shrinking of glaciers and underground water from space in the same way.

In Chapter 2 you treated $g$ as a constant $9.8\ \text{m/s}^2$. In this section you will see where that number comes from, use it to "weigh" the Earth, and find how $g$ changes with height above the surface and with depth below it. The law of gravitation and the shell result of Section 7.1 do all the work.

The acceleration due to gravity at the Earth's surface

Treat the Earth as a sphere of mass $M_E$ and radius $R_E$ whose density depends only on the distance from the centre. By Result 7.5 it attracts a body of mass $m$ on its surface as if all its mass were at the centre, a distance $R_E$ away: $F=\dfrac{GM_Em}{R_E^2}$. Dividing by $m$:

With $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$, $M_E=5.97\times10^{24}\ \text{kg}$ and $R_E=6.37\times10^6\ \text{m}$, $g=\dfrac{(6.67\times10^{-11})(5.97\times10^{24})}{(6.37\times10^6)^2}=9.81\ \text{m/s}^2$. Turned round, $M_E=\dfrac{gR_E^2}{G}$: measuring $g$ and the Earth's radius, and knowing $G$, gives the mass of the Earth. This is the sense in which Cavendish "weighed the Earth".

Variation of $g$ with height

At a height $h$ above the surface a body is $r=R_E+h$ from the Earth's centre, and the Earth still acts as a point mass at the centre. So $g$ simply follows the inverse-square law.

The approximation uses $(1+x)^{-2}\approx1-2x$ for small $x$ (the binomial approximation, which you will meet in Class 11 Mathematics). Check it with numbers: for $x=0.01$, $(1.01)^{-2}=0.98030$ while $1-2x=0.98$ — an error of only $0.03\%$. For $h=R_E$ the approximation would give $-g$, which is nonsense: there, use the exact form, $g(R_E)=\dfrac{g}{4}$.

Variation of $g$ with depth

Now take a body down a deep shaft to depth $d$, a distance $r=R_E-d$ from the centre. Picture the Earth as a solid inner sphere of radius $r$ surrounded by a thick outer layer — a set of concentric shells. The body is inside every one of those outer shells, and by Result 7.5(ii) each of them pulls on it with zero net force. Only the inner sphere of radius $r$ attracts it, as if that sphere's mass $M_r$ were at the centre.

Model the Earth as uniform, with density $\rho$. Then $M_r=\tfrac43\pi r^3\rho=M_E\dfrac{r^3}{R_E^3}$, and

Compare the two results for small distances. Going up by $h$ reduces $g$ by the fraction $\dfrac{2h}{R_E}$; going down by $d$ reduces it by $\dfrac{d}{R_E}$. So $g$ falls twice as fast with height as with depth near the surface: $g$ at a small height $h$ equals $g$ at depth $d=2h$.

Exercises

Take $G=6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}$, $M_E=5.97\times10^{24}\ \text{kg}$, $R_E=6370\ \text{km}$ and $g=9.8\ \text{m/s}^2$ at the Earth's surface, and treat the Earth as a uniform sphere, unless told otherwise.