Gravitational Field of Earth

Suppose that we model the Earth (not very accurately) as a uniform sphere of mass $M$ and radius $R$. Let the center of the Earth lie at the origin. We wish to determine the gravitational acceleration due to the Earth, ${\bf g}$, both inside and outside the Earth. By symmetry, we expect the gravitational acceleration at a general point whose displacement is ${\bf r}$ to be directed radially inward toward the center of the Earth, and to only depend on the radial distance, $r$, from the center of the Earth. In other words,

$${\bf g} = - a(r)\,\frac{{\bf r}}{r}.$$ (1.250)

Consider a spherical Gaussian surface, $S$, of radius $r$, whose center corresponds to the center of the Earth. The flux of gravitational acceleration out of this surface is

$$\oint_S {\bf g}\cdot d{\bf S} = - 4\pi\,r^2\,a(r).$$ (1.251)

Thus, Gauss's law, (1.246), yields

$$a(r) = \frac{G\,m(r)}{r^2},$$ (1.252)

where $m(r)$ is the mass enclosed by the surface. Now, if $r<R$ then $m(r)= (r/R)^3\,M$ (by proportion), but if $r>R$ then $m(r)=M$. Thus, we deduce that

$$a(r) =\frac{G\,M\,r}{R^3}$$ (1.253)

for $r<R$, and

$$a(r) =\frac{G\,M}{r^2}$$ (1.254)

for $r>R$. In other words, inside the Earth, the gravitational acceleration due to the Earth increases linearly with distance from the Earth's center, but, outside the Earth, it falls off as the inverse-square of distance from the Earth's center. In particular, the gravitational field outside the Earth is exactly the same as that of a point object whose mass is equal to that of the Earth, and that is located at the Earth's center. This is an example of an important result first derived by Newton; namely, that the gravitational field outside a spherically symmetric mass distribution is the same as that of a point object located at the center of the distribution whose mass is equal to that of the distribution.

It is clear from Equation (1.254) that the gravitational acceleration at the surface of the Earth is

$$g\equiv a(R) = \frac{G\,M}{R^2}.$$ (1.255)

Given that the measured (average) gravitational acceleration at the Earth's surface is $g=9.81\,{\rm m\,s^{-2}}$, and that the measured (mean) radius of the Earth is $R=6.371 \times 10^{6}\,{\rm m}$, we arrive at the following estimate for the Earth's mass,

$$M = \frac{g\,R^2}{G} =\frac{9.81\times(6.371 \times 10^{6})^2}{6.67430\times 10^{-11}}=5.965\times 10^{24}\,{\rm kg}$$ (1.256)

This estimate lies within $0.1\%$ of the correct value, which is $M=5.972\times 10^{24}\,{\rm kg}$