Surface Gravity of Earth

Making use of Equations (1.329), (1.349), (1.354), (1.364), and (1.365), the combined gravitational and centrifugal potential outside the Earth is

$${\mit\Phi}+\chi = -\frac{G\,M}{R}\left[\frac{R}{r} -\frac{3\,\zet... ...\sin^2\theta + \frac{3\,\zeta}{2}\left(\frac{r}{R}\right)^2\sin^2\theta\right].$$ (1.369)

According to Equations (1.328), (1.329), and (1.364), the surface of the Earth lies at radius

$$R_\theta(\theta) = R\left(1-\frac{5\,\zeta}{2} + \frac{15\,\zeta}{4}\,\sin^2\theta\right).$$ (1.370)

The effective gravitational acceleration at the Earth's surface is $g=\vert\nabla({\mit\Phi}+\chi)\vert(R_\theta,\theta)$, which reduces to

$$g(\lambda) = \frac{G\,M}{R^2}\left[1+\frac{\zeta}{2} - \frac{15\,\zeta}{4}\,\cos^2\lambda + {\cal O}(\zeta^2)\right],$$ (1.371)

where $\lambda = \pi/2-\theta$ corresponds to terrestrial latitude. The previous expression shows that the Earth's rotation, combined with its equatorial bulge, causes the acceleration experienced by objects close to the Earth's surface, that co-rotate with the Earth, to vary slightly with latitude. The acceleration is greatest at the poles ( $\lambda=\pi/2$), and weakest at the equator ($\lambda=0$). To be more exact, we predict that

$$g_{\rm pole} = \frac{G\,M}{R^2}\left(1+\frac{\zeta}{2}\right)= 9.826\,{\rm m\,s^{-1}},$$ (1.372)

and

$$g_{\rm equator} = \frac{G\,M}{R^2}\left(1-\frac{13\,\zeta}{4}\right)=9.783\,{\rm m\,s^{-1}}.$$ (1.373)

In fact, the true values of the polar and equatorial accelerations are $g_{\rm pole}= 9.832\,{\rm m\,s^{-1}}$ and $g_{\rm equator} = 9.781\,{\rm m\,s^{-1}}$, respectively.