Gravitation Near Earth's Surface — Question 5

Question 5

Rotation tilts effective gravity away from the radial direction

At latitude $\lambda$ on a spherical rotating planet, use outward radial unit vector $\hat r$ and northward tangent $\hat\lambda$. The centrifugal acceleration in the rotating frame has components $\Omega^2R\cos^2\lambda$ outward and $-\Omega^2R\sin\lambda\cos\lambda$ northward. Gravitational acceleration is $-g_0\hat r$.

1. Find effective gravity and the small angle it makes with inward radial at northern latitudes.

2. At $\lambda=45^\circ$ and $\Omega^2R/g_0=.004$, calculate the tilt and its direction.

Question 5 - Solution

Rotation tilts effective gravity away from the radial direction

(a) Derivation and calculation

Adding gravitational and centrifugal accelerations gives $$\boxed{\vec g_{\mathrm{eff}}=-(g_0-\Omega^2R\cos^2\lambda)\hat r -\Omega^2R\sin\lambda\cos\lambda\,\hat\lambda}.$$ For positive inward radial component, define $q=\Omega^2R/g_0$: $$\tan\beta=\frac{q\sin\lambda\cos\lambda}{1-q\cos^2\lambda},\qquad \beta\simeq q\sin\lambda\cos\lambda\quad(q\ll1).$$

(b) Result and interpretation

At $45^\circ$, both $\sin\lambda\cos\lambda$ and $\cos^2\lambda$ equal $1/2$. Thus $$\tan\beta=\frac{.004/2}{1-.004/2}=\frac{.002}{.998}.$$ $$\boxed{\beta=0.00200401\ \mathrm{rad}=0.114821^\circ}.$$ The negative northward component tilts the plumb direction toward the equator.

Physical check

The tilt vanishes at the equator and poles, and everywhere as $\Omega\to0$.