general relativity - Geodesic devation on a two sphere

So the geodesic deviation equation gives the relative acceleration between two geodesics in motion. But given a pair of geodesic (let's say on the two sphere) that start at the equator, separated by some distance. Is there a way to compute their separation as a function of time without using the geodesic equation? Let's say they're moving at northward toward the pole along a line of constant longitude at unit velocity.

Answer

Re your edited question, this is just simple spherical geometry. If the initial separation is $d$ then the separation at time $t$ is $d \cos(vt/r)$, where $r$ is the radius of the sphere, $v$ is the vehicle speed and $t$ is time.

The diagram shows a cross section through the poles. The vehicle is driving north at a velocity v, so the distance it drives in a time $t$ is just $s = vt$, so the angle $\theta$ is:

$$\theta = \frac{vt}{2\pi r} 2\pi = \frac{vt}{r}$$

Suppose the vehicles start out at a separation $d$. The angular separation along the equator $\Delta\phi$ is:

$$\Delta\phi = \frac{d}{2\pi r} 2\pi = \frac{d}{r}$$

As the two vehicles drive north the angular separation $\phi$ doesn't change, so we just need to calculate the circumference of the line of latitude at the angle $\theta$, $C_\theta$, and the separation will be $C_\theta\tfrac{\Delta\phi}{2\pi}$.

$$C_\theta = 2\pi r \cos\theta$$

So the separation $s$ is:

$$\begin{align} s &= 2\pi r \cos\theta \frac{\Delta\phi}{2\pi} \\ &= 2\pi r \cos\left(\frac{vt}{r}\right) \frac{d/r}{2\pi} \\ &= d \cos\left(\frac{vt}{r} \right) \end{align}$$