What is the metric of the grs 80 ellipsoid? So, similarly to how a $2$-sphere has the metric, and if I understand this correctly:
$$
g_{ij} = 
\begin{bmatrix}
R^2 & 0 \\
0 & R^2\sin^2\theta
\end{bmatrix}
$$
I was wondering, in geodesy, what is the metric for the GRS 80 ellipsoid currently in use with most datums?  I've been googling but can't really find this.  The Wikipedia page (https://en.wikipedia.org/wiki/GRS_80) describes a bunch of its properties like eccentricity, etc. but I can't really figure out how I would get the metric.  In theory, I should be able to use such a thing to calculate the distance between two points given by latitude and longitude, correct?
 A: It should be easy enough to derive it: we're dealing with an oblate spheroid. The difficulty comes in the choice of coordinates: this page notes the difference between the geocentric and geodetic coordinates (but for some reason the diagram showing the difference is on this page): essentially we have an ellipsoid parametrised as
$$ N(\phi) (\cos{\phi}\cos{\lambda},\cos{\phi}\sin{\lambda},b^2/a^2 \sin{\lambda}), $$
where
$$ N(\phi) = \frac{a^2}{\sqrt{a^2\cos^2{\phi}+b^2\sin^2{\phi}}} $$
is related the radius at angle $\phi$: we see that it varies from $a$ at the equator to $a^2/b$ at the poles. Now one can compute the metric in the usual way using inner products of tangent vectors: one tedious calculation later, we find
$$ ds^2 = \frac{a^4b^4}{(a^2\cos^2{\phi}+b^2\sin^2{\phi})^3} \, d\phi^2 + \frac{a^4\cos^2{\phi}}{a^2\cos^2{\phi}+b^2\sin^2{\phi}} \, d\lambda^2. $$
As a check, if we put $b=a$, we find this reduces to $a^2 \, d\phi^2 + a^2\cos^2{\phi} \, d\lambda^2$, which also serves as a valuable reminder that $\phi$ is latitude and $\lambda$ is longitude, not the usual coordinates we use in mathematics.
Since geodesics on an ellipsoid are horrible, you're probably better off looking up standard techniques of approximation: that flattening of 1/300 makes things surprisingly messy!
