# Gravitational field for oblate spheroid?

Consider a planet, described as an oblate spheroid. Assume that the spheroid is uniformly dense but not a point source. Outside of the object, do all vectors in the gravity field point through a single point inside the planet?

## 1 Answer

If you have a self gravitating sphere than all force vectors at the surface are normal to that surface (and other equipotentials interior to the surface). These force vectors only point towards the centre in the case of a sphere (or at the poles and equator of an arbitrary spheroid). However, even if you remove any centrifugal effects and just have a rigid, non-rotating spheroid, the gravity vector still does not point to the centre of the spheroid except at the poles or the equator and that is true either inside or outside the surface.

The gravitational potential outside of a uniform spheroid of mass $$M$$ can be expressed to high accuracy with $$\Phi \simeq -\frac{GM}{r} + \frac{kG}{2r^3}(3\cos^2\theta -1),$$ plus smaller terms depending on $$r^{-5}$$, where $$r$$ and $$\theta$$ are the usual spherical coordinates and $$k$$ is a constant equal to the difference in the moments of inertia about axes parallel and perpendicular to the rotation (symmetry) axis (note that the object would have to be rotating in order to be a self gravitating spheroid, but that the potential above is just the gravitational potential and does not include any centrifugal, non-gravitational component and is equally appropriate for a rigid oblate spheroid). $$k$$ can also be written as $$J_2 Ma^2$$ where $$J_2$$ is the mass quadrupole parameter and $$a$$ is the equatorial radius (see https://en.wikipedia.org/wiki/Geopotential_model and https://physics.stackexchange.com/a/146954/43351).

Taking the gradient of this potential (since gravitational field $$\vec{g} = -\nabla \Phi$$), you see there is a radial term in $$\hat{r}$$, but also a $$\hat{\theta}$$ component $$g_{\theta} = \frac{3kG}{r^4}\sin \theta\cos\theta$$ that is not directed towards the centre of the spheroid.

In the limit of an almost disk-like spheroid, then at large distances from the centre and reasonably close to the disk plane, this term will dominate and the gravity will tend to act towards the disk midplane rather than towards the centre of the disk.

The above analysis only includes the monopolar and quadrupole gravity fields. There are (smaller) higher order terms that can be included in the potential, but the result is the same -- that the gravitational field outside an oblate spheroid is not a central force (Hofmeister et al. 2018).

• @medley56 you have misunderstood. There is no centrifugal force in the potential I have given. This is appropriate for an object outside the spheroid (e.g. an object in orbit). If you were standing on the spheroid, the details would differ. Jan 17, 2020 at 14:32
• @medley56 by differ, I mean that although the gravitational potential at the surface is given by the same formula, the "effective gravity", reduced by co-rotation, would have an additional centrifugal component. Jan 17, 2020 at 14:39
• I understand you've derived for a rotating oblatespheroid, but let's go back to the question - what's the gravity vector field on the surface of a rigid oblate spheroid, no motion of any kind involved? Jan 17, 2020 at 16:28
• @CarlWitthoft Exactly what is in the answer. Jan 17, 2020 at 17:45
• The only reason I mentioned rotation is that this is the only way to get a self-gravitating spheroid. The different levels of oblateness are then just correlated with rotation rate. The gravitational field outside a spheroid has nothing to do with its rotation rate. Jan 17, 2020 at 17:55