<p>If you have a self gravitating sphere than all <em>force</em> vectors at the surface are normal to that surface (and other equipotentials interior to the surface). These <em>force</em> 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 <em>gravity</em> 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.</p>
<p>The gravitational potential <strong>outside</strong> of a uniform spheroid of mass <span class="math-container">$M$</span> can be expressed to high accuracy with
<span class="math-container">$$\Phi \simeq -\frac{GM}{r} + \frac{kG}{2r^3}(3\cos^2\theta -1),$$</span>
plus smaller terms depending on <span class="math-container">$r^{-5}$</span>, where <span class="math-container">$r$</span> and <span class="math-container">$\theta$</span> are the usual spherical coordinates and <span class="math-container">$k$</span> 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). <span class="math-container">$k$</span> can also be written as <span class="math-container">$J_2 Ma^2$</span> where <span class="math-container">$J_2$</span> is the mass quadrupole parameter and <span class="math-container">$a$</span> is the equatorial radius (see
<a href="https://en.wikipedia.org/wiki/Geopotential_model" rel="nofollow noreferrer">https://en.wikipedia.org/wiki/Geopotential_model</a> and <a href="https://physics.stackexchange.com/a/146954/43351">https://physics.stackexchange.com/a/146954/43351</a>).</p>
<p>Taking the gradient of this potential (since gravitational field  <span class="math-container">$\vec{g} = -\nabla \Phi$</span>), you see there is a radial term in <span class="math-container">$\hat{r}$</span>, but also a <span class="math-container">$\hat{\theta}$</span> component
<span class="math-container">$$g_{\theta} = \frac{3kG}{r^4}\sin \theta\cos\theta$$</span>
that is <em>not</em> directed towards the centre of the spheroid.</p>
<p>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.</p>
<p>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 (<a href="https://www.sciencedirect.com/science/article/pii/S003206331730257X" rel="nofollow noreferrer">Hofmeister et al. 2018</a>).</p>

CC BY-SA 4.0
2020-01-17T14:20:14.820Z
This is great as it applies to an observer moving at the same angular velocity at the object (in rotation) but what if the observer is stationary, i.e. not affected by centrifugal (centripetal?) force. My sense is that you'd still have to integrate the gravitational force created by each 'element' of the object and that in general, the integral does not point toward the center of mass (unless it's spherical and can therefore be reduced to a point mass). Is that intuition correct for an oblate spheroid?
medley56
CC BY-SA 4.0
2020-01-17T14:32:01.457Z
@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.
ProfRob
CC BY-SA 4.0
2020-01-17T14:39:27.410Z
@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.
ProfRob
CC BY-SA 4.0
2020-01-17T16:28:20.287Z
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?
Carl Witthoft
CC BY-SA 4.0
2020-01-17T17:45:23.520Z
@CarlWitthoft Exactly what is in the answer.
ProfRob
CC BY-SA 4.0
2020-01-17T17:55:37.290Z
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.
ProfRob
CC BY-SA 4.0
2020-01-18T02:31:06.313Z
more about that: https://space.stackexchange.com/a/29535/12102
uhoh
CC BY-SA 4.0
2023-11-03T00:39:13.047Z
@ProfRob what is the source of this expression for the gravitational potential outside of a uniform spheroid of mass M? Thank you....
Lucy
CC BY-SA 4.0
2023-11-03T10:58:32.023Z
@Lucy see edit.
ProfRob
CC BY-SA 4.0
2020-01-17T08:00:55.373Z
2023-11-03T10:58:09.367Z
2023-11-03T10:58:09.367Z
ProfRob
<p>Consider a planet, described as an oblate spheroid. Assume that the spheroid is uniformly dense but <em>not a point source</em>. Outside of the object, do all vectors in the gravity field point through a single point inside the planet?</p>

<gravity>
Gravitational field for oblate spheroid?
medley56
# Gravitational field for oblate spheroid?

(Asked by: username_0 on 2020-01-16)

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?

**Tags:** gravity

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## 1 answers

### Answer by username_1 on 2020-01-17. Score: 10 | Accepted 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](https://www.sciencedirect.com/science/article/pii/S003206331730257X)).

> **Comments:**
> **username_0**: This is great as it applies to an observer moving at the same angular velocity at the object (in rotation) but what if the observer is stationary, i.e. not affected by centrifugal (centripetal?) force. My sense is that you'd still have to integrate the gravitational force created by each 'element' of the object and that in general, the integral does not point toward the center of mass (unless it's spherical and can therefore be reduced to a point mass). Is that intuition correct for an oblate spheroid?  
> **username_1**: @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.  
> **username_1**: @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.  
> **username_2**: 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?  
> **username_1**: @CarlWitthoft Exactly what is in the answer.  
> **username_1**: 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.  
> **username_3**: more about that: https://space.stackexchange.com/a/29535/12102  
> **username_4**: @ProfRob what is the source of this expression for the gravitational potential outside of a uniform spheroid of mass M? Thank you....  
> **username_1**: @Lucy see edit.  

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