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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 (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).

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).

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 $$\Phi = -\frac{GM}{r} + \frac{kG}{2r^3}(3\cos^2\theta -1),$$ 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).

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).

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 (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).

If you have a self gravitating sphere than all force vectors point are normal to the surface (and other equipotentials). 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, 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 $$\Phi = -\frac{GM}{r} + \frac{kG}{2r^3}(3\cos^2\theta -1),$$ 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 does not include any centrifugal, non-gravitational component and is equally appropriate for some sort of hypothetical rigid oblate spheroid).

Taking the gradient of this potential you see there is a 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).

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 $$\Phi = -\frac{GM}{r} + \frac{kG}{2r^3}(3\cos^2\theta -1),$$ 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).

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).