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I have a question where I need to find the gravitational force between an ellipse of negligible thickness and a point mass with given coordinates in 3D space. Not being from a physics background I found this force from the centre of the ellipse since it is of uniform density, however people that I know from a physics background have said that this gives a different answer to using integration.

So would both ways give the same answer, and if the two methods are different then I would assume that the integration method gives the correct result so how would I do this?

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  • $\begingroup$ If your point is distant compared to the ellipse dimensions then your approximation is fine. $\endgroup$
    – JMLCarter
    Commented Nov 21, 2017 at 22:22
  • $\begingroup$ One way to see that they do not give the same answer is that close to the surface of the ellipse, the field is approximately uniform. (This is similar to the electric field of one plate of a capacitor.) The field of a point mass or a sphere is never uniform. $\endgroup$
    – user4552
    Commented Nov 21, 2017 at 23:16
  • $\begingroup$ @BenCrowell The gravitational field of the Earth is approximately uniform close to its surface. $\endgroup$ Commented Nov 22, 2017 at 17:56

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The force between the point mass and the ellipse is the integral of the force between each infinitesimal element of the ellipse and the point mass. Assuming the ellipse has a uniform density $\sigma$ per unit area, and assuming the ellipse is essentially a 2D object:

$$\mathbf{F}=-Gm\sigma\iint \frac{dxdy}{r^2}\mathbf{\hat{r}}$$

Where $r$ is the distance between the infinitesimal element and the point mass, and $\mathbf{\hat{r}}$ is the unit vector pointing from the infinitesimal element to the point mass.

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No, the method in your title cannot be used in your case, if you want the exact answer. Your physics colleagues are correct. Integration (as explained by probably-someone) gives the exact answer.

Replacing the distributed mass by a point mass at the centre of mass works for a spherically symmetric (3D) distribution, but not (in general) for any other distribution. For a spherically symmetric distribution this is called Newton's Shell Theorem. By extension this applies also to the force between two spherically symmetric masses. It does not work for a 3D ellipsoid, unless the ellipsoid is spherical.

As JMLCarter suggests, your method gives an approximate answer, and this approximation becomes better as the separation becomes much bigger than the dimensions of the ellipse. But this is trivial, I think, because in the limit of large separations all finite objects look like point particles; any departure from spherical symmetry becomes insignificant.

The Shell Theorem does not work in 2D, so it does not work for an elliptical disk or ring, not even if the disk or ring is circular. In order for it to work for circular distributions of mass, the force law would have to be prortional to $1/r$ instead of $1/r^2$. See Newton's Gravitation Force between two bodies and Gravitational field of thin 2D ring - numerical simulation. So it will work for the force between a line mass and a cylindrical shell or solid, provided that their axes are parallel and both are infinitely long - ie much longer than the distance between them.

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  • $\begingroup$ If you look at the multipole expansion, there should be a minimum distance away from the ellipse such that only the monopole moment is significant (the higher moments all have higher powers of $r$, so they get wiped out far away). In that limit, it can be treated as a point mass. $\endgroup$ Commented Nov 22, 2017 at 18:19
  • $\begingroup$ And, since you can't have negative mass, unlike in electromagnetism, zero monopole moment means all higher moments are also zero, since there's no mass anywhere. So there's no "pure dipole" in gravitation, and the above limit always exists except in the trivial case. $\endgroup$ Commented Nov 22, 2017 at 18:27
  • $\begingroup$ @probably_someone Yes. However, I think there is no strict cut-off point, the minimum distance to which you refer depends on your measure of significance. As JMLCarter implies, the approximation in the title improves with separation for any mass distribution. There is nothing special about the ellipse. $\endgroup$ Commented Nov 22, 2017 at 18:40
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Not for sure. Even perfect circle has different gravity. For the ring you have

$$ \varphi = \frac{GM}{{\rm agm}(r_1,r_2)} $$

Where $\varphi$ correspond to gravitational potential. $r_1$, $r_2$ is a distance from the nearest/farest point of the ring. ${\rm agm}(x,y)$ means "Arithmetic Geometric Mean". See https://en.wikipedia.org/wiki/Arithmetic%E2%80%93geometric_mean

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