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Consider spherical symmetric$^1$ masses of radii $R_1$ and $R_2$, with spherical symmetric density distributions $\rho_1(r_1)$ and $\rho_2(r_2)$, and with a distance between the centers of the spheres $d$. What is the exact force between them? I know point masses are a good approximation, but I'm looking for an exact formula. This would be useful for a gravity-simulation toy software.

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$^1$ Assume for simplicity the idealization where tidal or centrifugal forces do not deform the spherical symmetric, i.e., the various mass parts are held in place by infinitely strong and rigid bonds.

Qmechanic
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user20998
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4 Answers4

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For spherically symmetric bodies the point-mass approximation is actually exact as a description of the centre-of-mass force on the spheres. This is due to Gauss's law in its gravitational form: the flux of the gravitational field through a concentric spherical surface is proportional to the mass inside it, or mathematically $$\oint\mathbf{g}\cdot d\mathbf{S}=4\pi G M_\text{int}.$$ This can be seen to be equivalent to Newton's law of gravitation and it yields the point-mass formula when applied to a spherical mass.

Community
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Emilio Pisanty
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  • This result e.g. follows from repeated reduction to point masses via Newton's shell theorem. – Qmechanic Feb 15 '13 at 18:13
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    The formula for the field due to a sphere is exact. It does not give you an exact result for the force felt by the sphere experiencing the force. – Zo the Relativist Feb 15 '13 at 21:01
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    @JerrySchirmer Huh??? –  Feb 15 '13 at 22:02
  • @ChrisWhite: Gauss's law tells you about the field generated by a spherically symmetric mass distribution. But the force exerted on a mass is $\int dV \rho {\vec g}$, Gauss's law doesn't help you with that second integral, because there's no integral of a divergence anymore. Just a nasty vector valued integral. – Zo the Relativist Feb 15 '13 at 22:22
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    If you interpret "force between them" as the force that describes the motion of their center of masses, then the point mass description is exact. The hard integral is only required if the questioner wants to model tidal forces on the bodies. – Dave Feb 15 '13 at 23:00
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    @Dave, but, doesn't that integral give us the force on the center of mass? When I evaluate the integral (for a shell, not over a filled sphere) I get a logarithm in it... Series result for it shows that it's equal to $G M (4 \pi r^2 \rho)/x^2$ plus fourth order (and higher) terms of r. So, unless my original formula is wrong, it doesn't look like the force exerted by a gravitational field on a shell is the same as a single force acting on the point at the center of mass of the sphere! (x is distance, r is radius of the sphere) –  Feb 16 '13 at 01:21
  • Mathematica code used: int=Integrate[ 2 Pi r^2 G M \[Rho] Sin[\[Theta]]/(r^2 + x^2 - 2 r Cos[\[Theta]] x), \[Theta]] (integral), s=FullSimplify[(int /. \[Theta] -> Pi) - (int /. \[Theta] -> 0)] (from 0 to Pi), Series[s, {r, 0, 4}] (series expand). –  Feb 16 '13 at 01:21
  • Didn't mean to post that yet. clarification: The sphere is at the origin, x is the x-position of the gravitational field source, and r is the radius of the sphere. The formula uses $dV=r^2 Sin[\theta] d\phi d\theta dr$. Since the field doesn't depend on the coordinate $\phi$ (spherical coordinate on the YZ plane), integration gives that factor of $2 \pi$. Let me know if I'm unclear or incorrect. –  Feb 16 '13 at 01:30
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    @JerrySchirmer I believe you are wrong simple because the force on mass element $i$ (of one mass) due to mass element $j$ (of the other mass) is equal and opposite to the force on element $j$ due to element $i$, and of course the forces are linear in the fields. All of the above assuming that neither body can deform under the tidal influence of the other ... that changes things entirely. – dmckee --- ex-moderator kitten Feb 16 '13 at 02:38
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    @dmckee Oh. You're right. My mistake was that I didn't account for the vector nature of force, I was thinking of the force in the x direction, but I forgot to actually add in the correct factor. (of $(x-\cos(\theta))/d$ where $d$ is the distance, $\sqrt{r^2+x^2-2r x \cos(\theta)}$. int=Integrate[2 Pi r^2 G M \[Rho] Sin[\[Theta]](x-Cos[\[Theta]]r)/(r^2+x^2-2 r Cos[\[Theta]] x)^(3/2),\[Theta]]; s=Simplify[(int/.\[Theta]->Pi)-(int/.\[Theta]->0)]; gives the correct results. (though, the simpler correct argument is by symmetry) –  Feb 16 '13 at 03:23
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    @dmckee: the integral is symmetric between the masses. Each of them sets up a $\propto \frac{1}{$r^{2}}$ field, and this must be integrated to get the force. – Zo the Relativist Feb 16 '13 at 21:41
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The Gauss's Law answer is certainly right, but for a different take on things, you might want to have a look at the Principia itself. Long before Gauss came along, Newton proved the result you seek with geometry.

Basically, you consider how a test particle ("corpuscle") outside a spherical shell of mass is pulled by different sections of that shell, assuming the inverse-square law of gravity. This is Theorem XXXI in Book I. Theorem XXXIV extends this to a test particle outside a solid sphere of uniform density, essentially by integrating. Theorem XXXV generalizes again to the force between two disjoint spheres of uniform density. Finally, Theorem XXXVI tells us that in the general case you seek, the force is the same as though all the mass were concentrated at the spheres' centers, using the principle of superposition.

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In the event you are interested in some methodology or a transparent way on how to solve this problem, here is an analytic approach.

Let the two spheres be $S_1$ and $S_2$ radius $R_1$ and $R_2$ at centres C1 and C2 respectively. The distance $C1C2 =d\ge R_1+R_2$. Work in spherical coordinates with the frame at C1:

1)Consider two point P1$( r_1\cos\phi_1\cos\theta_1, r_1\cos\phi_1\sin\theta_1, r_1\sin\phi_1)$

and P2$( d+ r_2\cos\phi_2\cos\theta_2, r_2\cos\phi_2\sin\theta_2, r_2\sin\phi_2)$ in the volume of $S1$ and $S2$ respectively.

2) Now write the square of the distance P1P2

$r_{12}^2=(d+ r_2\cos\phi_2\cos\theta_2- r_1\cos\phi_1\cos\theta_1)^2+(r_2\cos\phi_2\sin\theta_2-r_1\cos\phi_1\sin\theta_1)^2+ (r_1\sin\phi_1- r_2\sin\phi_2)^2$

3) Write the differential volume elements for the two spheres at the points P1 and P2 in spherical coordinates:

$dV_1=\sqrt {g_1}dr_1d\phi_1d\theta_1$ and $dV_2=\sqrt {g_2}dr_2d\phi_2d\theta_2$

where $g_1$ and $g_2$ are the Jacobian determinants for spherical coordinates (not difficult to find, or look it up in a vector analysis book, or use $dV=r^2\cos\phi drd\phi d\theta$). Hence the differential masses of these volume elements are

$dm_1=\rho_1(r_1)dV_1$ and $dm_2=\rho_2(r_2)dV_2$

Now you can put all these together and write the total magnitude of the force

$F=G\int_{V_1}\int_{V_2}\frac {\rho_1(r_1)\rho_2(r_2)dV_1dV_2}{r_{12}^2}$

and you need to do this integral with the transformations given above.

This is a very general formula giving the total force in the problem. One can see that if $d>>R_1+R_2$ this equation reduces to the gravitational attraction between two point masses at large distance $d$. One could end up with simpler integrals by using the symmetry in the line C1C2, and considering discs instead the general volume elements we have done in the above method

JKL
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    $d$ can be any amount greater than $R_1+R_2$ and the integral will be exactly $M_1M_2/d^2$. Try it :) –  Feb 16 '13 at 17:29
  • @user20998 Please see new answer to the exact gravitational force... – JKL Feb 16 '13 at 18:12
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If you are looking only for the Newtonian gravitational force, the other answers to this question are correct. The spherically symmetric mass distributions can be replace by the total mass at the center of mass and then the Newtonian gravitational force can be computed for these two masses.

However, if you want EXACT calculations, you must use general relativity. In general relativity it is also true that a spherically symmetric mass distribution can be replaced with the Schwarzchild black hole of the same mass at the same center. However this is for a single spherical distribution. It is probably not true for two spherical distributions or for two Schwarzchild black holes. There is no know exact general relativity solution to the two black hole problem. All we have is very good numerical simulations of general relativity for two black holes.

FrankH
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