# How is gravity consistent when you split an object into multiple pieces?

Here's the picture:

In picture 1: Assume that we have 2 objects, namely A, and B. Object A with mass $2M$, and object B with mass $m$; and their distance is $r$. Then the gravitational force of A acting on B ($\vec{v}$)'s magnitude will be: $$G\times\frac{2Mm}{r^2}.$$

In picture 2: Now divide object A into 2 equal parts of mass $M$ each. The distance from the centroid of each part of A to B is $\dfrac{r}{\cos \alpha}$. The gravitational forces of 2 parts of A acting on B ($\vec{v}_1; \vec{v}_2$)'s magnitudes are: $$G\times \frac{Mm\cos^2\alpha}{r^2}.$$

Now, I'm pretty sure that if I take the sum $\vec{v}_1 + \vec{v}_2$, I wouldn't get $\vec{v}$. The direction is the same, but the magnitude isn't. They are off a factor of $\cos ^ 3 \alpha$. :(

What's going on here?

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As an aside, this is the argument by which Galileo deduced that objects of different mass fall at the same speed. Aristotle believed that heavy objects fall faster (which to be fair is not totally ridiculous considering objects in air, although Aristotle was misled by his philosophical assumptions not by experiment). Galileo considered that two objects loosely connected together would behave the same as one object with the same total weight, but also the same as each part independently. Hence gravity must affect all equally. –  Steve Jessop Apr 1 '14 at 8:40
@SteveJessop Perhaps another lesson that we should be blind to our philosophies to do good science? Or, at least, another argument for it. –  hadsed Apr 2 '14 at 3:38
@hadsed: yeah, I go back and forth on whether I personally blame Aristotle for his errors. I think this is one where even given that it didn't make sense to him to experiment, he could have reached the correct answer using pretty much the same methods by which he reached an incorrect answer. What he was doing is not what we would now call "science", that's for sure. Then again without some philosophy/theory you often lack motivation to do the right experiments. –  Steve Jessop Apr 2 '14 at 8:12

The problem is in your assumption that the force is $F = 2GMm/r^2$. This is true for the force on a point mass from a sphere or another point mass, but not otherwise. What you need to do is sum each the force on each particle from every other particle. For a continuum object, $$\vec F = \int \rho \vec g\, dV$$ where $\rho$ is the density and $\vec g$ the acceleration due to gravity. Both can vary over the volume. To find $\vec g$ you sum up the contribution from all points, $$\vec g(\mathbf x) = G\int\rho \frac{\mathbf y -\mathbf x}{|\mathbf x -\mathbf y|^3}\, dV$$ do as you can see it is significantly more difficult for bodies other than point masses (and spheres, it turns out). But from these expressions you can see that forces on parts of rigid bodies add, and so do the forces from parts of bodies.

Often you can pretend that you are dealing with point masses because from far away, anything looks like a point mass. But as you have discovered, this is an approximation and not exact.

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Great, thank you very much. I see where the flaw is now. Do you know any proof that the formula actually holds for the case when both object are spherical? I really want to broaden my mind, and since I'm studying for a master degree in maths, I think it should help me sometimes in the future. –  user49685 Mar 31 '14 at 10:55
Gauss' law gives that result inmediately. –  Davidmh Mar 31 '14 at 11:29
If you already know some calculus, there's a good proof at hyperphysics: "Gravity Force of a Spherical Shell" and more-or-less the same proof at Wikipedia: "shell theorem". –  David Cary Mar 31 '14 at 12:46
Sweet, thank you guys very much, I'll have a look at them. :* –  user49685 Mar 31 '14 at 12:48
If you know about Gauss law / Poisson's equation, you can also just say that a spherically symmetric harmonic function in 3 dimensions is up to an additive constant proportional to 1/r. So a spherically symmetric potential is the same as that from a point mass. –  Robin Ekman Mar 31 '14 at 15:31

The "intuitive" expectation is that both methods should give the same answer - and they do, if you do it right.

Were method 2 goes wrong is in calculating the force at distance $$d = \frac{r}{\cos{ \alpha}}$$ although this gives you the force between M and m, this is the wrong force. The force that is required, is the force along a line between m and the center of mass of the two Ms. This means that the distance to be used is $d {\cos{\alpha}}$, which is r! So the effective force between M and m is $$F1 = F2 = \frac{MmG}{r^2}$$ Since the two vectors have the same magnitude and direction, the resultant is$$F1 + F2 =\frac{(2)MmG}{r^2}$$ which is the same as the force between 2M and m $$F =\frac{(2M)mG}{r^2}$$

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,I think you should review your answer and originally should integrate in the same way as proposed by Robin Ekman as it seems your final answer is not correct. –  user22180 Apr 4 '14 at 19:53

In your picture 1, the gravitational force is calculated incorrectly. The formula you have used only applies between pointlike masses. You have to divide the object into elements, calculate the contributions of each element and sum up. The picture 2 is only the first step in the whole process, so actually not even your picture 2 is generally correct.

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So, I should divide the object A into really small (infinitesimal) parts $m_i$, then calculate the gravitational force of each $m_i$ acting on $m$, then sum the whole thing up? Is the formula used to calculate here still $G\times \dfrac{m_im}{r^2}$? –  user49685 Mar 31 '14 at 10:46
@user49685 Yes, you need to integrate all the infinitesimal elements. The formula for the point masses is unchanged. You can see that better in the answer given by Robin Ekman. –  mpv Mar 31 '14 at 10:55
Thank you very much. :* I know that the formula also holds in the case, when 2 objects are spherical. Do you know a proof of it? Thank you very much, :D –  user49685 Mar 31 '14 at 10:57
@user49685 The spherical case has the same formula due to the symmetry. It is better seen if you actually calculate that using the Gauss law - en.wikipedia.org/wiki/… –  mpv Mar 31 '14 at 11:03
Thank you very much mpv for your help :* –  user49685 Mar 31 '14 at 12:47