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Assume a point mass $m$ located at $\vec{x}$. Assume also a solid body whose coordinates $\vec{x}'$ belong to a connected subdomain $\vec{x}' \in \Omega$. The solid body has a non-uniform mass density $\rho(\vec{x}')$. We may thus define the mass of the solid object as

$$M = \int_{\Omega} \rho(\vec{x}')\,{\rm d}^3x'$$

and its center of mass

$$\vec{x}_c'= \int_{\Omega} \vec{x}' \rho(\vec{x}')\,{\rm d}^3x'$$

The infinitesimal force of gravity acting on a volume element of the solid object is given by

$$\,{\rm d}\vec{F} = Gm\,{\rm d}M\frac{\vec{x}' - \vec{x}}{|\vec{x}' - \vec{x}|^3} = Gm \rho(\vec{x}') \frac{\vec{x}' - \vec{x}}{|\vec{x}' - \vec{x}|^3} \,{\rm d}^3 x'$$

The total force can be obtained by integration.

$$\vec{F} = Gm \int_{\Omega} \rho(\vec{x}') \frac{\vec{x}' - \vec{x}}{|\vec{x}' - \vec{x}|^3} \,{\rm d}^3 x'$$

Question: How does one prove that the total force is

$$\vec{F} = GmM \frac{\vec{x}'_c - \vec{x}}{|\vec{x}_c' - \vec{x}|^3}$$

I am aware of the question Why does gravity act at the center of mass?. I am explicitly interested in the integral proof, which the solution does not provide

Now, I admit that I understood or learned it wrong. What I was assuming to be the general case, was in fact a special case for large distances and/or spherical bodies.

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    $\begingroup$ Possible duplicate of Why does gravity act at the center of mass? $\endgroup$ – sammy gerbil Apr 25 '18 at 14:25
  • $\begingroup$ You seem to be trying to prove that the gravitational force acts at the centre of mass. This is not true, so it cannot be proven. $\endgroup$ – sammy gerbil Apr 25 '18 at 14:27
  • $\begingroup$ @sammygerbil I disagree. The force acts on the entire volume, but it is equivalent to effectively acting on the centre of mass. All of solid body mechanics is based around this idea $\endgroup$ – Aleksejs Fomins Apr 25 '18 at 14:45
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    $\begingroup$ Consider a rigid body composed of the Earth and the Moon (connected by an invisible massless rod). The centre of mass is inside the Earth, but an astronaut standing on the Moon is attracted towards the centre of the Moon, not the centre of mass of the Earth and Moon. $\endgroup$ – sammy gerbil Apr 25 '18 at 15:04
  • $\begingroup$ Ok, I am starting to see that I am wrong, but I need another moment of thinking to understand why :D $\endgroup$ – Aleksejs Fomins Apr 25 '18 at 15:06
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You can't prove that because it isn't true.

For example, suppose the "solid body" is dumbbell - two point masses connected by a very, very long massless rod. Put one side of the dumbbell near the point mass. You'll get a pretty strong gravitational attraction, which you can make arbitrarily strong by just bringing them closer and closer together. But the other side of the dumbbell is very, very far away, so the center of mass is very far away from the point mass and your formula says the gravitational attraction should be very weak. This shows that the formula you want to prove isn't correct. You might have been thinking of the special case of a spherically-symmetric mass distribution.

Just try a couple of really simple examples before you try to prove a complicated general theorem. It will save a lot of time.

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  • $\begingroup$ I thought it was completely obvious, so I somehow did not bother checking it. I was quite wrong it seems :D $\endgroup$ – Aleksejs Fomins Apr 25 '18 at 15:19
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It seems to me that what you are stating is wrong.

Here is an easy counter-example to your statement. Consider your point particle $m$ at $x = 0$, and consider a uniform rod of mass M spanning from $x=1$ to $x=L-1$. Then, its center of mass is located at $x= L/2$, and thus, by your formula, we would expect the resultant force on the rod to have magnitude $GmM\frac{4}{L^2}$, where $M = \rho(L-2)$

Consider now the integral expression :

$$\vec{F} = Gm\rho\int_{1}^{L-1}\frac{x}{x^3}dx = Gm\rho\int_{1}^{L-1}\frac{1}{x^2}dx = Gm\rho \frac{L-2}{L-1} = \frac{GmM}{L-1}$$

Notice how the two expression don't match ? As a check, we can take $L\rightarrow 2$, which is when the rod becomes a point particle, and in that case we indeed recover that both forces have the same magnitude, $GmM$.

Note also (it is a good exercise), that if we put this rod at a distance $a>>L$, the 2 methods of computing the force should match, as, when you see the rod from very far, it behaves like a point particle.

So, summarizing, the theorem you state is not true in general. I think you may have been confused by the fact that, in newtonian mechanics, when we treat gravitational bodies we always act with the gravitational force at the center of mass. This is just because we are generally interested in the center of mass movement of the planet, and as you know it is then sufficient to put all external forces as acting on the center of mass.

However, this does not entirely describe the real action of the gravity on the planet. Indeed, there will be other effects like, for example, tidal forces.

Finally, it is interesting to note that conversely the gravitational field created by a spherically symmetric body will look exactly like one of a point particle, as stated by sammy, through the Shell Theorem.

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