I am going to ask a really stupid question here. It is a very well known fact that gravity is inversely proportional to the distance squared between two masses. I understand how he arrived at this idea using Kepler's law and centripetal acceleration. But I was wondering how he found out that gravity is proportional to the product of two masses. Is this something obvious that I am too dumb realize or is there other proof behind this?


2 Answers 2


If you assume that one mass does not inhibit the force from any other mass, then the result follows from the multiplicity of all pairwise interactions.

Without loss of generality* you can imagine partitioning each object into an integer number of small pieces, all of the same mass. Then the force of each piece in one object on each piece in the other object must all be equal. The number of these small pairwise interactions is the product of the number of pieces, which is proportional to the product of the masses.

multiplicity of pairwise interactions example

*If one objects mass is an irrational multiple of the other, use a sequence of successive rational approximations.

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    $\begingroup$ I understand your argument but one point seems to be missing. You say "Then the force of each piece in one object on each piece in the other object must all be equal." I don't quite follow this since the center of masses of pieces are different and thus the distance between the pieces of the 1. Object and pieces of the second object is different. Thus the forces are different. A generalisation of $m \to 0$ and $n \to \infty$ would make this a perfect answer. $\endgroup$
    – Gonenc
    Commented Jun 1, 2015 at 9:34
  • $\begingroup$ @gonenc You are right. If the distance is large compared to the size of the objects, that complication can usually be ignored. But at closer range it has to be taken into account. This is what is known as tidal forces. $\endgroup$
    – kasperd
    Commented Jun 1, 2015 at 11:46
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    $\begingroup$ The generalization you are speaking about is exactly why integration is needed if the objects are not uniform spheres. That's not the problem, though, because Newton's law talks about point masses anyway (the additivity described in this answer still holds - more mass, more force, and that's on both sides). $\endgroup$
    – orion
    Commented Jun 1, 2015 at 12:56

You can't figure it out just from Kepler's law. An electron orbiting a proton will follow the same path, even though force has nothing to do with the masses of the proton and the electron. However, there are a few basic assumptions that will make it clear.

If the force wasn't proportional to the falling mass, then different bodies would accelerate at different speeds. This would pull planets apart. Since this isn't happening, we know that proportionality works. Also, we know this from older gravitational theories. All objects accelerate at the same rate from gravity, so it must pull proportional to mass. If it didn't, two bowling balls tied together by a tiny string would fall at a different rate than each of them falling separately.

If you accept the law of action and reaction, or in Newton's case, if you discover that law, then gravity must also be proportional to the mass of the object doing the pulling. Another way to see it is to imagine the solar system having two suns in the middle. Planets would be pulled to each of them, and would experience twice the force. This won't tell you that it's proportional to mass rather than, say, electric charge, but if you've been paying attention this far, mass would be the obvious thing to guess.

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    $\begingroup$ An electron doesn't orbit a proton, or indeed follow any path. $\endgroup$
    – OrangeDog
    Commented Jun 1, 2015 at 12:47
  • $\begingroup$ Not in the case of an atom. If the electron was much further away from the proton, Newtonian physics would still be a good approximation, and it would orbit like a planet. $\endgroup$
    – DanielLC
    Commented Jun 2, 2015 at 17:57
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    $\begingroup$ If planets form a quantum probability cloud centred on their star, then maybe. $\endgroup$
    – OrangeDog
    Commented Jun 2, 2015 at 18:00
  • $\begingroup$ If you guys insist on using macroscopic masses, then just use charged bodies. $\endgroup$
    – DanielLC
    Commented Jun 2, 2015 at 18:57

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