# Why is Gravitational force proportional to the masses?.

We know that two mass particles attract each other with a force

$$F~=~\frac{G M_1 M_2}{r^2}.$$

But what is the reason behind that? Why does this happen?

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Is there a particular part of the equation that you are asking about, like are you asking where say G comes from? Or why all mass attracts? – DJBunk Jul 24 '12 at 20:35
I am not asking about the equation.I am asking why all mass attracts. – orange Jul 24 '12 at 20:46
If you have a physics background Zee in 'QFT in a Nutshell' gives a rough explanation of this in chapter I.5. And Peskin mentions this on pg 126. Otherwise I don't know of a good heuristic explanation for this. – DJBunk Jul 24 '12 at 20:58

One could explain "well, gravity is the curvature of spacetime due to the mass-energy". But that would only lead to "well, why does mass-energy curve spacetime?" And, should someone produce a proposed answer to that, the follow-up question would have to be "but why is that so?" etc.

At some point though, one must accept that there are genuine fundamentals, genuine primaries that cannot be explained in terms of something "more" fundamental, "more" primary.

Gravity is considered one of those fundamentals. But the question "what is the reason for gravity" presumes that gravity isn't fundamental. So, the only proper "answer" to your question is "to the best of our knowledge, gravity is fundamental".

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I wonder if you would like to have a crack at this question – Floris Apr 2 '15 at 13:53

I'll try to give some intuition... Ok, so we live in a 3-dimensional space. That may be a bit hard to visualise for you so let us start with a 2D space. Then, consider a gravitational field in this space with 2 different circles around it:

Now, it makes sense to say that the overall work on both circles (ok, that's misleading, what I mean is formally the sum of the work done on each point) should be the same. More Formally, $$\int_{C_1}\vec F\left(r_1\right)\cdot \mbox{d}\vec s=\int_{C_2}\vec F\left(r_2\right)\cdot\mbox{d}\vec s$$ Now, the gravitational force on any 2 points on the same circle should be the same (formally, this is SO(2) symmetry, which becomes an SO(3) symmetry in 3-dimensions but it is really just another way of saying that the two points are just the same distance away from the mass so the gravitational force is the same), right? So, that means that the condition can be simplified to:

$$2\pi r_1 F_1=2\pi r_2 F_2$$ $$r_1 F_1=r_2 F_2$$ $$F_2=\frac{r_1F_1}{r_2}$$

If we choose a constant $F_1$ and $r_1$, we see that this means that $F_2$ is inversely proportional to $r_2$.

Now, let us extend our approach to 3 dimensions. The line integral becomes a 2-Dimensional- surface integral. $$\iint_{S_1}\vec F\left(r_1\right)\cdot \mbox{d}\vec s=\iint_{S_2}\vec F\left(r_2\right)\cdot \mbox{d}\vec s$$

The same reasoning can be applied (this time an SO(3) symmetry): $$4\pi r_1^2 F_1=4\pi r_2^2 F_2$$ $$r_1^2 F_1=r_2^2 F_2$$ $$F_2=\frac{r_1^2F_1}{r_2^2}$$

If we once again set $F_1$ and $r_1$ as constant, we see that this means that $F_2$ is inversely proportional to the square of $r_2$.

In other words, we can construct our law of universal gravitation: $$F=\frac{f(m_1,m_2)}{r^2}$$

$m_1$ and $m_2$ we call the "gravitational charge". It happens to be (really, in Newtonian Gravity, gravity is assumed to be a force that acts on mass) that the gravitational charges are exactly equal to the inertial mass, so let us just call it $$Mass$$

Now, why is $f(m_1,m_2)$ $Gm_1m_2$?. It should make sense that the gravitational force is proportional to the masses. Experimental evidence shows that objects accelerate at the same speed towards the earth regardless of their mass. So, by newton's second's law, we have it that the force is proportional to the mass. It can't be just proportional to one mass, it has to be proportional to both the masses, right? It's a law/principle of symmetry and is also a consequence of Newton's third's law. So,

$$F=\frac{m_1m_2}{r^2}$$

Right? Actually, NO. Wrong! The right-hand-side has units of kilograms squared per meter squared whereas the left hand sides has units of Newton's, i.e. kg meter squared per second squared! I'll do a little attempt to see what happens if newton's were equal to the right hand side's units, i.e. kg^2/ meter^2. Ok, coming back to the problem, let us just put in a constant of units m^3/kgs^2 on the right-hand-side (RHS) to ensure that both sides have the same units: $$F=\mathcal G\frac{m_1m_2}{r^2}$$

This constant is called "Newton's constant' of Universal Gravitation" and this law is called "Newton's law of universal gravitation" because it holds everywhere in the universe!!!

**Note: the below is not correct. It is only to serve as an entertaining result if some entertaining statement were true.**

Ok, as promised earlier, let me play around and see what happens if the units were, in fact, actually, equal. Then, you would have:

$$N=kg^2/m^2$$
$$N=Pa kg^2/N$$
$$N^2=Pa kg^2$$
%$$m^2/s^2=Pa$$
Yay! Pressure = velocity squared! But there's more. Dividing Newton's by both sides of this the equation:
$$kg/(m^2/s^2)=m^2$$
$$kg=m^2/s^2$$
Yay! Mass = velocity squared! Recall earlier about Pressure=Velocity^2.! So, MASS = PRESSURE!

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Not to nitpick, but notice in the question's comments he said that he is not asking about the equation, but about why all mass attracts. Also, you listed a Newton as being a $kgm^2s^{-2}$ when this is a Joule. A Newton is $kgms^{-2}$. Also, G is not in units of $kgm^3s^{-2}=Nm^2$, it is $kg^{-1}m^3s^{-2}=Nm^2kg^{-2}$ – Jim Jun 6 '13 at 13:08
@Jim: I realised that after I answered but that is briefly (but not in any detail) covered in my post. But I think it is still relevant and I saw somewhere that the main goal of Physics SE is to create Q/A pages rather than to solely answer the OP so I don't think it is a good idea for me to remove it. As for the correction, I have now incorporated it in my post since it. It was a typo... – centralcharge Jun 6 '13 at 13:17
Dear @dimension10, for your information, it is frown upon to post nearly identical answers to similar posts. Please delete one of them. See also e.g. this meta discussion. – Qmechanic Jul 2 '13 at 7:28
@Qmechanic: Ok, I deleted my other one. – centralcharge Jul 2 '13 at 8:47

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