# Combining Proportions to get Newton's Law of Universal Gravitation

I've read a little on the history of Newton's Law of Gravitation and noticed that the formula can be separated into 3 distinct parts that lead to the end result of $F_g = G \frac{m_1 m_2}{r^2}$; the masses ($m_1 m_2$), the inverse square relation ($\frac{1}{r^2}$), and the gravitational constant ($G$). While looking through readings, however, I noticed that a considerable amount of focus was placed on the discovery and development of the inverse square law, with little or no focus on how the other parts of the formula were "glued" together to form the final result. Thus I felt I should open up a discussion to see if there is anyone out there that can shed some light on how these 3 distinct sections/parts of this formula came together to form the whole.

From what little I've read and through naive reasoning I just logically deducted the following to be what had happened for each part:

1.) $m_1$ and $m_2 -$ Related through proportioning techniques, equating ratios of say 2 different sets of masses between celestial bodies with similar orbits. (ie. $\frac{m_1}{M_1}=\frac{M_2}{m_2}$, generalizing the quantity $m_1m_2$)

2.) $\frac{1}{r^2} -$ Found empirically using Kepler's Laws and geometric techniques

3.) $G -$ Found much later by Henry Cavendish via his famous Cavendish experiment (though not numerically calculated, so I've read)

The mystery now is how all of these parts come together at the end.

• If I understand your question correctly, you may be interested in this article in my blog: densytics.com/about Sep 11, 2012 at 23:45

The other parts, other than the inverse square, were clear already before Newton, or at least were easy to guess. That the force of gravity is proportionality to mass of a small object responding to the field of another comes from Galileo's observation of the universal acceleration of free fall. If the acceleration is constant, the force is proportional to the mass. By Newton's third law, the force is equal and opposite on the two objects, so you can conclude that it should be proportional to the second mass too.

The model which gives you this is if you assume that everything is made from some kind of universal atom, and this atom feels an inverse square attraction of some magnitude. If you sum over all the pairwise attractions in two bodies, you get an attraction which is proportional to the number of atoms in body one times the number of atoms in body two.

So the only part that was not determined by simple considerations like this was the falloff rate. I should point out that if you look at two sources of a scalar field, and look at the force, it is always proportional to $g_1$ times $g_2$, where $g_1$ and $g_2$ are the propensity of each source to make a field by itself. Further, if you put two noninteracting sources next to each other, this g is additive, if the field is noninteracting, essentially for the reasons described above--- the independent attractions are independent. So that the proportionality to an additive body constant you multiply over the two bodies is clear. That for gravity, the g is the mass, this was established by Galileo.

### More mathemtically

Let's call the force law between the objects $F(m_1,m_2,r)$. We know that if we put the body m_1 in free fall, the acceleration doesn't depend on the mass, so

$$F(m_1,m_2,r) = m_1 G(m_2,r)$$

So that the mass will cancel in Newton's law to give a universal acceleration. This gives you the relation

$$F( a m_1 , m_2, r ) = a F(m_1,m_2,r)$$

We know that if we put body 2 in free fall, the same cancellation happens, but we also know Newton's third law: $F(m_1,m_2,r)= F(m_2,m_1,r)$ so that

$$F( m_1, a m_2, r) = a F(m_1,m_2,r)$$

So you now write

$$F( m_1 \times 1 , m_2 \times 1, r) = m_1 F( 1, m_2\times 1 , r) = m_1 m_2 F(1,1,r)$$

And this tells you that the force is proportional to the masses times a function of r. The form of the function is undetermined.

An independent argument for the scaling is that if you consider the object m_1 as composed of two nearby independent objects of mass $m_1/2$, then

$$F(m_1/2 , m_2 , r) + F(m_1/2 , m_2 , r) = F(m_1,m_2,r)$$

Then the same conclusion follows.

These types of scaling arguments are second nature by now, and they are automatically done by matching units. So if you have a force per unit mass, the force between two massive particles must be per unit mass 1 and per unit mass 2.

This general argument fails for direct three-body forces, where the force between 3 bodies is not decomposable as a sum of forces between the pairs bodies individually. There are no macroscopic examples, since the pairwise additivity is true for linear fields, but the force between nucleons has a 3-body component.

• Ah, Ron! Thanks again for the quick response. I don't think this is quite what I was looking for though, unless I'm not reading this properly. I was just curious about the math that brought these separate parts together. But regardless, thanks for clarifying how this concept was conceived! :] Jul 25, 2012 at 1:30
• @ProSteve037: There was no math, it's what I said. The math is obvious from the concept. Jul 25, 2012 at 3:29
• I'm definitely not reading this correctly :[ I think I understand what you mean by the last section though, where you explain why the masses are multiplied with one another. I'm not sure if what you wrote here though explains how the inverse square law and the quantity of masses ($m_1m_2$) were put together. Aug 4, 2012 at 1:27
• I expanded the answer. These types of arguments are now second nature, and are made without comment in the literature, but they aren't explained well in elementary books, because the arguments establishing it are too clear for the authors. Aug 4, 2012 at 3:56