Newton never said how he discovered the equations describing motion and gravity, to my knowledge. I am aware of two clues.
The most famous one is his allusion to seeing an apple fall, which may or may not have been serious, and his encoded trade secret

note-to-self, to look for the differential equation.

Let's follow first this last clue. Kepler had summarized his analyses of the orbits of the planets with three observations—they form ellipses with one focus at the sun, they sweep out areas along those ellipses at a constant rate, and that they all share a common value for an expression formed from the size of the orbit, and the time it takes to complete an orbit, to be described later.

Let's express these ideas algebraically. First we need to express an ellipse. The definition of an ellipse is that it comprises the set of point in a plane having some fixed average distance to some fixed pair of points, which are called the *foci* of the ellipse.

We choose coordinates so that $(0,0)$ is the location of the sun in the orbital plane, the other focus is at $(2a, 0)$, and we call the average distance $\rho$. So $$\rho = 1/2 (\sqrt{(x-2a)^2 + y^2} + \sqrt{x^2 + y^2}).$$

After a little algebra, we get $$\frac{(x-a)^2}{\rho^2} + \frac{y^2}{\rho^2 - a^2} = 1.$$

Now the rate of area swept out is given by $x\dot{y} - y\dot{x}$.
The total area of the ellipse is $\pi \rho \sqrt{\rho^2 - a^2}$, and let us define $\tau$ to be the total time it takes to go around the ellipse, the length of a year

.

So $$x\dot{y}-y\dot{x} = \pi \rho \sqrt{\rho^2 - a^2}/\tau,$$ which we will abbreviate as $K$.

Then notice first that $$ 0=\dot{x}\dot{y} + x\ddot{y} - y\ddot{x} - \dot{x}\dot{y} = x\ddot{y} - y\ddot{x},$$ so $$x\ddot{y} = y\ddot{x}.$$

This means that $\ddot{x}, \ddot{y}$ is proportional to $x, y$, which is to say that the acceleration is *radial*,
either towards or away from the origin, where is the sun is located. At this point, if you think about a falling apple, you
might equate falling towards the sun

with falling toward the earth

, and suspect that gravity is what is causing the orbits. We'll see stronger evidence of that before long.

So let's write $$\ddot{x} = \alpha x,$$ $$\ddot{y} = \alpha y,$$ and try to determine $\alpha$.

We have $$\frac{(x-a)^2}{\rho^2} + \frac{y^2}{\rho^2 - a^2} = 1,$$ so $$\frac{\dot{x}(x-a)}{\rho^2} + \frac{y\dot{y}}{\rho^2-a^2} = 0.$$

But then $$\frac{\dot{x}^2 + \ddot{x}(x-a)}{\rho^2} + \frac{\dot{y}^2 + y\ddot{y}}{\rho^2-a^2}=0.$$ So $$\frac{\dot{x}^2 + \alpha x(x-a)}{\rho^2} + \frac{\dot{y}^2 + y\alpha y}{\rho^2-a^2}=0,$$ or $$\frac{\dot{x}^2 + \alpha x(x-a)}{\rho^2} + \frac{\dot{y}^2 + \alpha y^2}{\rho^2-a^2}=0,$$ or $$\alpha\left(\frac{x(x-a)}{\rho^2} + \frac{y^2}{\rho^2-a^2}\right)= - \left( \frac{\dot{x}^2}{\rho^2} + \frac{\dot{y}^2}{\rho^2-a^2}\right).$$ Now $$\frac{x(x-a)}{\rho^2} + \frac{y^2}{\rho^2-a^2} = \frac{y^2}{\rho^2-a^2} + \frac{(x-a)^2}{\rho^2} + \frac{a(x-a)}{\rho^2} = 1 + \frac{a(x-a)}{\rho^2} = \frac{\rho + \frac{a(x-a)}{\rho}}{\rho},$$ so $$\alpha\frac{\rho + \frac{a(x-a)}{\rho}}{\rho}= - \left( \frac{\dot{x}^2}{\rho^2} + \frac{\dot{y}^2}{\rho^2-a^2}\right).$$

But $$x^2 + y^2 = x^2 + (\rho^2-a^2)(1-\frac{(x-a)^2}{\rho^2})=\frac{\rho^2x^2 + (\rho^2-a^2)(\rho^2-x^2+2ax-a^2)}{\rho^2} =\frac{\rho^2x^2 + \rho^4 - \rho^2x^2+2a\rho^2x -\rho^2a^2 - a^2\rho^2+a^2x^2-2a^4x+a^4}{\rho^2} =\frac{ (\rho^2 - a^2)^2 + 2a(\rho^2-a^2)x +a^2x^2}{\rho^2} =\left(\frac{ \rho^2 - a^2 + ax}{\rho}\right)^2 =\left( \rho + \frac{a(x-a)}{\rho}\right)^2 .$$ But then $$r = \sqrt{x^2+y^2} = \rho + \frac{a(x-a)}{\rho} .$$

Substituting into the equation for $\alpha$, we have $$\alpha= - \left( \frac{\dot{x}^2}{\rho^2} + \frac{\dot{y}^2}{\rho^2-a^2}\right)\frac{\rho}{r}.$$

At this point, it is completely clear that alpha is negative, meaning that the acceleration is, indeed, towards the sun. This is a little more confirmation that gravity is causing the orbit.

We calculate $$\frac{\dot{x}^2}{\rho^2} + \frac{\dot{y}^2}{\rho^2-a^2}.$$

Since $$x\dot{y}-y\dot{x}-K=0,$$ and $$\frac{\dot{x}(x-a)}{\rho^2} + \frac{y\dot{y}}{\rho^2-a^2} = 0$$ $$\left(x\dot{y}-y\dot{x}-K\right) \frac{x-a}{\rho^2} + y\left( \frac{\dot{x}(x-a)}{\rho^2} + \frac{y\dot{y}}{\rho^2-a^2}\right) = 0$$ so $$\left(x\dot{y}-K\right) \frac{x-a}{\rho^2} + y\left( \frac{y\dot{y}}{\rho^2-a^2}\right) = 0$$ or $$x\dot{y} \frac{x-a}{\rho^2} - -K \frac{x-a}{\rho^2} + y\left( + \frac{y\dot{y}}{\rho^2-a^2}\right) = 0$$ or $$\dot{y}\left( \frac{x(x-a)}{\rho^2} + \frac{y^2}{\rho^2-a^2}\right) = K \frac{x-a}{\rho^2} $$ or $$\dot{y}\frac{r}{\rho} = K \frac{x-a}{\rho^2} $$ so $$\dot{y} = K \frac{x-a}{r\rho} $$ so $$\frac{\dot{y}^2}{\rho^2-a^2} = K^2 \frac{(x-a)^2}{r^2\rho^2(\rho^2-a^2)} $$ Similarly $$\left(x\dot{y}-y\dot{x}-K\right) \frac{y}{\rho^2-a^2} - x\left( \frac{\dot{x}(x-a)}{\rho^2} + \frac{y\dot{y}}{\rho^2-a^2}\right) = 0$$ or $$\left(-y\dot{x}-K\right) \frac{y}{\rho^2-a^2} - x\left( \frac{\dot{x}(x-a)}{\rho^2}\right) = 0$$ or $$-y\dot{x} \frac{y}{\rho^2-a^2} - x\left( \frac{\dot{x}(x-a)}{\rho^2}\right) = K \frac{y}{\rho^2-a^2} $$ or $$\dot{x}\left( \frac{y^2}{\rho^2-a^2} + \frac{x(x-a)}{\rho^2}\right) = -K \frac{y}{\rho^2-a^2} $$ or $$\dot{x} \frac{r}{\rho} = -K \frac{y}{\rho^2-a^2} $$ so $$\dot{x} = -K \frac{y\rho}{r(\rho^2-a^2)} $$ and $$\frac{\dot{x}^2}{\rho^2} = K^2 \frac{y^2}{r^2(\rho^2-a^2)^2} $$ But then $$ \frac{\dot{x}^2}{\rho^2} + \frac{\dot{y}^2}{\rho^2-a^2}= K^2 \frac{(x-a)^2}{r^2\rho^2(\rho^2-a^2)} + K^2 \frac{y^2}{r^2(\rho^2-a^2)^2}= \frac{K^2}{r^2(\rho^2-a^2)} \left( \frac{(x-a)^2}{\rho^2} + \frac{y^2}{(\rho^2-a^2)} \right)= \frac{K^2}{r^2(\rho^2-a^2)} $$ But $$K^2 = \pi^2 \rho^2 (\rho^2 - a^2)/\tau^2$$ Therefore, $$\alpha= - \frac{K^2}{r^2(\rho^2-a^2)} \frac{\rho}{r} = -\frac{ \pi^2 \rho^2 (\rho^2 - a^2)/\tau^2}{r^2(\rho^2-a^2)} \frac{\rho}{r} = -\frac{ \pi^2 \rho^3 /\tau^2}{r^3} $$ So, remembering the definition of $\alpha$, we have $$\ddot{x} = - \frac{\pi^2\rho^3}{\tau^2}\frac{x}{r^3}$$ $$\ddot{y} = - \frac{\pi^2\rho^3}{\tau^2}\frac{y}{r^3}$$ and of course, because $z=0$ for the entire orbit, $$\ddot{z} = - \frac{\pi^2\rho^3}{\tau^2}\frac{z}{r^3}$$

But now
$\frac{\rho^3}{\tau^2}$ is exactly the expression from the third observation Kepler made. It's the same for *all* the planets,
in spite of their widely differing values for $\rho$ and $\tau$.

A key to interpreting it is given by some useful facts about Jupiter:

- Like the sun, it's a giant ball of gas
- It has orbiting moons
- Kepler's observations apply to those lunar orbits, with a
*different*common value for $\frac{\rho^3}{\tau^2}$

So the Kepler constant is proportional to the mass of the object being orbited, so $\frac{\pi^2\rho^3}{\tau^2}/M$ must be constant. We give it the name $G$. But from $\frac{\pi^2\rho^3}{\tau^2}/M =G$ we see that $\frac{\pi^2\rho^3}{\tau^2} =GM$

Thus we have the equations $$\ddot{x} = - GM \frac{x}{r^3}$$ $$\ddot{y} = - GM \frac{y}{r^3}$$ $$\ddot{z} = - GM \frac{z}{r^3}$$ where $M$ is the mass of the sun, or whatever object is being orbited.

The direction of the acceleration is towards the sun. Its magnitude is $$ \sqrt{ \ddot{x}^2+ \ddot{y}^2+ \ddot{z}^2 } = GM \sqrt{ \frac{x^2}{r^6}+ \frac{y^2}{r^6}+ \frac{z^2}{r^6} } = GM\sqrt{\frac{r^2}{r^6}} = \frac{GM}{r^2} $$

There are a couple of directions we can take this. A direction that, to my knowledge, Newton didn't take is to think of these equations as defining a property of space. The equations incorporate no information about the planet, but its location. We'll come back to that.

But in particular, if $r$ is very large, eventually the acceleration is very small, close to zero.
Far away, bodies have an acceleration of $0$, which means that their velocity is constant.
If they start out motionless, they stay that way. If they're moving, both the direction and rate of that motion stays constant.
*That's the first law of motion*.

The direction I suspect Newton did take, is to multiply both sides of the equation by the mass of the planet; let's call it $m$. This give us $$m\ddot{x} = - GMm \frac{x}{r^3}$$ $$m\ddot{y} = - GMm \frac{y}{r^3}$$ $$m\ddot{z} = - GMm \frac{z}{r^3}$$

Why would we do that? Here's really where the apple comes in. The planetary orbit is governed by acceleration towards the sun. What makes something accelerate towards the sun? What makes something accelerate towards Jupiter? What makes the moon accelerate towards the Earth? What makes an apple accelerate towards the Earth? That's the definition of the word, gravity. Once you connect the falling towards the sun, with the pull of gravity, you start to think about the weight of the earth. The weight should be proportional to the mass. So the driver of acceleration towards the sun should be proportional to the mass of the earth. The mass of the earth was conspicuously absent from the equations, but we can remedy that by multiplication, as we just did, and we suspect that the right hand side really is the force of gravity.

If we call both sides $F$, we get
$$F_x = m\ddot{x}$$
$$F_y = m\ddot{y}$$
$$F_z = m\ddot{z}$$
or simply $$F=ma$$
which is *the second law of motion*.
Applying this second law of motion to elastic forces,
you can readily solve them and confirm their validity for those as well.
This impels you to conjecture them to be true for all forces.

This is too facile, though.
There are, *a priori*, three definitions of mass:

- The source of the pull exerted by that body on other bodies, also called its gravitational mass
- The source of the pull exerted on that body by other bodies, i.e. its weight
- The resistance of that body to changes in its motion, also called its inertial mass

At this point, the hint is strong that all three kinds of masses are the same.
For the two kinds of *pull* and *be pulled* masses enter completely symmetrically in the expression
$$\frac{GMm}{r^2},$$

But it only makes sense to find the *inertial mass* multiplying the acceleration,
so it must also be the same.

That equality finally achieves some clarity with the equivalence principle. Fuller clarity arrives with the general relativistic calculation, that inertia is itself a gravitational phenomenon, a consequence of the gravitational equations alone.

The inverse square law gets some clarity from quantum field theory, where exchange particles mediate interactions. Particles of $0$ mass can travel out to mediate, but their density drops as the surface area of the sphere increases. The surface area of a sphere, recall, is $4\pi r^2$. Of course, a believable quantum theory of gravity has yet to be found, though of course the inertial terms in quantum expressions are presumably in fact gravitational.

Returning now to the original equations $$\ddot{x} = - GM \frac{x}{r^3}$$ $$\ddot{y} = - GM \frac{y}{r^3}$$ $$\ddot{z} = - GM \frac{z}{r^3},$$ we reiterate that these equations involve only the location of the orbiting body $x$, $y$, $z$, the mass of the sun $M$, or whatever body lies at the focus, and a universal constant $G$, which doesn't depend on the sun, or the planet. Thus the equation can be thought of as describing a geometric property of that location in space; that the focal mass has somehow affected the geometry at that orbital location in such a way as to cause an acceleration.

We can imagine an experiment where we enclose particles in a spherical shell, and let the sphere fall freely from a sufficient height that there is negligible friction from the atmosphere. Alternately, we could imagine doing the experiment over the moon, which has no atmosphere. What would a camera inside the sphere observing the free floating particles see?

A particle sitting at the center of the shell would continue floating at that central point—it would fall exactly as fast as the shell surrounding it, though it does involve a calculation to show that. It turns out that the extra pull on the bottom of the shell, where $r$ is smaller since it's closer to the earth (or moon), is exactly balanced by the lesser pull on the top of the shell, where $r$ is larger by the diameter of the shell. So the upshot is that the shell falls as if it were a tiny sphere with all of its mass at the center.

The same cannot be said for the other floating particles. Since they are not connected to one another, each falls independently. Consider one floating above the center. Since $z/r^3$ decreases with increasing $z$, it is smaller above the center of the shell, so a particle floating there is accelerating towards the ground more slowly that the particle at the center; it will appear to be accelerating up, away from the center. By the same token, a particle below the center will appear to accelerate down away from the center.
A particle to one side of the center will move towards the center, because straight down

from the one is not exactly parallel to straight down

from the other—both point towards the center of the body below, which is only a finite distance away.

To understand this in geometric terms, think about curvature. Picture two ants walking on a ball. They start out from two nearby points, and walk in the same direction, away from the arc that joins their starting points. (If the ball is a globe, picture them starting on the equator, and walking due north, or south.) While they start out parallel, they keep getting closer and closer to each other, eventually meeting at the pole. It is as though some force were pulling them together, but that force is an illusion created by the geometry of the ball.

The full geometric analysis of gravity is of course the general theory of relativity.

I'm Andrew Winkler. I hope you've enjoyed exploring these ideas with me. Thanks for your time. If you've got any questions, send an email to 4af502d7148512d4fee9@cloudmailin.net.

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