# Which of Kepler's laws gave Newton the idea about gravitational force and how it is related to distance?

Apr 29, 2017

The inverse-square idea can be got at pretty easily from Kepler's 3rd Law , if one assumes a circular orbit.

So if we say that there is some radial force $f = f \left(r , \ldots .\right)$ pulling a planet in our system back to the Sun and so keeping it in orbit, and we balance the forces along that radial direction, then:

$f = m {\omega}^{2} r \triangle$

We know from Kepler's 3rd third law : ${T}^{2} \propto {r}^{3}$ or ${T}^{2} = \alpha {r}^{3}$.

We also know that period, $T$, and angular speed, $\omega$, are related for circular motion as: $T = \frac{2 \pi}{\omega}$, so $\triangle$ becomes:

$f = m \frac{\dot{4 {\pi}^{2}}}{T} ^ 2 r = m \frac{\dot{4 {\pi}^{2}}}{\alpha {r}^{3}} r = \frac{\beta}{r} ^ 2$, ie $f \propto \frac{1}{r} ^ 2$

That of course completely ignores Kepler's 1st Law , namely that planetary orbits are elliptical . The math becomes more turgid with ellipses, and certainly would have been for Newton as he was making it all up it as he went along.

Using post-Newton maths, and actually assuming (naughty) an inverse law $m a t h b f F = - \frac{\mu m}{r} ^ 2 \setminus m a t h b f {e}_{r} , \text{ with } \mu = G M$, the Lagrangian in a polar co-ordinate plane tells us that:

• $\ddot{r} - {\dot{\theta}}^{2} r = - \frac{\mu}{r} ^ 2 q \quad \star$

• $\frac{d}{\mathrm{dt}} \left(m {r}^{2} \dot{\theta}\right) = \text{ const.} q \quad \square$

$\square$ is just a statement that angular momentum $m a t h b f L$ is preserved, and so we can say that $| m a t h b f L | = m {r}^{2} \dot{\theta} \implies \dot{\theta} = \frac{L}{m {r}^{2}} = \text{ const.}$; and we then pop that into $\star$ to get:

$\ddot{r} = - \frac{\mu}{r} ^ 2 + {L}^{2} / \left({m}^{2} {r}^{3}\right) q \quad \circ$

That's a non-linear DE; but the strategy I have borrowed - namely, we want a polar co-ordinate expression $r = r \left(\theta\right)$ which describes an ellipse - drives the solution, and that involves eliminating $t$ from the equations. (I found this approach on a pretty ancient site that I will link at the bottom, I needed to translate it into more modern language.)

First we introduce a new fictional variable, $u$ on terms that:

$r = r \left(u \left(t\right)\right) = \frac{1}{u \left(t\right)}$ and $u = u \left(\theta \left(t\right)\right)$

We can do that, this is physics not maths, and we assume everything is sufficiently smooth, as it is in the physical world. In maths terms, we are, I think, assuming that these inter-related functions are invertible, so the chain rule is operating on full-power.

What's truly mind-blowing is that, thanks to Newton et al, we are now just pushing letters about according to certain rules, because it follows from the chain rule that:

$\textcolor{red}{\frac{\mathrm{dr}}{\mathrm{dt}}} = \frac{\mathrm{dr}}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dt}}$

$= - \frac{1}{u} ^ 2 \cdot \frac{\mathrm{du}}{d \theta} \textcolor{b l u e}{\frac{d \theta}{\mathrm{dt}}}$

$= - \frac{1}{u} ^ 2 \cdot \frac{\mathrm{du}}{d \theta} \textcolor{b l u e}{\frac{L}{m {r}^{2}}} = \textcolor{red}{- \frac{L}{m} \frac{\mathrm{du}}{d \theta}}$

It then follows that:

$\frac{{d}^{2} r}{d {t}^{2}} = - \frac{L}{m} \frac{d}{\mathrm{dt}} \left(\frac{\mathrm{du}}{d \theta}\right)$

$= - \frac{L}{m} \frac{{d}^{2} u}{d {\theta}^{2}} \frac{d \theta}{\mathrm{dt}}$

$= - \frac{L}{m} \frac{{d}^{2} u}{d {\theta}^{2}} \frac{L}{m {r}^{2}}$

$= - \frac{{L}^{2} {u}^{2}}{m} ^ 2 \frac{{d}^{2} u}{d {\theta}^{2}}$

Then $\circ$ becomes:

$- \frac{{L}^{2} {u}^{2}}{m} ^ 2 \frac{{d}^{2} u}{d {\theta}^{2}} = - \mu {u}^{2} + {L}^{2} / \left({m}^{2}\right) {u}^{3}$

$\implies \frac{{d}^{2} u}{d {\theta}^{2}} + u = \frac{{m}^{2} \mu}{{L}^{2}}$

That's now trivial, it solves as:

$u \left(\theta\right) = A \cos \left(\theta - \phi\right) + \frac{{m}^{2} \mu}{{L}^{2}}$

Or

$r \left(\theta\right) = \frac{1}{A \cos \left(\theta - \phi\right) + \frac{{m}^{2} \mu}{{L}^{2}}}$

To finish this, from Wiki: "in the slightly more general case of an ellipse with one focus at the origin and the other focus at angular coordinate $\phi$:

$r \left(\theta\right) = \frac{a \left(1 - {e}^{2}\right)}{1 - e \setminus \cos \left(\setminus \theta - \setminus \phi\right)}$"

We can match that up and then we are guaranteed at least an elliptical orbit using specifically the inverse square law ..... and that orbit may even in the oddest of cases be circular.

The fact that elliptical orbits can keep the inverse square law happy is seminal.

And this is the brilliant resource I found, though I find it really hard to follow:

http://galileo.phys.virginia.edu/classes/152.mf1i.spring02/KeplersLaws.htm