I read Feynman's Lost lecture a while ago. In that spirit, I'm trying to do a more simplified version of coming up with the inverse square relation.

Using Kepler's 2nd,

$ A_1 = \frac{1}{2}r_1^2\theta_1 $

$ A_2 = \frac{1}{2}r_2^2\theta_2 $

$ A_1 = \frac{1}{2}r_1^2\omega_1 t $

$ A_2 = \frac{1}{2}r_2^2\omega_2 t$

$ m\omega_1 A_1 = \frac{1}{2}mr_1^2\omega_1^2 $

$ m\omega_2A_2 = \frac{1}{2}mr_2^2\omega_2^2 $

$ mv_1 A_1 = \frac{1}{2}r_1^2F_1 $

$ mv_2A_2 = \frac{1}{2}r_2^2F_2 $

Dividing one by the other, I get

$\frac{F_1}{F_2} = \frac{v_1}{v_2}\frac{r_2^2}{r_1^2}$

which is almost there, except for the $\frac{v_1}{v_2}$

Any suggestions on how to proceed from here?


1 Answer 1


Consider a circular orbit (Kepler's first law tells us this is possible, as circles are particular cases of ellipses). By Kepler's second law, the speed $v$ is constant along the orbit. We can obtain its dependence on $r$ using Kepler's third law: $T^2\propto r^3$. The result is $v^2\propto 1/r$.

For the orbit to be circular, the force should satisfy \begin{equation} F(r)=m\frac{v^2}{r}\propto\frac{1}{r^2}. \end{equation}

We can also find the direction of the force from Kepler's laws! We work in two dimensions because Kepler's first law tell us that the orbits stay in a plane. The acceleration in radial coordinates is \begin{equation} \vec{a} = (\ddot{r}-r\dot{\theta}^2)\hat{r} + (r\ddot{\theta}+2\dot{r}\dot{\theta})\hat{\theta}. \end{equation} Notice that the $\hat{\theta}$ component of the acceleration is just $\frac{1}{r}\frac{d}{dt}(r^2\dot{\theta})$, and that $r^2\dot{\theta}$ is the areal velocity, which is constant by Kepler's second law. Therefore, the acceleration has the direction of $\hat{r}$, and so does the force. The latter should then be of the form \begin{equation} \vec{F}(\vec{r})=-F(r)\hat{r}, \end{equation} where the dependence only on $r$ and not $\theta$ is a consequence of the isotropy of space.

  • $\begingroup$ I have gone through this way & the differential equation it leads to in our mechanics course. All I want to ask is that , is there any way to solve it in Cartesian ( x , y ) without polar ? $\endgroup$
    – Shashaank
    Feb 3, 2017 at 21:20
  • $\begingroup$ That is a great answer. I have always wondered how did Newton come up with his formula. Now, one can see a possible way he may used to get it. $\endgroup$ Mar 13, 2018 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.