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?


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.

| cite | improve this answer | |
  • $\begingroup$ 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 '17 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. $\endgroup$ – Fawzy Hegab Mar 13 '18 at 9:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.