Consider a particle of mass $m$ moving in a circular path in the field $F(r)=\frac{-k}{r^2}$. I would like to know how changes in values of $k$ alter the path of the particle. I'm examining the situation in polar coordinates. Newton's second law gives two equations $$m(\ddot{r}-r\dot{\theta}^2)=-k/r^2,$$ $$m(2\dot{r} + r\ddot{\theta})=0$$

where the second equation equals zero, since there is no force acting on the angular direction. The second equation implies $(r^2 \dot{\theta})'=0$, and therefore $r^2\dot{\theta} = h$, where $h$ is a constant.

By introducing a new function $u(\theta(t))=1/r(t)$, these first equation can be solved for $r(\theta)$. The process is long, I can give details but the path of the particle is described by $$r(\theta) = \frac{1}{(1/R - k/h^2m)\cos \theta+k/h^2m}$$

and to solve for this, I used the initial conditions $\theta(0)=0$, and $r(0)=R$, where $R$ is the radius of the circular path. Now I would like to know how the geometric nature ( whether it is an ellipse, parabola etc...) of the path is dependent on the force constant $k$. Do I need $\theta(t)$ as well? I have computed it, but I assumed it is only needed to solve the constants from the initial conditions.

Edit : I realized $u$ is a function of $\theta(t)$, so the argument $t$ is changed to the dependent variable $\theta$.

I changed my initial conditions to : $\dot{\theta}(t=0)=0$ and $\theta(t=0)=0$. This gives another version for $r(\theta)$, $$r(\theta)=\frac{1}{\frac{k}{h^2 m}(\cos\theta+1)}$$

which I believe describes a parabola. But this contradicts the fact that the initial path is circular.

  • 1
    $\begingroup$ Find the relation between $r$ and $\theta$. Given that you'll see what curve the trajectory corresponds to $\endgroup$
    – nwolijin
    Mar 29, 2021 at 11:07

1 Answer 1


In your first equation, if the orbit is a circle, the force causes the centripetal acceleration and there is no change in the radius.


Your Answer

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

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