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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.

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    $\begingroup$ 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

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

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