# Determining the nature of orbit from the path function of a particle in CF-motion

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.

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