Central force law An object has an orbit in polar coordinates as $r(\theta) = a\theta^2$ (where $a$ is constant).
Assuming the central force is directed towards the origin $r=0$, how can I know which central force law lead to such an orbit? And how to find $r$ and $\theta$ as function of time?
 A: I don't want to give away the answer directly. So I will provide some hints.
A central force in polar coordinates has to be of the form: $$\vec{F} = m\vec{a} = m(\ddot r - r \dot \theta^2)\hat r$$
Now try to mess around with your $r(\theta) = a\theta^2 $
I believe you need to specify $\dot \theta$ in order to solve the full equation of motion. So pick for your self. A linear equation $\theta = kt $ may be a good starting point.
Good luck!
A: Perhaps I can help.
For any object in orbit, the Earth exerts a force on the object and the object exerts a force on the Earth.
So we know from $F=ma$ that:
$$∑ F_θ=ma_θ$$
Using polar coordinates {r,θ}, this equation becomes:
$$ 0=2\left(\frac{dr}{dt}\right)\cdot \left(\frac{dθ}{dt}\right)+r\left(\frac{d^2θ}{dt^2}\right) \implies \frac{1}{r} \left(r\left(2\left(\frac{dr}{dt}\right)\cdot \left(\frac{dθ}{dt}\right)+r\left(\frac{d^2θ}{dt^2}\right)\right)\right) $$
$$ 0=\frac{1}{r} \left(2r\left(\frac{dr}{dt}\right)\cdot \left(\frac{dθ}{dt}\right)+r^2\left(\frac{d^2θ}{dt^2}\right)\right) \implies \frac{1}{r} \left(\frac{d}{dt}\left(r^2\dfrac{dθ}{dt}\right)\right)=0$$
Integrating yields: $$r^2\dfrac{dθ}{dt}=h$$ where $h$ is the constant of integration
Therefore:$$\dfrac{dθ}{dt}=\dfrac{h}{r^2}$$
This proves Kepler's Second Law that the aerial velocity of a particle subjected to central-force motion is constant.
Remember that this also accounts for orbital eccentricity.
I hope this answers your question.
