[Physics] Shape of an inverse cube orbit

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If I have a particle orbiting a central force $$F=-k/r^3$$ what is the shape of the orbit (the radius as a function of the angle)?

Best Answer

Using Lagrange's equation

$$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{-m r^2}{l^2}F\left(r\right)$$

And plugging in $$F\left(r\right)=\frac{-k}{r^3}$$

We get $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}=\frac{k m}{l^2 r}$$

Which simplifies to $$\frac{d^2}{dt^2}\left(\frac{1}{r}\right)+\frac{1}{r}\left(1-\frac{k m}{l^2}\right)=0$$

Assuming $$r\left(0\right)=r_p, r'\left(0\right)=0$$

We can solve equation (3) for r which gives us $$r\left(\theta\right)=r_p sec\left(\theta\sqrt{1-\frac{km}{l^2}}\right)$$

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