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Orbit Equation:

Using the knowledge from Appendix B, we can quite easily deduce the orbit equation:

\[\ddot{r}-\frac{h^2}{r^3}=\frac{F_r(r)}{m}\]

I’ll leave this as an exercise for the reader.

Hint 1: We're dealing with Central Force, only radial force matters

Hint 2: Use Newton's Second Law

I enjoy the physics of small perbuations, especially in classical mechanics. First, assume a small perbuation around a circle with radius \(a\) i.e. \(r(t)=a+\rho(t)\) with \(\rho(0) \ll a\). substitute into our orbit equation we have:

\[\ddot{\rho}-\frac{h^2}{(a+\rho)^3}=\frac{F_r(a+\rho)}{m}\]

Here we’ll use binomial expansion for negative coefficients. For more details view the wikipedia page for Negative Binomial Distribution

However there’s a nice trick by differentiating the taylor series.

\[\ddot{\rho}-\frac{h^2}{a^3}\left(1-3 \frac{\rho}{a}\right)=\frac{F_r(a)}{m}+\frac{d F_r(a)}{d r} \frac{\rho}{m}\]

Further rearranging will get you:

\[\ddot{\rho}+\left[\frac{3 h^2}{a^4}-\frac{1}{m} \frac{d F_r(a)}{d r}\right] \rho=0\] 
Right-Hand-Side surprisingly reduces to 0, can you spot why?

I don’t know if you have done Differential Equations before. Above describes a simple harmonic oscillator with period \(\omega\), in the form of \(\ddot{\rho}+{\omega}^2 \rho=0\). Now we’ll ensure \(\omega^2>0\), s.t. \(\omega\) is real, i.e.

\[\frac{3 h^2}{a^4}-\frac{1}{m} \frac{d F_r(a)}{d r}>0\]

Hence if the above stands we have:

\[\omega^2 \equiv \frac{3 h^2}{a^4}-\frac{1}{m} \frac{d F_r(a)}{d r}\]

This describes a simple harmonic motion with \(\rho(t)=A \cos (\omega t+\phi)\).

It will look like this:

We have used simple math to analyse the orbit equation! How cool is that!