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In what follows, dot notation is used when deriving with respect to time \(t\), and prime notation is used when deriving with respect to \(\theta\).

In polar coordinates, we can write:

$$ \mathbf{v} = \dot{r}\mathbf{e_r} + r\dot{\theta} \mathbf{ e_\theta} $$$$ \mathbf{a} = (\ddot{r}-r\dot{\theta}^2) \mathbf{e_r} + (2\dot{r}\dot{\theta} + r\ddot{\theta})\mathbf{e_\theta} $$

Newton’s second law states that:

$$ m\mathbf{a} = -\frac{GMm}{r^2}\mathbf{e_r} $$

therefore:

$$ r^2\dot{\theta} = C $$$$ \ddot{r}-r\dot{\theta}^2 = -\frac{GM}{r^2} = -\frac{\mu}{r^2} $$

let \(u = \frac{1}{r}\), we have:

$$ r = \frac{1}{u} $$$$ \dot{r} = -\frac{u'}{u^2}\dot{\theta} $$$$ \ddot{r} = -\frac{u'}{u^2}\ddot{\theta} + (\frac{2u'^2}{u^3}-\frac{u''}{u^2})\dot{\theta}^2 $$

Let’s replace:

$$ \dot{\theta} = C u^2 $$$$ \ddot{\theta} = 2 C u' u \dot{\theta} $$$$ -\frac{u'}{u^2}\ddot{\theta} + (\frac{2u'^2}{u^3}-\frac{u''}{u^2})\dot{\theta}^2 - \frac{\dot{\theta}^2}{u} = -\mu u^2 $$

Let’s replace \(\dot{\theta}\) and \(\ddot{\theta}\):

$$ -\frac{2u'^2C\dot{\theta}}{u} + \frac{2u'^2C\dot{\theta}}{u} -u''C^2u^2 - C^2u^3 = -\mu u^2 $$$$ u'' + u = \frac{\mu}{C^2} $$

From this we find that:

$$ u(\theta) = \frac{\mu}{C^2} + A \cos(\theta) + B \sin(\theta) $$

If we arbitrarily set the minimum distance from the massive object to be reached at \(\theta=0\), then:

$$ u'(0) = B = 0 $$

So only the term in \(\cos\) remains. Defining \(e\) such that \(A = \frac{\mu}{C^2}e\), we finally get:

$$ u(\theta) = \frac{\mu}{C^2}(1 + e \cos(\theta)) $$

And setting \(p=\frac{C^2}{\mu}\), we retrieve the well-known result:

$$ r(\theta) = \frac{p}{1 + e\cos{\theta}} $$

Depending on \(e\), the trajectory can be elliptic, parabolic or hyperbolic.