From Newton's 2nd law to orbit trajectory

Derivation of Orbital Trajectories Link to heading

Orbital mechanics is a fundamental aspect of astrophysics and aerospace engineering, dealing with the motion of objects in space under the influence of gravitational forces. One of the cornerstones of this field is Newton’s second law of motion, which provides a framework for understanding how forces affect the motion of objects. By applying Newton’s second law to a two-body system, we can derive the equations that describe the trajectory of an object in orbit. This derivation not only highlights the elegance of classical mechanics but also lays the groundwork for more advanced studies in celestial mechanics and space exploration.

Mathematical Formulation Link to heading

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.