Orbital energy in terms of semi-major axis

In order to compute the length of ${\bf L}$, we need to prove the following formula for the energy.

$$E = \frac{1}{2} v^2 - \frac{\mu}{r} = -\frac{\mu}{2a},$$

where $a$ is the semi-major axis. In particular, this provides a simple way to compute semi-major axis. We use polar coordinates $(r, \theta)$ in the plane of the orbit. As before we have that

$$v^2 = \dot{r}^2+r^2\dot{\theta}^2.$$

Now, consider the coordinate transformation

$$\begin{eqnarray*} x & = & r \cos \theta \\ y & = & r \sin \theta \end{eqnarray*}$$

It follows that

$$\begin{eqnarray*} \dot{x} & = & \dot{r} \cos \theta - r \dot{\theta} \sin{\thet... ...dot{y} & = & \dot{r} \sin \theta + r \dot{\theta} \cos{\theta} \end{eqnarray*}$$

which can be written as

$$\left( \begin{array}{c} \dot{x} \\ \dot{y} \end{array} ... ...in{array}{c} \dot{r} \\ r \dot{\theta} \end{array} \right)$$

This is the formula for rotation by the angle $\theta$, which preserves the length of vectors. Hence, the length of the left hand side $v$ is the same as the length on the right, i.e.

$$v^2 = \dot{x}^2+\dot{y}^2 = \dot{r}^2 + r^2 \dot{\theta}^2.$$

We return now to dealing with energy. We learned that the absolute value of the angular momentum momentum $A$ can be expressed as

$$A= r^2\dot{\theta}$$

and that

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

where $p$ is the parameter of the ellipse. We also have

$$A = \sqrt{\mu a (1-e^2)}, \;\;\; p = a(1-e^2), \;\;\; h^2 = p\mu.$$

From these, we get

$$\dot{\theta} = \frac{A}{r^2},\;\;\; \dot{\theta}^2 = \frac{A^2}{r^4}.\;\;\;$$

Now we compute the energy in polar coordinates. Since the energy is a constant, we compute its value at any time. We choose the time of pericenter passage for convenience. Then $\dot{r} = 0$, which can be proved by differentiating the polar equation for the orbit, and we also have from the same equation (with $\theta = 0$)

$$r = \frac{p}{1+e}.$$

It follows that

$$v^2 = r^2 \dot{\theta}^2 = \frac{h^2}{r^2} = \frac{h^2}{p^2}(1+e)^2,$$

and

$$v^2 = \frac{\mu p}{p^2}(1+e)^2 = \frac{\mu}{p}(1+e)^2.$$

Hence

$$\frac{1}{2}v^2 - \frac{\mu}{r} = \frac{\mu}{2p}(1+e)^2 - ... ... - 2(1+e)\right] = \frac{\mu}{2p}(e^2-1) = -\frac{\mu}{2a},$$

and the desired formula for the energy is shown.

Now we return to the computation of the length of ${\bf L}$. From the energy relationship above, we have that

$$v^2 = 2\left(\frac{\mu}{r}-\frac{\mu}{2a}\right),$$

and we also know that at pericenter passage

$$r = a(1-e).$$

Hence

$$rv^2 = 2\mu- \frac{\mu r}{a} = \mu\left( 2 - (1-e)\right) = \mu(1+e).$$

Finally,

$${\bf L}= (rv^2-\mu, 0,0 ) = (\mu e, 0,0)$$

Therefore, the length of ${\bf L}$ is $\mu e$.

Notice that this result can be useful when computing orbital elements. ${\bf L}$ provides both the direction of pericenter and the eccentricity.