How to calculate the energy in a hyperbolic orbit?

Question

I'm recently reading a book about rocket science which involved orbital mechanics. I know that in an elliptical orbit, the energy $ E=-\frac{GMm}{2a}$, and therefore can get the vis-viva equation: $ v^2=GM(\frac{2}{r}-\frac{1}{a}) $. But now I'm facing a hyperbolic orbit, and its energy and vis-viva equation are a bit different from the previous one. In this case, $ E=\frac{GMm}{2a}$ and the vis-viva equation becomes $v^2 = GM(\frac{2}{r}+\frac{1}{a}) $.

I want to know where the formula $ E=\frac{GMm}{2a}$ comes from. The book I'm reading, Introduction to Rocket Science and Engineering by Travis S Taylor, doesn't provide a satisfying explanation. Here's a screenshot: And I searched through the web, asked ChatGPT and still got no answers. I read https://users.physics.ox.ac.uk/~harnew/lectures/lecture20-mechanics-handout.pdf and find it hard to understand its contents(especially page 4). Understanding this involved angular momentum and vector cross products which I'm not familiar with(I can learn though!). Also I have a limited knowledge on calculus, so please don't get too many derivatives or integrals involved in the calculation.

I'm puzzled about how to calculate the angular momentum of the spacecraft when it's very far away from focus. Given the conservative energy and the angular momentum stays constant, I still failed combining these formulas together to get the desired $ E=\frac{GMm}{2a}$.

Can someone give me a not-so-hard explanation? I will be very thankful!

Answer 1

Let me derive the result for you here. When the rocket is very far away, the rocket's trajectory approaches the asymptote of the hyperbola. We can therefore draw a triangle whose vertices are the sun, the center of the hyperbola, and the rocket. If $P$ is the position of the rocket, $\theta$ is the included angle of $P$, $C$ is the center of the hyperbola, $F$ is the position of the sun, and $PF=r$, then $CF=c$, the focal length, by definition, as shown in the triangle below.

One of the properties of hyperbolas is that the asymptote has slope $\frac{b}{a}$, where $b$ is the semiminor axis, and $a$ is the semimajor axis. Therefore, $\tan{C}=\frac{b}{a}$. But, by the law of sines, $\frac{\sin{C}}{r}=\frac{\sin{\theta}}{c}$. If $\tan{C}=\frac{b}{a}$, then we can draw a right triangle with legs $a$ and $b$ to find $\sin{C}$. Since $a^2+b^2=c^2$ is another property of hyperbolas, we can see that the hypotenuse must be $c$ in this triangle.

Hence, we can see that $\sin{C}=\frac{b}{c}$. Plugging this back into $\frac{\sin{C}}{r}=\frac{\sin{\theta}}{c}$ yields $\sin{\theta}=\frac{b}{r}$.

Now, if $L$ is the conserved angular momentum of the rocket, then the speed of the rocket must satisfy $L=rp\sin{\theta}=rmv\frac{b}{r}=mvb$. Therefore, $v=\frac{L}{mb}$ in this case.

The energy of the rocket, $E$, at any time satisfies $E=\frac{1}{2}mv^2-\frac{GMm}{r}$. We have just found $v$, for when $r\to\infty$, so we can write $E=\frac{1}{2}m(\frac{L}{mb})^2-\frac{GMm}{\infty}=\frac{1}{2}m\frac{L^2}{m^2b^2}=\frac{L^2}{2mb^2}$.

Now, let's consider when the rocket is closest to the sun. $P$ is now at the vertex of the hyperbola, so the distance from the focus is now $r=c-a$. At the vertex, the velocity of the rocket is purely perpendicular to a line connecting the rocket to the sun, as depicted below.

Thus, all of the rocket's velocity will contribute to the angular momentum. Therefore, since angular momentum is conserved, $L=mvr$ implies $v=\frac{L}{mr}=\frac{L}{m(c-a)}$ in this case.

Plugging this back into $E=\frac{1}{2}mv^2-\frac{GMm}{r}$ yields $E=\frac{1}{2}m(\frac{L}{mr})^2-\frac{GMm}{r}=\frac{L^2}{2mr^2}-\frac{GMm}{r}=\frac{L^2}{2m(c-a)^2}-\frac{GMm}{c-a}$.

Since we found earlier that $E=\frac{L^2}{2mb^2}$, we can write the equation as $L^2=2mb^2E$, and plug it back into $E=\frac{L^2}{2m(c-a)^2}-\frac{GMm}{c-a}$ to get $E=\frac{2mb^2E}{2m(c-a)^2}-\frac{GMm}{c-a}=\frac{b^2E}{(c-a)^2}-\frac{GMm}{c-a}$

Isolating $E$ we get: $E-\frac{b^2E}{(c-a)^2}=-\frac{GMm}{c-a}$

Factoring E out: $E(1-\frac{b^2}{(c-a)^2})=-\frac{GMm}{c-a}$

We can simplify the left hand side as such:

$E(1-\frac{b^2}{(c-a)^2})=E(\frac{(c-a)^2-b^2}{(c-a)^2})=E(\frac{c^2+a^2-2ac-b^2}{(c-a)^2})= E(\frac{a^2+a^2-2ac}{(c-a)^2})=E(\frac{2a^2-2ac}{(c-a)^2})=E(\frac{-2a}{c-a})$, where we have used $c^2=a^2+b^2$ yet again.

Now we have the equation $E(\frac{-2a}{c-a})=-\frac{GMm}{c-a}$, which simplifies to $\boxed{E=\frac{GMm}{2a}}$, our final answer.

I hope this derivation helps!