~15.5817 -- Is there a name for this number?

Question

I've been playing with Hohmann transfers from one circular orbit to another.

I've been calling the radius of the departure orbit 1 and the radius of the arrival orbit $r$ with $r>1$.

There are two burns:

1. Departure burn to leave circular departure orbit and enter Hohmann transfer orbit.
2. Arrival burn to exit Hohmann transfer and enter circular arrival orbit (aka a circularize burn)

The total $\Delta V$ is the sum of the $\Delta V$s these two burns take.

As $r$ increases, total $\Delta V$ increases up until a certain point. Then total $\Delta V$ starts falling!

If my calculus weren't so rusty I'd try to solve for $r$ where $f'(r)=0$. But my brute force numerical efforts seem to indicate at the top of this hill $r$ is roughly 15.5817.

Is there a name for this number?

Answer 1

For reference, that number is:

$$5+4\sqrt 7 \cos\left({1\over 3}\tan^{-1}{\sqrt 3\over37}\right)$$

It is the positive root of:

$$x^3−15x^2−9x−1=0$$

If we take the equation for the total $\Delta V$ of a Hohmann transfer between two circular orbits, and express it in terms of the ratio of the radius of the larger to the radius of the smaller orbit, $x$, and without loss of generality setting $\mu$ and the smaller radius both to $1$, we get:

$$\sqrt{2x\over x+1}+\sqrt{1\over x}-\sqrt{2\over x\left(x+1\right)}-1$$

Taking the derivative with respect to $x$, we get:

$$x^{-{3\over 2}}\left({\frac{3 x+1}{\sqrt{2} (x+1)^{3/2}}-\frac{1}{2}}\right)$$

Setting that equal to $0$ to find the extremum:

$$6 x+2=\sqrt{2}(x+1)^{3/2}$$

Squaring both sides:

$$36 x^2+24 x+4=2 x^3+6 x^2+6 x+2$$

And finally, simplifying:

$$x^3−15x^2−9x−1=0$$

Answer 2

I don't think it has a particular name other than "worst-case Hohmann transfer".