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

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?

• $$5+4\sqrt{7}\cos\left({1\over 3}\tan^{-1}{\sqrt{3}\over 37}\right)$$ Feb 26 '15 at 2:39
• Thanks Mark! With that arctan in there I bet there's some cool geometrical drawings behind that number. Feb 26 '15 at 17:15
• Not that I'm aware of. It is an analytical solution to a cubic equation. I had to do some work on it to get rid of the imaginary part that cancels, and ended up with the trig functions you see. The $1\over 3$ in front of the arctangent is effectively part of a cube root. Feb 26 '15 at 17:31
– Paul
Oct 6 '17 at 18:33
• @uhoh At that time (Feb 2015), I edited the Hohmann Transfer Orbit entry in Wikipedia with that information, in order to preserve it for posterity. However I have also just added it here, with some of the derivation. Oct 1 '18 at 0:07

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

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:

$$x^3−15x^2−9x−1=0$$

• fyi you've been pinged
– uhoh
Oct 3 '18 at 15:51