# Geostationary orbits

### From Math

I was thinking about gravity and space elevators today. They still confuse me. So I decided to see if I could work out the height of a geostationary orbit in my head, while walking to school. This despite the fact that I don't remember any of the classical mechanics stuff I learned in college, but with the powerful ally of ... dimensional analysis, which is approximately the coolest thing ever. I got it badly wrong, and later figured out why.

## What we know

I chose to start with 10 and 6.4 (instead of, say, 9.8 and 6) because I thought I would want to take the square root of their product.

$C$ is the characteristic time of the Earth's rotation (how long it takes a point on the surface to travel the length of the radius). In my experience, the characteristic time (not the period) tends to be the quantity that gives you the right answer in dimensional analysis.

## The simple answer

We have too much information for a good dimensional analysis (too many ways of combining our quantities to get the right units). But there does seem to be a natural, straightforward way to do it.

$\sqrt{gr}=8\mathrm{k}\mathrm{m}/\mathrm{s}$ is a speed. This should have something to do with something. Multiply by 14ks to get 112Mm. This seems way too high, though, so I should think this through more carefully.

I wonder if $8\mathrm{k}\mathrm{m}/\mathrm{s}$ is orbital velocity or escape velocity and the dimensional analysis discovered that by accident.

## The right answer

If we're moving away from the surface of the Earth, we have to respect our knowledge that gravity goes as ${r}^{2}$ to make use of $g$, so we need to construct $K=g{r}^{2}$. Translating the prefixes back to km gives us one extra 1000, so we have $400,000\mathrm{k}{\mathrm{m}}^{3}/{\mathrm{s}}^{2}$.

The radius of the earth doesn't really directly affect the orbit. In fact, I used it only because I know it, and I don't know the mass of the Earth or the gravitational constant $G$. This means that the right answer must be made from $K$ and $C$, which means in turn that there's only one way to do it: ${\left(K{C}^{2}\right)}^{1/3}$.

This is a pain to calculate mentally.

The computer claims it's 43Mm, which still seems way too high, so what's up?

Checking wikipedia, it turns out that's what's *up* is the geostationary orbit, which really is 42Mm above the center of the Earth (or 36Mm above your head). Which is wild, but score one at least for dimensional analysis.