“Vinnie, those nifty-looking transfer orbits that Hohmann invented but didn’t get to patent — you left something out.”

“What’s that, Sy?”

“The geometry looks lovely — a rocket takes off tangent to its orbit around one planet or something and inserts along a tangent to an orbit around something else. Very smooth and I can see how that routing avoids having to spend fuel to turn corners. But that ignores speeds.”

“What difference does that make?”

“It makes a difference whether or not you can get into the orbit you’re aiming for. Any orbit is a trade-off between gravity’s pull and the orbiter’s kinetic energy. Assuming you’re going for a circular orbit, there’s a strict relationship between your final height and your approach speed when you’re finally flying on the horizontal. You don’t want to come in too fast or too slow.”

“First thing I learned in pilot school. But that relationship’s an equation, ain’t it?”

“A couple, actually, but they’re simple. Let’s back into the problem. Say your mission is to put a communications relay satellite into lunastationary orbit around the Moon—”

“Lunastationary?”

“Like geostationary, but with the Moon. The satellite’s supposed to hover permanently above one spot on the Moon’s equator, so its orbital period has to equal the Moon’s ‘day,’ <*pulling out Old Reliable, tapping*> which is 27.322 days. Your satellite must loop around the Moon in exactly that much time. Either it’s scooting at low altitude or it’s ambling along further up. If we knew the speed we could find the radius, and if we knew the radius we could find the speed. We need some math.”

“I knew it. You’re gonna throw calculus at me.”

“Relax, Vinnie, it’s only algebra and we’re only going to combine two formulas and you already know one of them. The one you don’t know connects the speed, which I’m calling *v*, with the radius, *R*. They’re tied together by the Moon’s mass, *M* and Newton’s gravitational constant *G*. The formula is *v ^{2}=2G×M/R*. You can handle that, right?”

“Lessee … that says if I either double the mass or cut the distance by two, the speed has to be four times larger. Makes sense ’cause that’s about being in a deeper gravity well or getting closer in. Am I on track?”

“Absolutely. Next formula is the one you know, the circumference of a circle or in this case, the distance around that orbit.”

“That’s easy, *2πR*.”

“And that’s also speed times the time, *T* so I’ll set those equal. <*tapping on Old Reliable*> Okay, the first formula says *v ^{2}* so I square the circumference equation and solve that for

*v*. You still with me?”

^{2}“You’re gonna set those two *v*-squareds equal, I suppose.”

“You’re still on track. Yup and then I gather the *R*s on one side and everything else on the other. That gives me something in *R ^{3}* but that’s okay. Plug in all the numbers, take the cube root and we get that you need to position that satellite 111 megameters out from the Moon’s center, flying at 296 meters per second. Think you can manage that?”

“Given the right equipment, sure. Seventy thousand miles out from the Moon … pretty far.”

“It’s about ¾ of the way to the Moon from Earth.”

“Cool. Does that *R ^{3}* formula work for the planets?”

“Sure. Works for the Sun, too, but that’s so massive and spins so fast the sol‑stationary orbit’s half way to Mercury. An orbiter would have to fly 205 000 miles an hour to keep up with an equatorial sunspot. Flying something‑stationary over other planets offers problems beyond targeting the orbit, though.”

“Besides how long the trip would be?”

“Well, that, yes, but here’s another one. Suppose you’re going to Mars, aiming at an ares‑stationary orbit. It’ll be 20 megameters, 12500 miles from the center. You need to make your tangential injection at a Mars‑relative speed of 1439 meters per second. Problem is, you left Earth from a geostationary orbit at 3075 m/s relative to Earth. At the classic Hohmann positions, Earth’s going 5710 m/s relative to Mars, Somehow you’re going to have to shed 7346 m/s per second of excess speed.”

~~ Rich Olcott