# How can the Hohmann transfer work?

When you want to move from a lower circular orbit to a higher one, one may decide to boost their velocity, which causes the satellite to require more centripetal force than gravity can provide and thus spirals outward. This makes sense, however, the issue I'm having is, how can this satellite ever go back into a circular orbit? At further distances, the satellite must travel slower to stay in orbit given by $$v=\sqrt\frac {GM} {r}$$ however the satellite is traveling very fast, so how could it reattain circular orbit? In all the demonstrations I've seen, the rocket merely boosts again once at the new orbit, but wouldnt this only increase it's velocity further?

• For an elliptical orbit, the satellite will be slower at the apoapsis than it would be in a circular orbit at that distance. – HiddenBabel Sep 29 '18 at 2:29

Starting from a circular orbit, increasing your speed will make the orbit elliptical (with periapsis at the point when you fired your engines, and apoapsis directly opposite), meaning that the speed will no longer be the same at all points. When the craft reaches apoapsis, it will have a slower speed than the new required speed for a circular orbit at that altitude (by conservation of energy). Your equation $$v=\sqrt\frac {GM} {r}$$ only holds for circular orbits with constant speed. For elliptical orbits, it is replaced by $$v=\sqrt{ GM \left(\frac {2}{r} -\frac{1}{a} \right)}$$, where $$a$$ is the semi-major axis.

The semi-major axis is particularly important for elliptical orbits, and it is the average of the apoapsis and periapsis.

• Sorry, could you explain how the conservation of energy comes into ensuring that lower speed? Do you mean by the conservation of total orbital energy (U + KE) in the same orbit? – John Hon Sep 29 '18 at 2:41
• @John Hon: The total energy (which is the sum of kinetic and potential) is conserved at all points along the new orbit. For elliptical orbits, it is no longer $-\frac{GMm}{2r}$, instead it is replaced with $-\frac{GMm}{2a}$, where $a$ is the semi major axis. – user7777777 Sep 29 '18 at 2:43
• Oh sorry, just missed your comment as I updated mine! Thanks!! – John Hon Sep 29 '18 at 2:44
• @John Hon: It works exactly in reverse, two retrograde boosts are needed as it will be at a higher speed upon reaching the new orbit. – user7777777 Sep 29 '18 at 2:55
• @John No. If you're doing a Hohmann transfer from a higher orbit to a lower one, you first do a slowdown burn to begin falling, but the fall makes you go faster. When you reach the periapsis, you will be going too fast and you will need to perform another slowdown burn. – Emilio Pisanty Sep 29 '18 at 2:58

The slowdown occurs during the elliptical orbit, between its periapsis and apoapsis. During that time, the craft is climbing a gravitational hill: it has a nonzero radial velocity, which means that the central gravitational force does perform work on the craft, slowing it down. Thus, the craft starts off going too fast to stay at a circular orbit, but when it reaches its apoapsis it has been slowed down so much (it will in fact be going slower than its initial speed in the initial circular orbit) that it would fall right back down to its periapsis if it didn't re-boost its speed at the apoapsis.

This is a good example of a broader trend in orbital mechanics: if you try to speed up by boosting in the direction of your velocity, then that will only raise your orbit and slow you down. Thus, if you are trying to rendezvous with a craft that's ahead of you in your same orbit, you need to boost away from it, so you can drop down to a faster orbit where you can catch up.