# Maneuver to the optimal point of a given orbit

There is a spacecraft on an initial orbit. There is a target orbit. The Keplerian elements are given for the orbits.

The problem is: when and what $\Delta V$ to apply to the SC to move it to the target orbit in the optimal way (with minimum fuel consumption). Impulsive maneuvers are considered.

The problem would be easier, if we know the target point (i.e rendezvous problem), in this case I could solve the Lambert problem. What to do in my case (when any point of target orbit may be chosen)?

The question is related to arbitrary orbits. However, if it's required to make the question more specific, we can consider the following orbits: the initial orbit is sun-synchronous, the target orbit is elliptic, with inclination of 63, apogee at 8000km and perigee at 600km.

• @Leeloo I'm guessing he just means that sometimes a trip through the Van Allen belts can endanger satellites that are not as radiation-hard, so that maneuver would not be seen as an option for those satellites. Probably for the purposes of your question it can be ignored. – uhoh Jul 2 '18 at 1:05
• I've added a bounty, let's see what happens next. – uhoh Jul 5 '18 at 9:21
• How many of elements of the target orbit are you targeting? If six orbital elements are fixed, then you can convert that to a set of radius and position vectors in the inertial frame and apply a Lambert solver. If you have only some elements fixed (e.g. all but the true anomaly), then you best bet is to use a suboptimal control (cf. space.stackexchange.com/a/19805/1391) and search for the least dV through iteration on the free elements (using a Jacobian, a secant method, or brute force). For the true optimal answer, cf. space.stackexchange.com/a/21513/1391 . – ChrisR Jul 6 '18 at 4:25
• @ChrisR The true anomaly is not set- any point of target orbit may be chosen – Leeloo Jul 6 '18 at 7:01
• Idle intuition: If the orbits are close, then 1. burn at periapsis, 2. plane change at ascending/descending node, 3. burn to match orbits; or 'tother way around to match at periapsis if the target orbit is lower. Or; if not close, then bi-elliptical transfer 1. Burn co-planar at ascending/descending node to high orbit (how high? dunno; how long have you got?) descending/ascending node, 2. combined plane change/transfer burn at high point, 3. match orbits. I may well be missing something in the latter; I'll have to give it a try. None of this helps in proving what's optimal, of course. – Slow Dog Jul 6 '18 at 12:38

You can't escape the energy required for a plane change. In general, you would want to do the plane change first to set your inclination, then adjust perigee, then apogee.

For timing, it would be most efficient to make changes when the current orbit and the new orbit intersect. If they don't intersect, then make the plane change anyway, then adjust the perigee first from the apogee, then the apogee from the perigee.

I can't prove that's the most efficient, but it should be close. If this was for real mission planning purposes, I would simulate a few different scenarios before deciding on the one I like best (fuel use is rarely the only criteria).

• What sw do you suggest to simulate on? GMAT? – Leeloo Oct 7 '18 at 11:10
• I haven't used GMAT myself, but it seems like a reasonable place to start. – RickNZ Oct 7 '18 at 11:44

Assuming you're staying around Earth: A Hohmann transfer is always going to provide the least energy transfer - burning at perigee of initial orbit is the most optimal initial burn and burning at apogee of target orbit is the most optimal final burn. Everything else in between is sub - optimal (when you think about it in laymans terms, you are travelling fastest at perigee where you need speed to change your orbit, and slowest at apogee where you need to be close to target speed to match. All other points in an ellitptical orbit are either slower than perigee or faster than apogee requiring greater burns)

So the general answer to your question is - no matter what, the most optimal way is orbit changing at perigee of initial and orbit matching at target.

• Bi-elliptic or, more generally, bi-tangential transfers can be more efficient than Hohmann transfers in certain situations. In any case, they are only most efficient for co-planar orbits and the question specifies an inclination change. – Jack Aug 7 '18 at 8:14

One could discretize the model and define a system of linear equations that could then be solved to create a list of various maneuvers that would achieve the desired orbit from a starting orbit. The list can then be searched to find the maneuver plan that uses the least amount of fuel.