Two orbits are given with the identical orbital elements, despite the RAAN, which differs by 90 degrees. The inclination is 80 deg, AOP is 310 deg, apogee is on 10000km, perigee on 1000km altitude.

The transfer should be done within a few days. The launcher will place 2 satellites on these orbits, the true anomalies of the satellites will differ by 180 degrees.

What is the optimal transfer trajectory to minimize the fuel consumption? How to calculate the $\Delta V$? Impulsive maneuvers are assumed.


AoP is the angle from the ascending node to the periapsis.

RAAN is the angle from the reference direction to the ascending node.


I'm going to solve the Lambert problem (boundary value problem), considering 2 impulses. Is this a good idea, or there is a better way?

  • 1
    $\begingroup$ That will largely depend on the inclination, and the optimal maneuver will vary wildly depending on that, ranging from a pair of inclination change maneuvers, through waiting 3 months for orbital precession to do your job, up to a lunar fly-by followed by aerobraking into the right orbit. The usual approach will be two separate launches though. $\endgroup$
    – SF.
    Oct 29, 2018 at 1:20

1 Answer 1


That's a very expensive scenario in general, and especially with your constraint of 'a few days'. And in your case it's best performed using three burns - the maneuver of orbital plane change performed near distant apoapsis.

First, let's adjust the parameters - since the perigee is at 1000km, it means you used altitudes, not radii (or your satellite would fly underground), so let's adjust that by average Earth radius (6371km).

Pe = 7371km
Ap = 16371km

a = (Ap+Pe)/2 = 11871km (let's mark this semi-major axis as a0)

Geometric proof that 2a = Ap+Pe (accounting for planet radii): enter image description here

Let's have a look at what our orbits should look like: enter image description here

The direct orbit change there would be a 90 degrees turn, that's original speed times $\sqrt{2}$

geometric proof: diagonal of a square:

enter image description here

Using Vis Viva equation

$$ v = \sqrt{GM ( {2 \over a} - {1 \over r}) } $$

we can find speed at the apogee: sqrt(G (Mass of Earth) ( 2 / 11871km - 1 / 16371km)) = 3888m/s. Multiplied by the aforementioned square root of 2, our delta-V is 5498m/s. That's way too expensive to do in a single launch.

The moon flyby scenario is right out due to timing constraints and inclinations that aim nowhere towards the moon.

1000x10000 is not a great eccentricity but let's see what we can work out with degrading the (actually optimal) scenario of 'plane change at infinity' to fit our timing constraints. I'll assume the 'few' days to be 10.

For the orbital period to be 10 days, semi-major axis (a1) must be 196060km. Pe+Ap = 2a, so Ap = 2*196060-7371 = 384749km.

  • So, first, a burn prograde at intersection of the orbital planes near periapsis to raise apoapsis. The initial speed won't be much different from the periapsis speed, so I'll approximate it with it.

Initial speed, Vis Viva with r=Pe, a=a0: 8636m/s
Final speed, r=Pe, a=a1: 10301m/s

$\Delta v_1$ = 10301 - 8636 = 1665m/s.

  • At the intersection of the target orbital plane (which should happen very near the new apoapsis) perform a normal burn to adjust your orbital plane. Our apoapsis speed will be 197m/s Times $\sqrt{2}$ for the 90 degrees turn, $\Delta v_2$ = 279m/s.

The actual orbit may be a little higher or lower to make the arrival provide the correct true anomaly - make the period be the nearest (n+0.5) times the period of the satellite, instead of 10 days sharp. The change in delta-V will be quite small.

  • Another 1.664km/s burn ($\Delta v_3$), retrograde, at periapsis to bring apoapsis back.

A total of 2*1664+279 = 3607m/s and that's about the least delta-V you can get it done with.

In reality, this will be done through two separate launches, because it will be cheaper than using a launcher that can bring up a transfer stage of over 3.5km/s and three reignitions besides the satellites.

The whole procedure would look kinda like this:

enter image description here

enter image description here

  • $\begingroup$ What value for the argument of perigee you took? Actually, I consider the AOP = 310 deg. Another question, could you explain 10.3 - 8.636 = 1.664km/s. calculation? And why do you multiply by $\sqrt{2}$? $\endgroup$
    – Leeloo
    Oct 29, 2018 at 13:50
  • $\begingroup$ @Leeloo: 1. "Such that the perigee falls on intersection of the two orbital planes". It will be somewhere around +/-80 degrees from respective RAAN. The delta-V was found with two speeds obtained through Vis-Viva for r=Pe and the two respective a pre- and post-burn. Sqrt(2) is diagonal of a square; sum of two perpendicular (velocity) vectors, or otherwise the vector required to rotate your current velocity 90 degrees. $\endgroup$
    – SF.
    Oct 29, 2018 at 14:01
  • $\begingroup$ @Leeloo: I noticed my solution violates your requirement of 'other parameters being the same' - argument of periapsis of the two orbits differs by the same 90 degrees as RAAN. OTOH, that makes the orbits actually similar, just rotated around Earth axis. Same precession, same latitude range etc. Quite similar to Molinya constellation actually. With arg. of pe. the same but RAAN 90 degrees off you'd make them very different. One's apogee above antarctic, the other's - in equatorial zone. And they'd differ in orbital precession so much the 90 degree difference in RAAN would be quite short-lived. $\endgroup$
    – SF.
    Oct 29, 2018 at 17:08
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    $\begingroup$ Sorry, I'm sick currently (over a month now) and this would take more focus than I have in the state I'm in... the edit will need to wait some more. $\endgroup$
    – SF.
    Nov 8, 2018 at 21:08
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    $\begingroup$ @Leeloo: You're turning 90 degrees. You must cancel out the entirety of the original apoapsis speed, and then rebuild it to the exact same value in perpendicular direction. By combining the two operations you're paying $v \sqrt{2}$ instead of $2 v$, Exactly the same situation as in the earlier "direct change", the picture with perpendicular vectors - except now the apoapsis is very high and the speed is low. $\endgroup$
    – SF.
    Nov 21, 2018 at 14:59

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