There are 100 circular orbits with different radii, inclinations and RAAN. The nearest orbit by definition is the one, which requires less $\Delta V$ for a transfer.

Assumed that 2 impulsive maneuvers will be applied for a transfer. From an initial orbit, any point of the orbit may be chosen as the final destination.

Is there an analytical way to determine the nearest orbit for a transfer, without numerically calculating the $\Delta V$? Even 90% accuracy is acceptable.


We may also consider transfer to a specific point of the orbits, in this case the arguments of the analytical formula may be [x1, y1, z1, Vx1, Vy1, Vz1], [x2, y2, z2, Vx2, Vy2, Vz2]

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    $\begingroup$ What an intriguing question! +1 $\endgroup$
    – uhoh
    Commented May 18, 2019 at 17:41
  • $\begingroup$ What about orbital radii? (also, did you mean RAAN - Right Ascension of Ascending Node, or something else?) $\endgroup$
    – SF.
    Commented May 21, 2019 at 7:05
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    $\begingroup$ If you want to find the nearest orbit with minimal delta v, how should it be possible without calculation of delta v? $\endgroup$
    – Uwe
    Commented May 21, 2019 at 9:00
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    $\begingroup$ @uhoh Added some additional info $\endgroup$
    – Leeloo
    Commented May 21, 2019 at 9:38
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    $\begingroup$ Is the "2 impulsive maneuvers" a hard constraint, or just an assumption that two should be enough? Because sometimes you can save delta-v by using three maneuvers (essentially like a bi-elliptic transfer), especially if you need to make a major inclination change. $\endgroup$ Commented May 21, 2019 at 10:06

1 Answer 1


Determining the required delta-V for such a transfer can perfectly be done analytically, as long as some assumptions are made. For this answer I am assuming hohmann transfer for the orbit raising or lowering manoeuvre coupled with a plane change manoeuvre taking place at the same time as the injection into the transfer orbit. More efficient manoeuvres would split up the plane change between the two manoeuvres but this would be done numerically. Furthermore the injection into the transfer orbit is performed at the point where the second orbit would intersect with the final orbit in case the radius would be identical.

The calculation is split in three parts:

1) Orbit raising/lowering only.

This is just the calculation for a hohmann transfer but for completeness i'll add it here.

  • First burn

$V_1 = \sqrt{\frac{\mu}{r_1}}\sqrt{\frac{2r_2}{r_1 + r_2}}$

The $\Delta V$ for this manoeuvre will be calculated in the third part, where this velocity is combined with the required plane change

  • Second burn:

since the transfer orbit is already in the correct plane, the $\Delta V$ is simply:

$\Delta V_2 = \sqrt{\frac{\mu}{r_2}}(1-\sqrt{\frac{2r_1}{r_1 + r_2}})$

2) Plane change only.

We need the angle between the two orbits. This derivation involves some spherical geometry but at the end you get the following expression for the angle $\theta$ between the two orbits:

$\theta = \arccos{(a_1 b_1 + a_2 b_2 + a_3 b_3)}$


$\begin{split} a_1 &= \sin{i_i}\cos{\Omega_i}\\ a_2 &= \sin{i_i}\sin{\Omega_i}\\ a_3 &= \cos{i_i}\\ b_1 &= \sin{i_f}\cos{\Omega_f}\\ b_2 &= \sin{i_f}\sin{\Omega_f}\\ b_3 &= \cos{i_f} \end{split}$

Where the subscripts $i$ and $f$ refer to initial and final orbit respectively.

3) Combine both to get to the total $\Delta V$ required.

The first manoeuvre combines the plane change with the injection into the transfer orbit. The $\Delta V$ is calculated using the law of cosines:

$\Delta V_1 = \sqrt{V_i^2 + V_1^2 - 2V_iV_1\cos{\theta}}$

Here, $V_i$ is the velocity in the initial circular orbit: $V_i = \sqrt{\frac{\mu}{r_1}}$, $V_1$ is the velocity calculated in part 1, and $\theta$ is the plane change angle from part 2.

Finally adding the $\Delta V$ calculated in part 1 gives the total $\Delta V$:

$\Delta V = \Delta V_1 + \Delta V_2$

  • Input is given as two states

The method for the second part of your question is not much different than the first, as long as the same restriction of circular orbits only is maintained and the two states correspond to circular orbits. If you have these states as input, simply calculate the kepler orbit associated with these states and follow the procedure described above.

The calculation of the corresponding kepler orbits is explained in the Wikipedia page on kepler orbits.

  • $\begingroup$ Thank you, i'll try this. Do you have any comments about the transfer to a specific point (added as an update). $\endgroup$
    – Leeloo
    Commented May 23, 2019 at 14:17
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    $\begingroup$ I'll add that to my answer. $\endgroup$ Commented May 23, 2019 at 14:21
  • $\begingroup$ Thank you. İn the second part the true anomaly also figures. Could you please, clarify in more detail about that. $\endgroup$
    – Leeloo
    Commented May 23, 2019 at 15:57
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    $\begingroup$ Not sure if I understand correctly. The true anomaly just determines where you are in your current orbit. If this is not the correct location to perform the burn then you just wait until you are at the correct location. $\endgroup$ Commented May 24, 2019 at 6:26

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