How can you calculate a Hohmann transfer orbit but ending at the same orbit, just having changed the time?

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    $\begingroup$ For a true Hohmann transfer, going from one circular orbit to another, your transfer duration is fixed as half the period of your transfer orbit. If you need to arrive at the destination orbit at a different time, you need to leave your source orbit at a different time, or do some other type of transfer. $\endgroup$
    – notovny
    Dec 9, 2022 at 16:55
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    $\begingroup$ Rereading the question, are you asking "How do I calculate a rocket maneuver that brings me back to the same point on my original orbit, but at different time than my current orbit would require?" $\endgroup$
    – notovny
    Dec 9, 2022 at 21:44
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    $\begingroup$ @notovny it sure looks like it - "...just having changed the orbital phase" perhaps? $\endgroup$
    – uhoh
    Dec 9, 2022 at 21:57
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    $\begingroup$ Yes, exactly that: “...just having changed the orbital phase”. English is not my first language, sorry. $\endgroup$ Dec 9, 2022 at 22:00

1 Answer 1


Looks like you want a two-impulse phasing manoeuvre, first lifting your apoapsis with a burn, then going one or more orbits around, and then lower the apoapsis back down.

The orbital period can be calculated from the semi-major axis, which for the initial orbit is equal to its radius $a_1 = r$ and for the boosted orbit is the average of the apoapsis and periapsis distance $a_2 = \frac{r + r_A}{2}$.

Thus, we have the orbital periods:

$T_{circular} = 2\pi\sqrt{\frac{r^3}{\mu}}$

$T_{boosted} = 2\pi\sqrt{\frac{(r+r_A)^3}{8\mu}}$

The time difference after some $n$ orbits is then simply taking the remainder:

$\Delta T = nT_{boosted} \text{ mod } T_{circular}$

(or possibly as $\Delta T = n(T_{boosted} -T_{circular})$ if you care for counting the revolutions)

or as an angle

$\theta = \frac{2\pi\Delta T}{T_{circular}}$

The cost of this can be calculated from the vis-viva equation.

At the location of the impulse, the velocity of the boosted orbit is:

$v_{boosted} = \sqrt{\mu\left(\frac{2}{r} - \frac{2}{r + r_A}\right)}$

We need to burn the difference to the circular velocity twice, so:

$\Delta v = 2\left(\sqrt{\mu\left(\frac{2}{r} - \frac{2}{r + r_A}\right)} - \sqrt{\frac{\mu}{r}}\right)$


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