How can you calculate a Hohmann transfer orbit but ending at the same orbit with a time difference between orbits?

Question

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

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:

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