I am studying the drag make up maneuver. In this maneuver, I need to go from point A to point B keeping the same phase angle as shown schematically in this picture

Drag make up. Reach an higher orbit keeping the phase.

When I've approached this problem, I've thought about bi-elliptic transfers but my professor said that if we need to have a symmetric jump (from ๐‘Žโˆ’ฮ”๐‘Ž to ๐‘Ž+ฮ”๐‘Ž), while keeping the same phase, in ideal condition, also a Hohmann transfer would be the solution. I have made some math, but I've found that the only way to keep the phase with Hohmann is to keep the ratio between the perigee and the apogee equal to one...

Any solution will be appreciate. Thanks.


1 Answer 1


To see why a Hohmann transfer is applicable for a symmetric jump preserving phase angle, let's consider what the semi-major axis of the transfer ellipse is:

$$a_{H} = \frac{periapsis_H + apoapsis_H}{2}$$

$$a_{H} = \frac{a - \Delta a + a + \Delta a}{2}$$

$$a_{H} = a$$

That is, the semi-major axis of the Hohmann transfer orbit is equal to the nominal orbit!

Since orbital period depends only on semi-major axis, the nominal orbit and the Hohmann transfer orbit must have the same orbital period.

After the transfer half-orbit has been completed, the nominal orbit must also have completed half an orbit. With the same angle (180 degrees) covered in the same time, the phase angle is preserved.

  • 1
    $\begingroup$ Thank you very much! I was focusing on the wrong angle because the sketch made by professor was a little bit confusing. The phase angle to keep is the one between the real and nominal satellite! Thanks a lot, I was going mad! $\endgroup$ Nov 2, 2021 at 14:29

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