There are 2 coplanar orbits (same $i$,$\Omega$,$\omega$):

  1. Apogee and perigee on 200km
  2. Perigee on 700km, apogee on 8000 km

I used the izzo.lambert function from the poliastro (python) package to solve the problem.

I used the transfer time in the interval [30, 60*24*10] minutes and the true anomaly of the first maneuver in the interval [0,360]. The spacecraft mass is approximately 8 tons, İSP is about 330 s. The optimal transfer became:

  1. 0.14 km/s impulse on 168 deg true anomaly of initial orbit
  2. 1.32 km/s impulse on the perigee of destination orbit

Due to the last impulse, I spend almost 3 tons of fuel! Is this fine? Is it practically feasible to spend 3 tons in 1 impulse?

Is there a better way?

  • 6
    $\begingroup$ This feels like a homework problem, and one that is missing a bit of detail. To known the fuel, we would need to know the mass of the spacecraft, and the ISP of the fuel. I'm assuming you are going from orbit 1 to orbit 2, but there is no inclination. The dela-v numbers sound reasonable. $\endgroup$
    – PearsonArtPhoto
    Nov 13, 2018 at 16:09
  • $\begingroup$ @PearsonArtPhoto Added mass and İSP $\endgroup$
    – Leeloo
    Nov 13, 2018 at 16:36

1 Answer 1


If I'm understanding your problem statement correctly, your maneuver is equivalent to a complete Hohmann transfer from 200km x 200km to 700km x 700km, followed immediately by the first burn (but not the circularization burn) of a second Hohmann from 700km to 8000km. The online Hohmann calculator here confirms your ∆v calculations.

The Tsiolkovsky equation can be rearranged to solve for final mass like so:

$$ \frac {m_0} { e ^ {\frac {\Delta v} {g_0 \cdot I_sp}} } = m_f $$ Assuming your 8 ton figure is the fully fueled mass of the spacecraft, the final mass after those burns is indeed just over 5 tons; nearly 3 tons of fuel is expended.

Whether that is feasible depends on the details of the spacecraft. The transfer ∆v calculations are assuming the ∆v is applied instantaneously, which is never the case in practice; somewhat more ∆v or fuel needs to be expended if it's spread over a significant amount of time. Another option is to do a series of short burns at the perigee of each orbit, raising the apogee a little bit at a time; the total ∆v applied will be the same.

  • 3
    $\begingroup$ Would it be better to raise the aphelion all the way first (ie boost to a 200 x 8000 elliptical orbit, Then boost at aphelion to raise the perihelion to 700? $\endgroup$ Nov 13, 2018 at 17:42
  • 1
    $\begingroup$ @RusselBorogove I don't think so. In units where $GM=1$ and distances are in $km$ the delta-V for going via 700x700 is 0.0341, while going via 8000x200 is 0.0289. The point is to do the biggest useful burn at 200km. $\endgroup$ Nov 13, 2018 at 18:45
  • $\begingroup$ Ah, yeah, sorry. $\endgroup$ Nov 13, 2018 at 21:03
  • $\begingroup$ Is it important to make burns at the perigee? Is it possible, that a different value for True Anomaly would be more optimal? $\endgroup$
    – Leeloo
    Nov 14, 2018 at 10:51
  • 2
    $\begingroup$ In this case, starting from a circular orbit, perigee burn is best. Going from one ellipse to a differently oriented ellipse, ask Lambert. $\endgroup$ Nov 14, 2018 at 15:36

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