I'm trying to calculate the delta-v needed to get from a 200 km circular orbit to a geostationary orbit. GSO is at an altitude of 42164140.1029448 m and at a speed of 3074.6611762 m/s. Because $\frac{42164140.1029448}{200000} = 210.82070051472402 > 15.58$, any bielliptic transfer should be more efficient than a Hohmann transfer.

Here is my Hohmann transfer math: $$ v_{pe_1}=\sqrt{\mu\Bigg(\frac{2}{r}-\frac{1}{a}\Bigg)}=\sqrt{3.9860044188\cdot10^{14}\Bigg(\frac{2}{6571000}-\frac{1}{6571000}\Bigg)}\approx 7788.48798575474 \\ v_{pe_2} = \sqrt{3.9860044188\cdot10^{14}\Bigg(\frac{2}{6571000}-\frac{1}{21182070.0514724}\Bigg)}\approx 10245.15814427758578 \\ \Delta V_1=2456.67015852284578 \\ v_{ap}=\sqrt{3.9860044188\cdot10^{14}\Bigg(\frac{2}{6571000}-\frac{1}{21182070.0514724}\Bigg)} \approx 1596.639561525085 \\ \Delta V_2 = 1478.0216146749153 \\ \sum \Delta V = 3934.69177319776108$$

Here is my bielliptic transfer for a 500,000 km apogee kick: $$v_{pe_2} = \sqrt{3.9860044188\cdot10^{14}\Bigg(\frac{2}{6571000}-\frac{1}{256471000}\Bigg)}\approx 10943.80722915345372 \\ \Delta V_1 = 3155.31924339871372 \\ v_{ap} = \sqrt{3.9860044188\cdot10^{14}\Bigg(\frac{2}{506571000}-\frac{1}{256471000}\Bigg)}\approx 139.8084419739 \\ v_{ap_2} = \sqrt{3.9860044188\cdot10^{14}\Bigg(\frac{2}{506571000}-\frac{1}{274367570.0514724}\Bigg)}\approx 347.739448805 \\ \Delta V_2 = 207.9310068311 \\ v_{pe_3}=\sqrt{3.9860044188\cdot10^{14}\Bigg(\frac{2}{42164140.1029448}-\frac{1}{274367570.0514724}\Bigg)} \approx 4177.83262958787 \\ \Delta V_3 = 1103.1714533878694 \\ \sum \Delta V = 4466.42170361768312$$

However, this contradicts the previous statement that any bielliptic transfer from a 200 km orbit to GSO is more efficient than a Hohmann. What's wrong with my math?

  • 4
    $\begingroup$ You need to work with absolute orbital height, not height above ground. The actual ratio of the orbits is about 6.5, not 210.8 $\endgroup$ Oct 19 at 13:47

At 200km altitude, the orbital radius is 6,378km (earth radius) + 200m, = 6,578km.

Geostationary orbit is at an orbital radius of 42,164km

That's a ratio of only 6.4, not above the limit of 11.94 for some bi-elliptic transfers becoming a viable option. So a Hohmann transfer will be both faster and cheaper.

Altitude must not be confused by distance to the centre of mass. Had your initial orbit been at 200km radius, well inside the core of the Earth, a bi-elliptic transfer would indeed (any, since it's >15.58) have been the best options. Ignoring all the liquid rock.

  • $\begingroup$ You'd have to ignore the explosion caused by the planet violently objecting to being compressed to a sphere less than 200km in radius. $\endgroup$
    – notovny
    Oct 19 at 14:17
  • $\begingroup$ I think OP actually uses the correct orbits in their calculations, its just in the initial test for efficiency of bi-elliptic that the 200km orbit was used. But this threw a spanner in the whole thought process. $\endgroup$ Oct 19 at 15:21

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