# What's wrong with my bielliptic transfer calculation?

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?

• 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 Oct 19, 2021 at 13:47