New answers tagged

2

What you are seeing here is the result of the inefficiencies in the orbital transfer strategy employed, in the transition region between direct transfers and bi-elliptic transfers. When transferring between two circular orbits with some inclination, eventually the "degenerate" bi-elliptic transfer (with infinite apoapsis) will become optimal, ...


10

(mostly recycled from What are the benefits of supersynchronous transfer orbits?) Excuse the wall of plots but I really do think they describe it better than my words ever could :) The total $\Delta V$ costs for the standard (Hohmann like) method (inclined 250 km parking orbit to geostationary orbit) are: While the total $\Delta V$ costs for the ...


20

This is a partially copied answer from this closely-related question: The other answerer focuses on the straight-up dV savings which occur when you're launching from a very inclined launched site. I'm going to focus on a second reason you might want to do a supersynchronous (new-to-me term) transfer, but first, let me detail how a traditional GTO is used ...


8

As the comments discuss, inclination is the missing variable in the equation. The standard geostationary launch that you are thinking of looks something like this: geostationary transfer orbit (GTO) injection burn: perigee @ parking orbit height (e.g. 250 km) apogee @ geosynchronous height (35786 km) inclination is ~launch site latitude (i.e., ~28.5° for ...


Top 50 recent answers are included