# Why does the delta-V for LEO-KSC to LEO-Equatorial orbit seem unusually high?

Why does going from a KSC-launched LEO to an orbit like one launched from the equator seem to require so much $\mathrm{\Delta V}$? $\mathrm{4.24~\frac{km}{s}}$ is almost half of what's required just to get to LEO to begin with, and that's factoring in a prograde launch!

Isn't this just a change in orbital inclination? KSC is "only" 28.57$^{\circ}$ N. latitude.

Related question: Is there a general formula for determining $\mathrm{\Delta V}$ given the initial and final (or change of) inclination? My intuition says that some trig functions would be involved.

• I can tell you from playing KSP that plane changes are very expensive. You have to cancel a lot of vertical velocity. – Javier Aug 12 '16 at 2:36

Isn't this just a change in orbital inclination?

"Just" is one of the most dangerous words in science and engineering.

Plane changes are very expensive in terms of delta V costs. Changing orbital inclination is a plane change. The cost to make an inclination change from a circular orbit to another circular orbit, with no change in orbital altitude, is $\Delta v = 2v\sin(\Delta i / 2)\$.

For example, the delta V needed to change from a geosynchronous orbit with an inclination of 28 degrees to a geostationary orbit is 1.4 km/s. The delta V cost to successively go from a low Earth orbit inclined at 28 degrees to a geostationary orbit inclined at 28 degrees and then to a geosynchronous orbit is 5.3 km/s.

That three burn sequence (one in LEO to transfer to geosynchronous altitude, a second at geosynchronous altitude to circularize the orbit, and a third at the node to transfer to geostationary) is not how it's done. Delta V costs are nonlinear, thanks to the nonlinear way in which gravitation works. The result is a savings of about one km/s compared to that three burn sequence, but it's still more expensive than launching from the equator.

Regarding the wikipedia page on delta V budget, you have to take that page with a huge grain of salt.

• Should I assume that the word "only" (in the next sentence) is equally dangerous? Note that I did put scare quotes around it. ;-) – pr1268 Aug 12 '16 at 21:14
• Also, my original question assumed a change in orbital inclination but remaining a low-Earth orbit. Does an orbital plane change remaining at LEO require a transfer to geosynchronous orbit and back? – pr1268 Aug 12 '16 at 21:18