Say I want to change the inclination of the ISS by one degree. The ISS orbits at 7660m/s so the manoeuvre will cost me 133.7m/s of $\Delta v$:

orbital plane change

From time to time, the station must be reboosted because of orbital decay. Let us just assume that this is done like a Hohman transfer, with the first burn being 10m/s to change the velocity of the station from 7660m/s to 7670m/s. But what if I do a slight plane change at the same time, say 0.1 degrees?

orbital plane change at reboost

Well, that costs me an additional 6.7m/s on top of the original 10m/s. But wait! If I do this on 10 consecutive reboosts, I have changed the inclination by one degree, but only spent 67m/s of $\Delta v$!

It only gets better. With 100 steps of 0.01 degrees, the total $\Delta v$ spent is only 8.9m/s, at a 1000 0.001 degree steps, only 0.9m/s. It is approaching zero.

Is this really a free propulsive inclination change? I think a way to summarize this odd effect is something like this:

spent and gained


1 Answer 1


Your initial maneuver is changing inclination and slowing down station

It's ok, for staying on the orbit (same altitude), but you are using reactive mass.

In second case slowing down done by air drag, instead of using reactive mass. So yes, second variant is more efficient then first one, in therms of reactive mass, if you wish to change inclination.

Is that super cheat, no it's not, it's just friction - sometimes it helps, some times not.


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