Why are efficient hohman transfer orbit altitudes higher than lowest-safe orbit to avoid atmospheric drag, given the Oberth Effect would imply such would be most efficient?

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The Oberth effect says that due to higher relative velocity deep in a gravity well, the highest change in orbital energy for the least propellant can be achieved by burning at as low as an altitude as safe for a given body. E.g.: If earth had no atmosphere and mountains and we had infinite TWR, the best way to launch a rocket to the moon would be to burn horizontally from sea level.

However, this table for KSP implies that you can actually save propellant by burning from higher orbits for interplanetary transfers – assuming of course that you can refuel at these altitudes to avoid the cost of raising your orbit to these from sea level.

What’s the mechanism behind this? I thought the only reason to burn from higher orbits for interplanetary launches was if you were using a low TWR interplanetary stage so that the long maneuver makes up less of the orbit’s duration (e.g.: a 2 hour burn in low orbit that has a period of 90 minutes is impossible, but in synchronous orbit it
s just a small fraction of the period, so steering losses are minimized)

Edit: Table in question – https://forum.kerbalspaceprogram.com/topic/33699-efficient-hohmann-transfer-altitudes/

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Anonymous 0 Comments

I think the important bit is “assuming you can refuel at these altitudes.” The burns aren’t more efficient, they are just smaller because you are starting from a more energetic orbit. Essentially you did part of the injection burn to get to the higher orbit, refueled, and are now doing a smaller injection burn.

I suppose there could be efficiencies gained since refueling means you would need less tank dry mass, but How much would depend heavily on the craft and fuel type.

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