How does efficiency of a burn of given duration change with altitude of periapsis?

Am I right, that for Earth (equatorial radius 6,371km), for orbits with 100 and 200km periapsis altitude respectively, it would be something along the lines of ${6472^{\ 2}}\over{6572^{\ 2}}$ or only about 3% ? Or less, or more?

  • $\begingroup$ Probably not, due to atmospheric drag. There's an optimal altitude that is somewhere in that range for the Oberth effect. $\endgroup$
    – PearsonArtPhoto
    Commented Jun 19, 2015 at 17:59
  • $\begingroup$ By the way, the equatorial radius is about 6378 km. 6371 km is the mean radius. $\endgroup$
    – Mark Adler
    Commented Aug 25, 2015 at 17:02

2 Answers 2


The proportional change in $\Delta V$ is about $\Delta r\over 2 r$. So going from 6472 km to 6572 km is a proportional increase of about 0.8%.

This can vary some depending on your starting and ending specific energies. Also at these small differences you need to be careful to make sure you compare apples to apples. You would need a particular example, to take into account the fact that you are ending up in or starting from an orbit with higher or lower energy to begin with. You should construct examples with identical starting and ending trajectories, which would mean more than one impulsive burn, and then comparing doing the burns in different places.

By the way, for a single pass at 100 km altitude at Earth, the atmospheric density is not a factor.

  • $\begingroup$ That's even less than I thought! Thank you very much :) $\endgroup$
    – SF.
    Commented Aug 26, 2015 at 6:31

There is an optimal height for the maximum safe oberth maneuver. This will depend on a few factors, namely the atmosphere of the object, the mountains and gravitational anomalies on said object, uncertainty of the aimpoint, etc. Bottom line, to get the maximal speed boost, aim for the lowest point that won't cause you to have significant atmospheric drag, and won't put you in any danger of hitting the surface.


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