# Oberth effect with regard to relativistic velocity

So the Oberth effect is a little application of kinetic energy in classical mechanics. Essentially, since KE is proportional to $$v^2$$, at higher velocities a change in kinetic energy produces a greater change in velocity. This has applications in orbital mechanics, especially Hohmann transfers.

How does the derivation of the Oberth effect change in a relativistic setting? I know that one effect of relativity is an asymptote at $$c$$ where as you approach, more energy to further approach, but does the Oberth effect still hold in its entirety?

• An interesting read; gsjournal.net/Science-Journals/… Commented Jan 30, 2020 at 3:24
• 'Interesting' is one word for it. Not sure if the author has ever read a style guide, but their ideas are pretty good. Was the mentioned experiment ever completed? Commented Jan 30, 2020 at 4:52
• I think you got it backwards; at higher velocities, a given change in velocity (or momentum) produces a greater change in kinetic energy. Commented Jan 30, 2020 at 5:53
• @JakobLovern To my knowledge, with a fair degree of certainty, the proposed experiment was never done. Commented Jan 30, 2020 at 8:13
• This is related to the question physics.stackexchange.com/questions/321835/… where we tried to figure out what happens to the effect near a black hole. But your question doesn't require a (very) curved spacetime to be interesting. Commented Jan 30, 2020 at 11:22

The relations we're goint to use are $$\frac pE = \frac{\gamma mv}{\gamma mc^2} = \frac v{c^2}$$ and $$E^2 = (pc)^2 + (mc^2)^2$$

For convenience's sake, let's ignore that chucking out propellant will change our spacecraft's mass. Then, by differentiating the last equation, $$2E\,dE = 2 pc^2dp$$ $$\implies dE = \frac pE c^2 dp = v\,dp$$

So as in classical mechanics, for the same impulse, the energy will grow more at higher velocities.

Eventually, this will have diminishing returns as far as velocity is concerned. However, increasing kinetic energy is still useful insofar that it will decrease travel time as the Lorentz factor will continue to grow.

• Nice answer, but it'd be even better if you justify why it's ok to ignore the change of the spacecraft's mass. ;) Commented Jan 30, 2020 at 17:13

Oberth effect: Rocket approaches a planet for time t at average speed v, recedes for time t' at average speed v + delta v. As t' < t , rocket is pulled forwards by planet and planet is pulled backwards by rocket.

Relativistic Oberth effect: Rocket with mass m approaches a planet for time t at average speed v, recedes with mass m' for time t' at average speed v + delta v , where delta v is quite small. As t' < t , rocket is pulled forwards by the planet, but only a little bit because of how small the change of time is.

The change of mass does not actually matter, we can see that if we take the point of view of passenger on the rocket, according to whom the time difference is the only difference between the free fall acceleration phase and the free fall deceleration phase. Or maybe it does actually matter after all. :) I don't know. I guess I have to think some more about this.

• That is not quite how it works, firing the rockets produces greater KE at greater speed, at periapsis in a flyby the craft has the greatest speed. Commented Jan 30, 2020 at 8:09
• @AdrianHoward As my effect occurs in a very similar situation as the Oberth effect, and is useful the same way as the Oberth effect, it is very likely that my effect is the Oberth effect. Commented Jan 30, 2020 at 10:09
• You are sort of describing a gravity assist maneuver, not an Oberth maneuver Commented Jan 30, 2020 at 10:33