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From a specified circular orbit it is often said that it is easier to go out than in, referring to the fact that achieving escape velocity from a circular orbit takes less delta-V than achieving zero velocity.

I originally thought this would apply as well to transfers between orbits, i.e., given two orbital radii $A$ and $B$, with $A<B$, it would take less delta-V to perform a Hohmann transfer from a circular orbit with radius $A$ to radius $B$ (with circularization at $B$) than for the inverse. But then I realized these two transfers are time-reversible and so the delta-Vs should be same in magnitude while inverted in sign.

I just wanted to check that my second thought is correct: delta-V requirements for Hohmann transfers, with circularization at the end, is the same whether going out or in between two specified circular orbits.

NeutronStar
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  • This seems to directly answer you, but I don't fully understand the answers. Let me know if you can marry your question to this question? https://space.stackexchange.com/questions/14435/question-about-the-hohmann-transfer-why-does-delta-v-go-down-when-transferring – IDNeon May 04 '21 at 23:55
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    @IDNeon It's possibly related, but not the same. It's asking why Hohmann transfer delta-V changes the way it does going from r1 to r2 when the value of r2 is changing; I'm asking, when r1 and r2 are fixed, if the Hohmann transfer delta-V is the same going from r1 to r2 as r2 to r1. – NeutronStar May 05 '21 at 14:27
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    "these two transfers are time-reversible and so the delta-Vs should be same in magnitude while inverted in sign." Correct. An equivalent way to think about it is in terms of the changes to the specific orbital energy. I can write a proper answer, if you like. I don't think this question has a close duplicate, but I'm not yet familiar with all the Hohmann-transfer questions on this site. – PM 2Ring May 05 '21 at 16:03
  • @PM2Ring Correct me if I'm wrong, but it's possible to have equivalent changes to specific orbital energy that use different delta-Vs, for example, going from r1 to r2 with a Hohmann transfer and then deciding you want to go back to r1 right fast so you use a maneuver that's not fuel-usage optimized to change orbits more quickly. – NeutronStar May 05 '21 at 17:00
  • Sure. An ideal Hohmann tranfer assumes you are going from one perfect circle to another perfect circle, via an ellipse. It also assumes the delta-Vs are instantaneous. And of course it assumes that there's a single central gravitational source. So if you aren't (approximately) following that scenario, it's not really Hohmann. When you aren't burning fuel, your trajectory is purely ballistic, and it has constant specific energy. – PM 2Ring May 05 '21 at 17:22

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