Computing a delta-V budget from one heliocentric orbit to another can be done as a Hohmann transfer.

How does one calculate delta-V for transferring from a heliocentric orbit to be captured by another planet? Is it as simple as computing the Hohmann delta-V and subtracting the velocity of the orbit around the destination planet?

This question is inspired by comments in Is space mining and development as shown in the television show “The Expanse” realistic?.

Bonus: Is the reverse process symmetrical? (I.e. same delta-V)

  • $\begingroup$ Check out my answer to this question, it presents a patch conics solution for an Earth to Mars transfer. $\endgroup$ Dec 21, 2015 at 19:36

1 Answer 1


Your capture delta-v is calculated by


Where $v_e$ is the escape velocity at your target periapsis, and $v_{inf}$ is the velocity at infinity. For any given target orbit, replace the $-v$ after the square root with the velocity of your orbit.

A minimal capture is when your periapsis is as low as possible, and the apoapsis just within the sphere of influence. If you do not know the SOI, using $v_e$ for $v$ gives a good aproximation.

This is a symmetrical process. You can also be captured by passing through the atmosphere of the planet, if it exist. In that case, it is not a reversible process.

  • $\begingroup$ You want an apoapsis within the planet's sphere of influence else the sun can tear the space craft away and it's not captured. Periapsis of this capture ellipse is as close to the planet as possible. Generally periapsis velocity of this capture ellipse is slightly less than escape. It is this velocity you would subtract, not Ve. $\endgroup$
    – HopDavid
    Dec 22, 2015 at 3:17
  • $\begingroup$ See hopsblog-hop.blogspot.com/2012/06/inflated-delta-vs.html $\endgroup$
    – HopDavid
    Dec 22, 2015 at 3:18
  • $\begingroup$ Am upvoting because the expression you give is very close. Only a tad bit more delta V would be needed. $\endgroup$
    – HopDavid
    Dec 22, 2015 at 4:02
  • $\begingroup$ @HopDavid I simplyfied it too much, fixed it now. $\endgroup$ Dec 22, 2015 at 8:48

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