the two equations

I'm wondering if it's appropriate to use the two equations in order to solve for m0/m1 to find the mass of propellant needed to perform a hohmann transfer. If it isn't, can someone suggest an alternative?

Edit: I'm basically happy with any assumptions/simplifications (I don't have much knowledge on this subject so I'm open to any as long as they are explained)

  • 2
    $\begingroup$ This is perfectly fine. As long as you're happy with the assumptions (instantaneous burn) $\endgroup$
    – ThePlanMan
    Mar 31 '15 at 23:48
  • 1
    $\begingroup$ Sure. For very large orbit changes, or if there is an inclination change, there may be more efficient approaches. E.g. a bi-elliptic transfer. $\endgroup$
    – Mark Adler
    Apr 1 '15 at 1:18
  • 3
    $\begingroup$ Please update your question with details of what assumptions/simplifications you are happy with, then we can say what method fits within your problem statement. $\endgroup$
    – ThePlanMan
    Apr 1 '15 at 11:13
  • $\begingroup$ Something to keep in mind here: Oberth effect. While it has no meaningful effect on the exact case you are describing (orbit to orbit) a Hohmann transfer is usually performed between objects, not merely in space. Doing your transfer burn in low orbit can save fuel. $\endgroup$ Apr 2 '16 at 19:54
  • $\begingroup$ So long as you know the dry mass, the absolute fuel mass is easy to calculate this way, if you're happy with the assumptions. Without the dry mass, all you can get is the required mass ratio, $mo/m1$ $\endgroup$
    – Leliel
    Jun 2 '16 at 4:47

Nothing is wrong with,


assuming that the orbits of the planets are circular and coplanar, and your parking orbits too.

This is basically combining the equation for the $\Delta v$ required for the Hohmann transfer:

$$\Delta v =\sqrt{\frac{\mu_S}{r_1}\left(\sqrt{\frac{2r_2}{r_1+r_2}}-1\right)^{2}+\frac{2\mu_1}{a_1}}-\sqrt{\frac{\mu_1}{a_1}}+\sqrt{\frac{\mu_S}{r_2}\left(\sqrt{\frac{2r_1}{r_1+r_2}}-1\right)^{2}+\frac{2\mu_2}{a_2}}-\sqrt{\frac{\mu_2}{a_2}}$$

And the inverse form of the rocket equation:

$$\frac{m_0}{m_1}=e^{\frac{\Delta v}{v_e}}$$

In some cases, a bi-elliptic transfer might give a better result. For a bi-elliptical transfer of the same type, use:

$$\Delta v=\sqrt{\left(3-2\sqrt{2}\right)\frac{\mu_S}{r_1}+\frac{2\mu_1}{a_1}}-\sqrt{\frac{\mu_1}{a_1}}+\sqrt{\left(3-2\sqrt{2}\right)\frac{\mu_S}{r_2}+\frac{2\mu_2}{a_2}}-\sqrt{\frac{\mu_2}{a_2}}$$

That can be combined with the rocket equation in the same way.


  • $\begingroup$ Do you suppose you could pull that exponent (or the top half of the fraction) out into a separate equation to make it more readable? $\endgroup$ Mar 31 '16 at 20:30
  • $\begingroup$ @NathanTuggy More readable form of the equation added. $\endgroup$ Mar 31 '16 at 22:29
  • $\begingroup$ That is such a tiny $v_e$. $\endgroup$
    – user
    Apr 1 '16 at 13:02

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