tl;dr: Hohmann Transfer appears to be the optimal way to achieve a circular-to-circular orbit, but is it possible to lower the periapsis in order to achieve a more elliptical orbit with apoapsis at the same distance, using less delta-v than a Hohmann transfer?

I've been playing an awesome game called Kerbal Space Program, which models to quite an accurate degree the spacecraft mechanics that you may encounter within a solar system.

I was reading about how the Oberth effect dictates that rocket burns are much more efficient when orbital velocity is higher.

I want to minimize fuel consumption, which is a noble and practical endeavor. The situation is this. A stable circular (or nearly circular) orbit has been established around a planet. The goal is to travel to a target location in orbit around the same planet that is in a much larger orbit. Being already in a circular low orbit makes it easy to adjust the inclination to match that of the target body, so I'm currently most interested in the case where I am already in a near-circular orbit in the same plane as that of my target body.

It is the next step that I'm confused about. I see two approaches.

  1. Fire engines in prograde direction at periapsis to gain enough energy to reach the vicinity of target body.

  2. Fire engines in retrograde direction at apoapsis to lower the periapsis to the lowest point possible before entering atmosphere. (this step is "undoing" some of the work done in entering the original circular orbit; If the planet we're orbiting is the one we launched from, it is possible to angle the launch to match target inclination and produce this elliptical orbit from the beginning. But let's not assume this is possible for now.) At the new lowered periapsis, the prograde engine burn is more efficient.

I am ignoring the fact that these will have us arriving at the target at different times; going back around a few times should eventually produce a capture opportunity.

So I believe what this means is that approach #2 may be preferable if it can be established that the same target distance can be reached by #2 using less total engine delta-v than with approach #1.

I started out by thinking that perhaps the most mathematically simple way to attack this is by observing the rate of orbital energy change introduced by these burns. How much delta-v does it take to lower periapsis to half its current location from a circular orbit? This must be balanced with the amount of extra energy rate gained by engine burn at periapsis, but does halving the periapsis mean that the delta-v here produces 4 times as much energy because orbital speed is 4 times as high? Or is it 16 times?

But then I realized that a circular orbit of radius N has a lot more orbital specific energy than a highly elliptical orbit with apoapsis N. I'm not sure I understand the proof, but the Hohmann transfer (e.g. approach #1 forms the first half of a Hohmann transfer) is apparently the most efficient way to gain a larger circular orbit. However I'm not interested in the second burn. The second burn is going to be for establishing an orbit around the target body and that should hopefully be negligible. It would not be predictable anyway.

I can't really even make up my mind as to whether it's always, sometimes, or never worth it to do this!

Update: I thought about it a little more, and it is possible that a Hohmann transfer is more or less necessary because having very low orbital speed at apoapsis for target body encounter is not conducive to establishing an orbit around it: the speed must be a close match in order to achieve gravitational capture. In this case the answer to my question would be "Probably, but your assumptions are flawed". It is possible that if we are in an orbit that is actually retrograde to your target body that this sort of high-eccentricity attack would reduce the backwards transverse motion upon arrival at apoapsis, so that's one thing I've learned (stumbled upon) today.

The original question still stands, can we get our orbit to swing out farther by intentionally reducing periapsis?


1 Answer 1


Let's generalize your idea and see if it can be more propellant-efficient, at least in principle.

Call your two orbits $\mho_1$ and $\mho_2$. Both have semi-major axis $a_2$ and $a_2$ and inclinations $i_1$ and $i_2$ respectively. As per the problem,

$$a_1<a_2$$ $$i_1=i_2$$

and any phasing issues may be ignored. Also,

$$r_{p1} = r_{a1}$$ $$r_{p2} = r_{a2}$$

i.e., the apocentre distance $r_a$ and pericentre distance $r_p$ are equal, since $\mho_1$ and $\mho_2$ are circular orbits.

Let's also make standard assumptions:

  • Changes in speed are instantaneous
  • Thrusts are given exactly parallel to the flight direction

Now let's establish the $\Delta V$ requirements for the Hohmann transfer. The Hohmann transfer requires two burns:

  1. One burn anywhere to insert the spacecraft into the elliptical Hohmann transfer orbit (the HTO), with pericentre at altitude $a_1$ and apocentre at altitude $a_2$.
  2. One burn at HTO apocentre to equalize the spacecraft's speed with that of $\mho_2$.

Given that

$$r_p^{HTO}=a_1$$ $$r_a^{HTO}=a_2$$

and that generally, for any orbit,

$$a = \frac{r_p+r_a}{2}$$ $$V = \sqrt{\mu\left(\frac{2}{r}-\frac{1}{a}\right)}$$

(which is the vis viva equation), it is straightforward to deduce that

$$V_p^{HTO} = \sqrt{2\mu\left(\frac{-a_1+a_2}{a_1(a_1+a_2)}\right)}$$ $$V_a^{HTO} = \sqrt{2\mu\left(\frac{a_1-a_2}{a_2(a_1+a_2)}\right)}$$

Together with the circular speeds of $\mho_1$ and $\mho_2$,

$$V_{c1} = \sqrt{\frac{\mu}{a_1}}$$ $$V_{c2} = \sqrt{\frac{\mu}{a_2}}$$

we can derive the magnitudes of the $\Delta V$ required:

$$\Delta V_1^{HTO} = \left| V_p^{HTO}-V_{c1} \right| = \sqrt{\frac{\mu}{a_1}}\left(\sqrt{\frac{2a_2}{a_1+a_2}}-1\right)$$ $$\Delta V_2^{HTO} = \left| V_a^{HTO}-V_{c2} \right| = \sqrt{\frac{\mu}{a_2}}\left(1-\sqrt{\frac{2a_1}{a_1+a_2}}\right)$$

and of course,

$$\Delta V_{HTO} = \Delta V_1^{HTO} + \Delta V_2^{HTO} $$

Now we do the same sort of analysis for your other idea. Call your idea a three-burn transfer (TBT). The TBT follows essentially the same steps, save for the fact that there's now 3 burns instead of 2:

  1. One burn anywhere to insert the spacecraft into an elliptical transfer orbit, with pericentre at altitude $r_3 < a_1$ and apocentre at altitude $a_1$. The semi-major axis $a_3 = (r_3+a_1)/2$. Call this orbit $\mho_3$.
  2. One burn at the pericentre of $\mho_3$, to insert the spacecraft into another elliptical transfer orbit, with pericentre altitude $r_3$ and apocentre altitude $a_2$. The semi-major axis is $a_4 = (r_3+a_2)/2$. Call this orbit $\mho_4$.
  3. One burn at the apocentre of this transfer orbit, to equalize the spacecraft's speed with that of $\mho_2$.

The analysis follows precisely the same steps as before, to arrive at the following:

$$\matrix{ \Delta V_1 &=& \left| V_a^{\mho_3}-V_{c1} \right| &=& \sqrt{\frac{\mu}{a_1}}\left(1-\sqrt{\frac{2r_3}{a_1+r_3}}\right)\\ \Delta V_2 &=& \left| V_p^{\mho_4}-V_p^{\mho_3} \right| &=& \sqrt{\frac{2\mu}{r_3}}\left(\sqrt{\frac{a_2}{r_3+a_2}}-\sqrt{\frac{a_1}{r_3+a_1}}\right)\\ \Delta V_3 &=& \left| V_a^{\mho_4}-V_{c2} \right| &=& \sqrt{\frac{\mu}{a_2}}\left(1-\sqrt{\frac{2r_3}{r_3+a_2}}\right)}$$

also as before,

$$\Delta V_{TBT} = \Delta V_1 + \Delta V_2 + \Delta V_3 $$

Rearranging all this, your question boils down to solving the following inequality for $r_3$:

$$\Delta V_{HTO} > \Delta V_{TBT} $$

$$ \rightarrow V_{c1}\left(\sqrt{\frac{2a_2}{a_1+a_2}}-1\right) + V_{c2}\left(1-\sqrt{\frac{2a_1}{a_1+a_2}}\right) > \\V_{c1}\left(1-\sqrt{\frac{2r_3}{a_1+r_3}}\right) + V_{\mathrm{esc},3}\left(\sqrt{\frac{a_2}{r_3+a_2}}-\sqrt{\frac{a_1}{r_3+a_1}}\right) + V_{c2}\left(1-\sqrt{\frac{2r_3}{a_2+r_3}} \right)$$

Making a few substitutions for brevity:

$$ V_{c1} A + V_{c2} B > V_{c1} C + V_{\mathrm{esc},3} D + V_{c2} E $$

Observe that

$$ V_{\mathrm{esc},3}D \rightarrow V_{c1}A$$ $$ E \rightarrow B$$ $$ C \rightarrow 0$$

when $r_3 \rightarrow a_1$. It is also easy to see that

$$ D \rightarrow 0 $$ $$ E \rightarrow 1 $$ $$ C \rightarrow 1 $$

for $r_3 \rightarrow 0$. Since $0<A<1$ and $0<B<1$, that means $\Delta V_{TBT} > \Delta V_{HTO}$.

It is also easy to see that the transition between these two extreme states is smooth and monotonous.

In other words, the value of the right hand side always exceeds the value of the left hand side. Therefrore, the three-burn transfer with $r_3 < a_1$ you propose, is never more efficient than the Hohmann transfer.

It is tempting to think that the same equations extend directly to cases where $r_3 > a_1$. If this were true, it would result in more efficient transfers than Hohmann. However, then you're forgetting that the absolute signs have been replaced by a proper ordering of the terms, which only works if the first term is greater than the second. This ordering is reversed when $r_3 > a_1$, flipping all the signs so that the total $\Delta V_{TBT}$ increases with increasing $r_3$. It is only at $r_3 = a_1$ that the minimum energy is encountered, i.e., the Hohmann transfer.

Three-burn-transfers still have their uses. For instance, multi-burn transfers can be used to facilitate that little "phasing" problem you stepped over in the beginning :) Or execute the transfer with less acceleration per burn, useful for acceleration-sensitive payloads. Or, as other calculations will show, it is way more efficient to use a three-(or more)-burn-transfer if the two orbits $\mho_1$ and $\mho_2$ have significantly different inclinations. It might also be beneficial if the other orbital elements are different. But whole books have been written on this subject, so let's leave that for the next question :)

  • $\begingroup$ > "it is way more efficient to use a three-(or more)-burn-transfer if the two orbits $\mho_1$ and $\mho_2$ have significantly different inclinations." True, but more generally, three burns can be more efficient if the two orbits are in significantly different planes - or, even more generally, if the two orbits have significantly different vector angular momentum. $\endgroup$ Commented May 5, 2018 at 16:05
  • $\begingroup$ @JimVanZandt "Different inclinations" equals different orbital planes & different angular momentum vectors; these are all different wordings of the same underlying concept. $\endgroup$ Commented May 6, 2018 at 2:57

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