Suppose that there are two planets of roughly the same volume and mass are orbiting each other. What would be the most efficient way to travel from one to the other? In other words, what kind of orbits, and connecting rocket-engine burns, would require the least amount of $\Delta v$ to do this transfer?

It seems unlikely, but could a rocket positioned at the center of the near side of one of the planets just simply travel upwards? Or is it pretty much the same maneuver as getting to the Moon? If so, would having the center of rotation closer to the center of the two bodies have an effect on fuel consumption?

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    $\begingroup$ Keep in mind that two such planets would likely become rotation locked, so the same side of each planet is always facing the other. This is just like one side of the moon is always facing the earth. Launching near the equator on the far side of one planet therefore sounds like the lowest energy way to get to the other planet. $\endgroup$ Commented Oct 7, 2013 at 12:49
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    $\begingroup$ Why do you care about $\Delta v$? If you are interested about efficiency you should ask about ways to minimize the necessary fuel and maybe also minimize the flight time. $\endgroup$
    – Noumeno
    Commented Dec 15, 2020 at 14:03
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    $\begingroup$ @Noumeno $\Delta v$ is the correct measure of fuel consumption. This is elementary material - if it isn't clear then start with Wikipedia or ask a separate question. $\endgroup$ Commented Dec 16, 2020 at 7:52

5 Answers 5


I worked out the outline of thee solution to this problem using Newtonian mechanics and vectors. This is not an easy problem because it is a three body problem. For simplicity I have made a few assumptions: 1. Mass of the spaceship is negligible and the orbits are perfect circles.

Working out the final solution requires some tedious calculation. Hopefully, the solution gives you an idea about how to solve it rather than the exact answer.

I believe that this question can be solved using Lagrangian and Lagrange multiplayer in an easier way.

The outline of the solution is here.(https://i.sstatic.net/lMGKJ.jpg)


I'm going to give an intuition-based answer, because - based on how the question was worded - it appears that is what the OP wants. I deleted my previous answer because there are too many changes in this version to really qualify as edits.

If the planets are the same mass, tidally locked and in circular orbits, there are two points - point A1 on Planet A and point B1 on Planet B - that are closest together. There are also two points that are farthest apart: A2 on Planet A and B2 on Planet B. The points A2 and B2 are moving slightly faster than the orbital velocity of the planets. Let's call the points A1 and B1 the "north poles" of their respective planets. A point on the equator of A1 is moving at precisely the orbital velocity of its planet.

If the planets have a size but no mass, then the lowest delta-v trajectory would begin on the equator of A, go straight away from the center of A, and would use practically no delta-v at all. It would be a nearly circular elliptical orbit that, after many orbital cycles of the two planets, would reach planet B. This is because its orbital period would be slightly longer than the periods of the circular orbits of the planets, so its orbit would "kiss" the orbital paths of the two planets, with the "kiss" point moving a bit closer to planet B after each orbit. With the right timing, it will eventually kiss planet B's orbital path at the same moment and location when and where planet B is there.

If the planets have mass, there would have to be enough of an initial boost to reach escape velocity from Planet A. The exact answer, in a case where the planets have a significant mass, is probably substantially different. I suspect that it would involve a trajectory that goes through Lagrange points L4 or L5 (a condition that is at least very nearly met by the massless solution).

Launching from a point on planet A somewhere south of the equator would take advantage of the excess speed at the south pole; launching from somewhere north of the equator would take advantage of the speed shortage at the north pole. Either would effectively reduce the escape velocity from the planet, but only by a small amount. Depending on the size of the planets and the orbital radii, the optimum launch angle is likely to be affected.

Other factors are the atmospheres of the planets and the acceleration, mass, efficiency, etc., of the rocket itself.

Last note: The massless solution above assumes you don't care how long the journey takes; so to make the problem well-formulated, you should impose a maximum journey duration in terms of orbital periods.


Okay, so we have a quantity to minimize, namely the total $\Delta v$, and we need to choose a function that minimizes it. This sounds like a problem for the calculus of variations. By analogy with the Action and the Lagrangian in classical mechanics, we'll refer to the total $\Delta v$ as $S$, and the $\Delta v$ per unit time as $L$. With no gravitational field, the $\Delta v$ per unit time is just the magnitude of the acceleration, so we have:

$$ S = \int |\ddot {\vec x}| dt $$

If there's a gravitational field, $\vec g$, then we might be fighting it, or it might be helping us out. So that will contribute to the $\Delta v$ that our rocket must expend in the following way (we subtract off the part of the acceleration that is due to gravity):

$$ = \int |\ddot {\vec x} - \vec g| dt $$

Note that unlike the Lagrangians commonly encountered in classical physics, this one depends on the second time derivative of position, rather than just the first. Since I'm too lazy to work out the Euler-Lagrange equations for a Lagrangian of this kind, I just took it from wikipedia:

$$ \frac{\partial L}{\partial x_i} - \frac{d}{dt}\frac{\partial L}{\partial\dot x_i} + \frac{d^2}{dt^2}\frac{\partial L}{\partial\ddot x_i} = 0 $$

Let's do some derivatives to find out what each of the terms in these equations is:

$$ \frac{\partial L}{\partial x_i} = \frac{\partial |\ddot {\vec x} - \vec g|}{\partial x_i} = \frac{\partial \sqrt{\ddot {\vec x}^2 - 2\ddot{\vec x}\vec g+\vec g^2}}{\partial x_i} $$ $$ =\frac{2\left(\frac{\partial}{\partial x_i}\vec g\right) \cdot \vec g - 2\left(\frac{\partial}{\partial x_i}\vec g\right) \cdot \ddot{\vec x}} {2L} =\frac{\left(\frac{\partial}{\partial x_i}\vec g\right) \cdot (\vec g - \ddot{\vec x})} {L} $$

$$ \frac{\partial L}{\partial \dot x_i} = 0 $$

$$ \frac{\partial L}{\partial \ddot x_i} = \frac{\partial |\ddot {\vec x} - \vec g|}{\partial \ddot x_i} = \frac{\partial \sqrt{\ddot {\vec x}^2 - 2\ddot{\vec x}\vec g+\vec g^2}}{\partial \ddot x_i} $$ $$ =\frac{2\left(\frac{\partial}{\partial \ddot x_i}\ddot{\vec x}\right) \cdot \ddot{\vec x} - 2\left(\frac{\partial}{\partial \ddot x_i}\ddot{\vec x}\right) \cdot \vec g} {2L} =\frac{\ddot x_i - g_i} {L} $$

So, multiplying both terms by $L$, our Euler-Lagrange system of equations is now:

$$ \left(\frac{\partial}{\partial x_i}\vec g\right) \cdot (\vec g - \ddot{\vec x}) + \frac{d^2}{dt^2}\left(\ddot x_i - g_i\right) = 0 $$

That's a hell of a 4th order PDE; I'd recommend a numerical solution. Remember that since the planets are moving around each other, $\vec g$ is a function of both position and time. Also, you'd likely want to specify the starting and ending points with the desired positions and velocities, so you'll need some kind of self consistent equation solver. It may actually be easier to just minimize $S$ directly using gradient descent on a discretized path.


You can go through or near the Lagrange points (https://en.wikipedia.org/wiki/Lagrange_point), where you sorta move like Tarzan from one swing (gravitational pull of one planet) to another swing (gravitational pull of another planet). You can even connect them to form the so-called interplanetary highways (https://en.wikipedia.org/wiki/Interplanetary_Transport_Network).


I would say that the most efficient way of completing this travel is to launch the rocket in the opposite direction of the departure planet, to get to the gravitational field of the arrival planet as soon as possible.


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