The slingshot effect or gravity assist maneuver produces a boost on the speed of the object which attempts to approach a planet head-on at a speed V while the planet is moving directly toward us at a speed U (both speeds defined relative to the "fixed" Solar frame).

Moon slingshot effect (if possible):

A spaceship is traveling at a speed of v1 at an angle θ towards the moon, which is traveling an an orbital velocity (U) of 1 km/s. Then the spaceship gets away at a much higher speed of v2.

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Could we actually use the moon for a gravity assist maneuver in order to boost up the speed of a spaceship?

  • 1
    $\begingroup$ tl;dr: yes, but it's not very efficient. Your equation lacks mass of the body and distance of flyby; the effect is a speed definitely not "much" higher, and the headaches of rigid timing and trajectory frequently outweigh the benefits of delta-V. $\endgroup$
    – SF.
    Aug 20, 2017 at 15:31
  • $\begingroup$ And thanks a lot for stealing energy from our tides. $\endgroup$ May 5, 2021 at 16:04

1 Answer 1


Your equation is correct, but it does a few assumptions that are seldom applicable to reality. For instance, it ignores the mass of the planet and also how close the fly by is. The later parameter is limited by the radius of the body, or the outer edge of its atmosphere. If we indeed treat the planet as a point mass like that, we can always achieve a "perfect turning angle". That is, the hyperbola of the flyby can always turn completely around to change the direction of the $\vec{v_{\infty}}$ vector. This would a speed boost once the coordinate system is transformed back (for the head-on case, this simplifies to $v_2=v_1+2u).$

The turning angle is however often very limited! It can be calculated in the following way: $$\delta=2\sin^{-1}{\left(\frac{1}{1+\frac{r_p v_{\infty}^2}{\mu}}\right)}$$ We can see directly from this that the turning angle diminishes with low planet mass, high flyby altitude and especially high entry velocity.

Moon example

Head-on, a Hohmann transfer towards the Moon has a relative velocity of around 850m/s. Using the radius of the Moon as the lowest possible flyby radius, the turning angle is then 106°. Not bad, but definitely not a complete 180°. Transforming the coordinate system back, that means the velocity is now ~1500m/s. That is approximately the escape velocity of the Earth system at that altitude.

But. The difference between a Moon transfer and a complete escape is only 90m/s in LEO. That is not much saved. To make matters worse, this is for barely escaping. Usually, you would like to have an injection into a interplanetary transfer from this manoeuvre. That would require a higher initial velocity and therefore a lower turn angle, ultimately decreasing the benefit further. A delta-v saving of less than 90m/s is not worth substantially narrowing launch windows for.

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    $\begingroup$ You're understating it. If you want to leave on an interplanetary trajectory Oberth is worth more than a slingshot around the moon. If you're going to do a big burn do it deep in a gravity well. $\endgroup$ Aug 20, 2017 at 18:04
  • $\begingroup$ I am sorry but I don't really understand the final outcome of all this. That formula is for calculating what angle? I don't get how you get 1500km/s when using the first formula (the one that I brought) it gives 1945km/s. And are you saying it isn't worth using the moon for a slingshot effect? $\endgroup$
    – Matthew
    Aug 21, 2017 at 11:16
  • $\begingroup$ @Matthew Yes, he's saying what you get out of a slingshot from the moon isn't worth much. His numbers show that you get about 90 m/s in benefit in the best case. This is being very optimistic as one normally wants to go somewhere, not merely escape the Earth. Once you take that into account the Oberth effect becomes more important. Do your whole burn in low Earth orbit. $\endgroup$ Aug 22, 2017 at 19:55
  • $\begingroup$ @Matthew one way thinking of the slingshot maneuver is that relative to the body you are slingshotting around you simply turn through a certain angle without changing your speed. So, you might go from approaching Jupiter from "in front" at 10 km/s to leaving it, also in front at 10 km/s, having turned through almost 180 degrees. Seen from the Sun's viewpoint you have gone from going 10 km/s slower than Jupiter to 10 km/s faster. To get something of the Moon's mass to turn you through 180 degrees at useful speed, you need to be closer to its center than the radius of the Moon. $\endgroup$ Sep 27, 2018 at 20:47

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