When playing around with the NASA trajectory browser, I noticed that practically every mission to the Jupiter system followed the mold (order of magnitude of $\Delta V$):

  1. Start in Earth Orbit (of course), into Orbit around the Sun (5 km/s)
  2. Deep-Space Maneuver (DSM) to change course for an Earth flyby (500 m/s)
  3. Earth flyby with further change in velocity (100 m/s)
  4. Onward!


Now, I understand that acceleration is more efficient when done in a gravity well, and that changes in direction require less fuel when the craft is moving fast - as is the case during the flyby.

But I can't figure out why the $\Delta V$ applied in the flyby can't be applied the same way with the first acceleration, as the position in the gravity well and the speed relative to Earth will be almost the same.

So, why Earth flybys?

  • $\begingroup$ Do the $\delta V's$ refer to the on-board fuel requirements needed to achieve a slingshot trajectory, or to the "free" change in velocity that the slingshot produces? Or both? $\endgroup$
    – DJohnM
    Commented Nov 15, 2013 at 0:31
  • $\begingroup$ @DJohnM There's no speed to be gained with respect to the Earth during passive Earth flybys, all you can do passively is vector direction change, not its amplitude (see Pearson's answer, and incidentally, that's also why orbits work, if that helps you visualize it). What speed gain there is relative to the flyby body, is done propulsively and with the help of Oberth effect, which makes any $\Delta v$ burns more efficient. Conversely, this works the other way around for orbital inclination/eccentricity change (easier with less momentum wrt focus). $\endgroup$
    – TildalWave
    Commented Oct 29, 2015 at 14:56

2 Answers 2


I believe what's confusing you is that the NASA trajectory browser is applying your query parameters more literally than you're likely reading them. You see, your query has, among other parameter values, two values that would likely make it seem that your observations are indeed correct:

  • Max duration: 20 years (Maximum acceptable total mission duration.)
  • Minimize: $\Delta V$ (Parameter to minimize if more than one trajectory per destination is found to meet above constraints.)

I.e. because your query selection was to optimize for $\Delta V$, the tool took as many maneuvers into account to help lower $\Delta V$ requirements as it fit the mission profile (duration), even if the benefits were really small.

This appearance that the additional Earth flybys are necessary changes drastically, if you simply change the Minimize query parameter to Duration. It also seems that the query results are limited in number (I wasn't able to get more than 13 results with any query), so that would further complicate matters, by not even listing more direct trajectory options with your example query.

In fairness to NASA Ames Research Center though, they do provide the Trajectory Browser User Guide that seems easy enough to read. ;)


The way to think about this is the way that all gravity slingshots work. NASA has provided a nice graphic to illustrate what happens:

enter image description here

In order for speed to be gained, you have to have an eternal reference, otherwise speed simply cannot be gained. As you can see, the relative speed of the ball will be 130 mph in this example. If you throw the ball from the train, you will have a similar speed. If you could imagine a situation by which you then bounce off of a train going the opposite direction, and you could have one go even further. But you have to start from a solar orbit first.

  • $\begingroup$ To keep with the analogy, of the train and the ball ... when I'm starting at earth, it's like I'm throwing from the train anyway. $\endgroup$
    – mart
    Commented Nov 15, 2013 at 8:15
  • 3
    $\begingroup$ ... but after 6 months, the train is going the same speed in the opposite direction $\endgroup$ Commented Sep 17, 2015 at 0:42

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