This question is inspired by Do you need 0 km/s velocity to crash into the sun? which brings up the peculiarity (at least to the layman) of Solar System escape from 1 AU taking less $\Delta V$ than a direct transfer into the Sun.

At least one answer mentions using Jupiter to reduce the $\Delta V$ requirement. I have some experience working with the wonders of a Jupiter gravity assist (here), even showing an example trajectory that impacts the Sun.

I am curious if such an Earth-Jupiter trajectory exists that, depending on the direction of gravity assist at Jupiter, would put a spacecraft on either a Solar System escape trajectory (heliocentric $C3 > 0$) or a Sun impact trajectory.

  • 2
    $\begingroup$ This will make for an interesting story - the aliens from Europa put the heroin's ship on a trajectory into the Sun and escape back to Europa. Faced with limited options, she figures out a way to get enough delta-v (an all-or-nothing option) to instead escape the solar system - hoping the Europans will still see her destined to die, only difference being freezing rather than burning up. However, what the Europans don't know is that... $\endgroup$
    – uhoh
    Commented Jul 5 at 3:47
  • 3
    $\begingroup$ My gut tells me the answer must be yes, but I'd be curious to see someone else prove it, so +1. $\endgroup$
    – Erin Anne
    Commented Jul 5 at 4:24
  • 1
    $\begingroup$ My gut contradicts and says it's unlikely because you need to approach Jupiter from very different angles for the two types of trajectory. Let's see what math has to say about it... $\endgroup$
    – asdfex
    Commented Jul 5 at 10:45
  • 1
    $\begingroup$ @asdfex a slight change in velocity could change the timing of closest approach with Jupiter enough that in one case you approach "from behind" and get a speed boost that takes you out of the solar system, and in other you approach "from in front" and get a speed reduction that sends you crashing into the Sun. $\endgroup$ Commented Jul 5 at 13:27
  • 1
    $\begingroup$ @ScienceSnake That would work if you approach Jupiter radially, but in reality the common swing by maneuvers "chase" Jupiter from behind, which is bad for a dive towards the Sun. So my feeling is a radial approach doesn't generate enough change in velocity. But as I said, just a feeling, we need someone doing the math. Is the anywhere a sketch of the originally planned trajectory of the Parker Solar Probe? That could give us some hints. $\endgroup$
    – asdfex
    Commented Jul 5 at 13:36

1 Answer 1


I did a search for Earth-Jupiter-Sun trajectories launching in the next 3 years or so (1200 days from 01-Jan-2024). Filtered to a $C3 < 150$ $km^2/s^2$ & a viable flyby at Jupiter (flyby close approach distance above 1 Jovian radius, $R_J$):

viable Leg 1 Earth-Jupiter porkchop plot

(Personal work, 2024 launch window subset of total search)

The required flyby close approach distance & deflection angles for the flyby of Jupiter are as follows:

Jupiter flyby close approach distances Jupiter flyby deflection angles

(Personal work, individual viable trajectories)

Next I used the first leg of these viable trajectories and assumed a gravity assist at Jupiter that would direct the outgoing asymptote in line with Jupiter's prograde motion in order to maximize the heliocentric velocity (and increase chances of achieving Solar System escape).

For all of these trajectories we see a positive heliocentric $C3$:

heliocentric C3 post Jupiter flyby

(Personal work)

All of these Solar System escape trajectories have a viable, albeit close, gravity assist at Jupiter:

flyby distance of Jupiter

Therefore all Earth-Jupiter trajectories in this search can both escape the Solar System & impact the Sun depending on the gravity assist at Jupiter.

Interestingly, the required flyby distances overlap between the two trajectories at around 2.5 $R_J$:

similar flyby distances

(Personal work)

The flybys of course occur on more or less opposite sides of Jupiter, but it is neat nonetheless. A particular Earth-Jupiter trajectory:

  • Launch Date: 14-Aug-2024
  • Jupiter Flyby Date: 09-Nov-2025 (452 day transfer time-of-flight)
  • Launch C3: 140 $km^2/s^2$, Flyby $v_{\infty}$: 14.5 km/s
  • Flyby Deflection: 104°, Flyby Close Approach: 175,200 km (2.45 $R_J$) (same for both!)
  • Sun Impact Date: 01-Apr-2027
  • Heliocentric Orbit C3 (Earth-Jupiter): -76 $km^2/s^2$ (not Solar System escape)
  • Heliocentric Orbit C3 post flyby: 278 $km^2/s^2$

Here is a video showing these trajectories:

(Personal work)

You may argue that these are two different Earth-Jupiter trajectories, and it's technically true; however, the $\Delta V$ to move from one side to the other of Jupiter is only about $20$ $m/s$ to a first order approximation (moving 2 times to impact parameter, 900,000km, over the course of the 452 day flight time requires a linear velocity of 23 m/s).

  • 2
    $\begingroup$ lovely video. Makes me think of a real tandem launch. $\endgroup$
    – Erin Anne
    Commented Jul 7 at 5:09
  • 1
    $\begingroup$ well done good answer $\endgroup$
    – Slarty
    Commented Jul 7 at 8:51
  • 1
    $\begingroup$ It seems current launchers can get about 2t on such a trajectory and SLS block 2 with Centaur could send about 7 t. Not too bad. $\endgroup$
    – asdfex
    Commented Jul 7 at 11:52
  • 1
    $\begingroup$ I wonder if there is another solution: Pass Jupiter on the same side, but aim for missing the Sun slightly. Given the high C3 this should be a escape trajectory. $\endgroup$
    – asdfex
    Commented Jul 7 at 15:45
  • 2
    $\begingroup$ kudos on the self-answer too. I figured you would, given what you'd shown in the question, but it's a nice Q+A pair. $\endgroup$
    – Erin Anne
    Commented Jul 7 at 17:08

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.