Comments below How much less delta-v would it take to reach the Sun using Venus and Earth flyby's compared to direct? tend to suggest that one way to send a spacecraft into the Sun (either straight in or in an orbit so tight that it touches the Sun's photosphere) would be to get to Jupiter and use a gravitational assist there.

Ulysses used Jupiter to enter into a polar heliocentric orbit, but that was basically a "right angle turn". Traveling from Earth or Venus to Jupiter, then heading into the Sun is a lot closer to 180 degrees.

Question: Is this trajectory possible, or is the turn so tight that it intercepts the surface of Jupiter? Please use math or show a believable simulation or an analogous actual trajectory; don't just say "yes, it is". Thanks!

Orbits of Earth (green, small), Jupiter (red, large), and Ulysses (blue, vertical) from JPL's Horizons.

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    $\begingroup$ Pre-coffee, sorry. $\endgroup$ Commented Sep 7, 2019 at 16:06
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    $\begingroup$ The key would be to inject to an aphelion somewhat beyond Jupiter and make the gravity assist on the inbound leg (contrary to the traditional Voyager or New Horizons outbound-assist maneuver) but I haven’t found the resources to quantify that. Saturn’s aphelion is only 7.3km/s from LEO, and I suspect that’s more than enough. $\endgroup$ Commented Sep 7, 2019 at 16:11
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    $\begingroup$ @RussellBorogove I want to try some of that coffee, it seems to work quite well! $\endgroup$
    – uhoh
    Commented Sep 7, 2019 at 16:20
  • $\begingroup$ @RussellBorogove ha! my own question. Yep, those answers definitely answer this question. I think it's fine to start closing as duplicate. Good find, thanks! (signing off) $\endgroup$
    – uhoh
    Commented Sep 7, 2019 at 16:34
  • $\begingroup$ Note that making a right angle turn is more expensive than killing your orbital velocity. Ulysses thus proves the orbit is possible. $\endgroup$ Commented Sep 8, 2019 at 0:28

1 Answer 1


Sample trajectory.

We want a final orbit with a perihelion touching the Sun, and an aphelion touching Jupiter's orbit. Using vis-viva, the velocity at aphelion is 0.55 km/s

Jupiter itself is travelling at 13.07 km/s, so that means that we want a $v_{\infty}$ of 12.52 km/s, escaping from Jupiter in retrograde direction.

To get this $v_{\infty}$, we can launch towards Jupiter in a transfer orbit with perihelion at Earth orbit, and aphelion at 1.66 times Jupiter's orbital radius. This will have a horizontal component of 7.67 km/s and vertical component of 9.89 km/s relative to the Sun when arriving at Jupiter.

Transforming this into a velocity relative to Jupiter, we will arrive at exactly 12.52 km/s, 28.6 degrees from the prograde direction.

Is a turning angle of 28.6 degrees possible at 12.52 km/s $v_{\infty}$? Yes.

The turning angle ($\alpha$), given flyby periapsis ($P$), mass parameter ($\mu$) and $v_{\infty}$ is:

$$\alpha = -2\sin^{-1}\left(\frac{-1}{1 + \frac{Pv_{\infty}^2}{\mu}}\right)$$

At a 100,000km periapsis (above the surface with some margin) and the given $v_{\infty}$ , the maximum turning angle is 125.7 degrees, so much more than this flyby requires. The example trajectory thus exists.

The $\Delta v$ cost of the initial transfer orbit is somewhere between a Jupiter transfer and a Saturn transfer, so around 6.3 - 7.3 km/s if using a direct transfer. If using an initial Venus and Earth flyby pattern, the cost is closer to a Venus transfer at 3.5 km/s

  • $\begingroup$ Note that this question was from my comment on anther question and I was suggesting using Venus and Earth flybys to get the boost to Jupiter. Thus you only need to leave with a Venus transfer orbit (and probably some small deep space burns), not anything like the numbers you give. $\endgroup$ Commented Sep 8, 2019 at 0:26
  • $\begingroup$ Thank you for the interesting answer! This one doesn't "wrap around Jupiter" "then (head) into the Sun" turning "(close) to 180 degrees." So it looks like the 100,000km perijove providing a maximum turn angle of 125.7 degrees really means that the orbit I've asked about is impossible, and the turn is "so tight that it intercepts the surface of Jupiter". $\endgroup$
    – uhoh
    Commented Sep 8, 2019 at 2:25
  • $\begingroup$ So I think that if you add something like "the 'wrap-around-Jupiter' orbit you've requested isn't possible because it would indeed intersect the planet, but that's not what's needed from Jupiter to reach the Sun" then this will be a perfect answer. $\endgroup$
    – uhoh
    Commented Sep 8, 2019 at 2:41
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    $\begingroup$ 180 degree turns don't exist. For all intents and purposes, this "wraps around Jupiter and heads straight towards the Sun", why would you expect that angle to be 180 degrees? $\endgroup$ Commented Sep 8, 2019 at 6:17
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    $\begingroup$ I think the Ulysses diagram has caused a fundamental misunderstanding. Ulysses uses Jupiter to go 90 degrees out of the plane, so you are assuming that using Jupiter to slow down means wrapping 90 degrees more, "going in the opposite direction". That's not the case. a flyby to slow down means not leaving the plane at all, leaving the inclination angle at 0 through the whole manoeuvre. How much the velocity vector turns in the plane while flying past Jupiter is not the same as how much the velocity vector turns relative to the Sun, and you seem to be mixing up the two. $\endgroup$ Commented Sep 12, 2019 at 12:17

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