5
$\begingroup$

Imagine arriving at Saturn, jettisoning your passenger cargo, and then utilizing the massive gravitational pull of the gas giant to slingshot back towards Jupiter.

Due to the immense mass of these celestial bodies, trajectories can be significantly reoriented, potentially sending you back in the direction you came from.

This approach could save considerable time, as opposed to waiting for a cycler to complete an orbit around the sun and reintercept with Saturn in 30 years. Instead, you could springboard back off Jupiter, and meet back up at Saturn in just 4 years, all without using a drop of fuel.

Have there been any theoretical studies examining potential trajectories for this type of maneuver?

$\endgroup$
6
  • 6
    $\begingroup$ Where's the four year timing coming from? $\endgroup$
    – Erin Anne
    Apr 11, 2023 at 22:15
  • 1
    $\begingroup$ How about this: A ping-pong cycler that uses hydrogen propulsion and an aerobrake? You jettison you cargo after dropping into the very top of the atmosphere of saturn, and then use the trace hydrogen you've collected to run through a fission reactor that doubles as crew power support, accelerate to jupiter and then aerobrake and repeat. would probably shorten transfer time considerably. $\endgroup$ Apr 12, 2023 at 23:51
  • $\begingroup$ @ErinAnne Time estimates from voyager 2 mission. This should give an idea of the trajectories and time frame frames I’m looking for. $\endgroup$
    – Enoch
    Apr 14, 2023 at 13:13
  • $\begingroup$ @RegenerativelyCooledAstronaut Ah, see thats the catch! The cycler is too massive for propulsion, too fragile for aerobraking, hence the smaller passenger cargo. We need to meet up with the large cycler again in the future, but we have no way of bleeding off its momentum, so we sling it around and catch up with it when it comes back. $\endgroup$
    – Enoch
    Apr 14, 2023 at 13:46
  • $\begingroup$ why is that necissary? nuclear fission can give some really high twr for a high efficiency, and not only that, your assumption of four years, from your note about voyager 2, is not feasible without propulsion: voyager probes didn't have to turn around, the gas giant slingshots were one-off $\endgroup$ Apr 15, 2023 at 0:56

1 Answer 1

25
$\begingroup$

In a sense, this "ping pong" is what cycler orbits attempt to do, you are just overestimating the power of flybys.

To make the distance between Jupiter and Saturn in just 4 years (one way?), you're doing the trip often enough that you also have to a bounce when the planets are on opposite sides of the Sun, almost 15 AU. Even in a straight line, that requires travelling at about 70km/s.

What limits flybys is the turning angle $\theta$.
If one slowly enters a system, one can bend the trajectory nearly 180º around. But as one enters the system with some velocity $v_{\infty}$, this angle shrinks, until one just flies past the planet so fast that the trajectory is barely bent.

This angle also depends on the mass parameter of the planet $\mu$ and the flyby periapsis distance $r_P$ at the very least limited by the radius of the planet

$$\theta =2\sin ^{ -1 }{ \left( \frac { 1 }{ 1+\frac { r_P{ v }_{ \infty }^{ 2 } }{ \mu } } \right) } $$

Entering Jupiter and Saturn at 70km/s means bending the trajectories no more than 31º and 13.5º respectively, far too little to turn the spacecraft around to nearly the same direction as it came from. Especially the Saturn end is limited, requiring entry velocities no higher than 16km/s to do a 90º degree turn, which even then is still too small an angle for anything near a rapid bounce between the planets.


Cycler orbits at best do a cycle for every synodic period of the two planets they serve

$$T_{synodic} = \frac{1}{\frac{1}{T_1} - \frac{1}{T_2}}$$

Or some integer multiple of this. For Jupiter-Saturn that's close to 20 years.

$\endgroup$
1
  • $\begingroup$ No, this is for one-body assists. I’m asking about multibody studies which have been numerically velocity stitched, hence the use of the word “ping pong”. You get much better results then. $\endgroup$
    – Enoch
    Apr 14, 2023 at 13:35

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.