3
$\begingroup$

Many papers mention that the compact and complex dynamics of the Jupiter and Saturn systems makes them ideal low energy transfer. My question is how?

How does the dynamics of the Jupiter and Saturn system make them be well suited for low-energy transfer?

For reference: https://repositories.lib.utexas.edu/handle/2152/68928

$\endgroup$
2
  • $\begingroup$ Can you link to one of the many papers? $\endgroup$ Commented Feb 4, 2020 at 19:50
  • $\begingroup$ @OrganicMarble Sure! I have edited the question and added a snippet. $\endgroup$
    – John
    Commented Feb 4, 2020 at 19:53

2 Answers 2

3
$\begingroup$

If the orbiting body's mass is a significant fraction of the central body's mass, the weak stability boundaries can be more dramatic.

Call the mass of the central body + orbiting body 1. Call the orbiting body's mass µ. Then the central body would have mass 1-µ.

Here are pairs arranged in order of µ

Pluto/Charon      1.043E-01  
Earth/Moon        1.216E-02  
Sun/Jupiter       9.545E-04  
Sun/Saturn        2.856E-04  
Saturn/Titan      2.374E-04  
Jupiter/Ganymede  7.789E-05  
Jupiter/Callisto  5.684E-05  
Sun/Neptune       5.153E-05  
Jupiter/Io        4.700E-05  
Sun/Uranus        4.366E-05  
Jupiter/Europa    2.526E-05  
Saturn/Rhea       4.046E-06  
Sun/Earth         3.039E-06  
Sun/Venus         2.448E-06  
Saturn/Dione      1.935E-06  
Saturn/Tethys     1.091E-06  
Sun/Mars          3.229E-07  
Saturn/Enceladus  1.935E-07  
Sun/Mercury       1.659E-07  
Saturn/Mimas      7.037E-08  
Mars/Phobos       1.682E-08  
Sun/Pluto&Charon  7.149E-09  
Mars/Deimos       2.803E-09  
Sun/Ceres         4.741E-10

Jupiter and Saturn have some big moons. You'll find a lot of the gas giant moons near the top of the list when arranged by µ.

For more on this see my mass parameter and ITN

$\endgroup$
3
  • $\begingroup$ I thought I might be able to get something done today, but then I saw your linked post. I'm looking forward to finding out what a "dramatic weak stability boundary" is ;-) $\endgroup$
    – uhoh
    Commented Feb 5, 2020 at 5:51
  • 1
    $\begingroup$ @uhoh I've been wanting to model Pluto and Charon using my orbital shotgun sims as well as the Galilean moons. I get mesmerized watching the paths unfold. How'd you arrange the pairs in neat columns? 4 spaces at the beginning of each line does it? $\endgroup$
    – HopDavid
    Commented Feb 5, 2020 at 12:52
  • $\begingroup$ yep, four spaces in the beginning provides a "code block" with equal space font, which is also commonly used for convenient display of tabular data as well. Those plots are excellent btw! $\endgroup$
    – uhoh
    Commented Feb 5, 2020 at 13:34
3
$\begingroup$

This is all about gravitational maneuvers. They allow to obtain huge accelerations/deceleration/velocity changes almost without using any energy. More heavy moons - more opportunities for maneuvers.

General idea is that if satellite trajectory at some point goes near heavy body (one of moons), by very small early adjustments from long distance before descending, it is possible to choose at which side of body to pass by and how close to it to descend. Closer to moon -> bigger effect of gravitational maneuver -> bigger amount of free velocity correction (=energy) extracted from maneuver.

By the way, to use this, it is necessary to have entrance hyperbolic trajectory to Saturn/Jupiter systems which will be near as much as possible moons. If there were no moons at all, trajectory would be always hyperbolic, and satellite will always leave gravitational field of Sat/Jup after some time. If on some point of this trajectory to pass by heavy moon, it is possible to use gravitation maneuver for deceleration and to stay in grav field of Sat/Jup just for free.

$\endgroup$
2
  • 2
    $\begingroup$ "More heavy moons - more opportunities for manures." I hope you ment manoeuvres there, I don't see horses being sent out to Jupiter any time soon. $\endgroup$ Commented Feb 7, 2020 at 7:35
  • 2
    $\begingroup$ Oh deer, of course not them. Thanks. $\endgroup$
    – halt9k
    Commented Feb 9, 2020 at 3:14

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.