How would one caclulate the G forces of the slingshot effect?

  • Context: Undergrad 
  • Thread starter Thread starter zeromodz
  • Start date Start date
  • Tags Tags
    Forces Slingshot
Click For Summary

Discussion Overview

The discussion centers around calculating the G forces and speed gained from the slingshot effect when using a planet's gravity, specifically considering scenarios involving Jupiter. Participants explore the theoretical aspects of this gravitational maneuver, including centripetal acceleration, velocity vectors, and the implications of changing direction during a fly-by.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants suggest that centripetal acceleration is relevant for calculating G forces and speed gained during a slingshot maneuver.
  • Others argue that G-force is effectively zero during the fly-by since the projectile remains in gravitational free-fall.
  • A participant proposes that the maximum speed gain is a multiple of the planet's orbital speed, depending on the trajectory and direction of the fly-by.
  • Some contributions emphasize that a change in direction is necessary to achieve a fly-by boost, with the conservation of energy being a key consideration.
  • One participant presents a formula for calculating final velocity based on the change in direction and the planet's orbital velocity.
  • There are discussions about the need for visual aids or diagrams to clarify the mechanics of the slingshot effect.
  • Some participants express uncertainty about the applicability of their calculations to different scenarios, such as atmospheric interactions or varying initial trajectories.
  • One participant reflects on the implications of achieving escape velocity after a fly-by, questioning the relevance of initial conditions and orbital mechanics.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the calculations or implications of the slingshot effect. Multiple competing views remain regarding the mechanics of the maneuver, the significance of G forces, and the conditions necessary for effective speed gain.

Contextual Notes

Limitations include assumptions about initial velocities, the dependence on specific trajectories, and the potential impact of atmospheric effects on the calculations. The discussion also highlights the complexity of the two-body problem in the context of gravitational assists.

zeromodz
Messages
244
Reaction score
0
You know how you can use a planets gravity to speed up. How exactly would you calculate the G force and speed you would gain when say using this effect on Jupiter? Wouldn't you use centripetal acceleration?
 
Physics news on Phys.org
G-force is actually a measure of acceleration, so you cannot gain or lose a significant amount after an object is far away from a planet. The "slingshot effect" fully relies on the velocity of the planet in the view of an observer. This maneuver is only useful if the planet is moving in the general direction of the intended final direction of the object (>90˚ away), otherwise the object decelerates. So to calculate the final velocity, you would add the velocity vectors of the object and the planet, of course with both measurements in the same frame of reference.
 
G-force: zero. Because the projectile stays in gravitational free-fall.

Speed gain: The idea is that the projectile speed (not direction) is practically unchanged relative to the planet. By drawing some very simple diagrams I think you'll find, relative to the solar system, the maximum addition to the speed is a basic multiple of the planet's orbital speed.

These answers presume your projectile needn't skim the atmosphere, I haven't checked how applicable that is for Jupiter. And that's aside from issues like which initial trajectories are practical for launches from earth.
 
Last edited:
You can't get a fly-by boost without changing direction. In fact, the greater the change in direction, the greater the boost.

To a good approximation, fly-by is a two-body problem. Go into coordinate system attached to the planet you are using for fly-by. In that system, due to conservation of energy as the simplest argument, the speed of the object going in and out is the same. Only direction can change. If the direction didn't change, then the velocity is exactly the same, and so it is after we go back to the star system coordinates. No boost.

The absolute maximum boost is if the satellite turns around. In that case, it gains exactly twice the orbital speed of the target planet. In general:

v_f = \sqrt{v_i^2 + 2(v_iv_p+v_p^2)(1-cos\theta)}

Where \theta is change of direction in planet's coordinate system, and v_p is planet's orbital velocity.
 
K^2 said:
You can't get a fly-by boost without changing direction.
Though the maximal direction change in the planet's frame may be no direction change in the external frame.
 
How? I think I might need a picture, or a detailed explanation for that one.

By the way, the formula I derived above assumes initial velocity to be collinear with planet's. It might have to be a 2-angle formula. I'll have to think about it a bit more.
 
K^2 said:
How? I think I might need a picture, or a detailed explanation for that one.

By the way, the formula I derived above assumes initial velocity to be collinear with planet's. It might have to be a 2-angle formula. I'll have to think about it a bit more.

Imagine you are looking "down" at the solar system. You have a planet moving in its orbit to the "left" at 10 km/sec. You have launched a probe from an inner planet so that it arrives at its perigee just a little ahead of the planet and traveling at 3 km/sec. From our viewpoint, both probe and planet are moving to the left. The probe is placed at just the right distance from the planet that it takes a parabolic orbit around it. From the perspective of the Planet, the probe falls in going left to right with a starting velocity of 7 km/sec, whips around the planet and ends up going 7 km/sec right to left when it reaches its starting distance again.

From our viewpoint, the probe ends up with a velocity of 10 +7 = 17 km/sec moving to the left. It has gained 14 km/sec without a net change in direction.
 
cesiumfrog said:
G-force: zero. Because the projectile stays in gravitational free-fall.
This bears repeating. Gs experienced are zero.

Unless of course, you are not satisfied with http://en.wikipedia.org/wiki/Neutron_Star_(short_story)" to slingshot around...

In this story, the target body was so compact that tidal forces came into play. i.e. the front half of the ship fell inward while the back half of the ship was flung outward.
 
Last edited by a moderator:
Janus said:
Imagine you are looking "down" at the solar system. You have a planet moving in its orbit to the "left" at 10 km/sec. You have launched a probe from an inner planet so that it arrives at its perigee just a little ahead of the planet and traveling at 3 km/sec. From our viewpoint, both probe and planet are moving to the left. The probe is placed at just the right distance from the planet that it takes a parabolic orbit around it. From the perspective of the Planet, the probe falls in going left to right with a starting velocity of 7 km/sec, whips around the planet and ends up going 7 km/sec right to left when it reaches its starting distance again.

From our viewpoint, the probe ends up with a velocity of 10 +7 = 17 km/sec moving to the left. It has gained 14 km/sec without a net change in direction.
Hmm... Alright, I did not consider situation where planet overtakes the probe. That's an interesting way of making it work. Thank you.

Though, the maximum change still seems to be when the probe does a 180° in the system fixed to the star. Simple change of the initial probe direction in the above yields 23km/s.
 
  • #10
K^2 said:
Hmm... Alright, I did not consider situation where planet overtakes the probe. That's an interesting way of making it work. Thank you.

Though, the maximum change still seems to be when the probe does a 180° in the system fixed to the star. Simple change of the initial probe direction in the above yields 23km/s.

Yes but what use is that? Is this maximum change simply in terms of its own velocity (eg. reversing), or is it relative to the final target of interest?
 
  • #11
That's a good question. I mean, yes, this is useful in itself, because final heading is where I want to go to reach outer system, but I also have to consider where the satellite was launched from. Inner planets have higher orbital velocity, so launching something in retrograde is tricky. I'd have to see if I can reverse direction first elsewhere, without loosing just as much.

Though, in Janus' example, the object is already past escape velocity for the star after the fly-by (Virial thrm) so I'm not sure there is even a point.
 
  • #12
K^2 said:
Though, in Janus' example, the object is already past escape velocity for the star after the fly-by (Virial thrm) so I'm not sure there is even a point.
Could you explain what you mean? I would have thought that if a planet is "in orbit", then anything at the same place moving in the same direction with only a third of that planet's speed cannot already have escape velocity.
 
  • #13
After the fly-by it's going 17km/s, which is significantly higher than escape velocity.
 

Similar threads

  • · Replies 47 ·
2
Replies
47
Views
5K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 22 ·
Replies
22
Views
3K
  • · Replies 21 ·
Replies
21
Views
4K
  • · Replies 8 ·
Replies
8
Views
3K
Replies
4
Views
2K
  • · Replies 4 ·
Replies
4
Views
845
  • · Replies 19 ·
Replies
19
Views
2K
  • · Replies 25 ·
Replies
25
Views
4K