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

1. Jun 14, 2010

### zeromodz

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?

2. Jun 14, 2010

### Vortico

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.

3. Jun 14, 2010

### cesiumfrog

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: Jun 14, 2010
4. Jun 15, 2010

### K^2

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.

5. Jun 15, 2010

### cesiumfrog

Though the maximal direction change in the planet's frame may be no direction change in the external frame.

6. Jun 15, 2010

### K^2

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.

7. Jun 15, 2010

### Janus

Staff Emeritus
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.

8. Jun 15, 2010

### DaveC426913

This bears repeating. Gs experienced are zero.

Unless of course, you are not satisfied with http://en.wikipedia.org/wiki/Neutron_Star_(short_story)" [Broken] 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: May 4, 2017
9. Jun 15, 2010

### K^2

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. Jun 15, 2010

### DaveC426913

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. Jun 15, 2010

### K^2

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. Jun 16, 2010

### cesiumfrog

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. Jun 16, 2010

### K^2

After the fly-by it's going 17km/s, which is significantly higher than escape velocity.