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

In summary, the slingshot effect relies on the velocity of the planet in the view of an observer and is only useful if the planet is moving in the general direction of the intended final direction of the object. To calculate the final velocity, you would add the velocity vectors of the object and the planet in the same frame of reference. The maximum boost can be achieved when the probe does a 180° in the system fixed to the star, resulting in a velocity of 23km/s. However, this may not be practical for launches from Earth and the direction of the probe must change in order to gain a boost.
  • #1
zeromodz
246
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?
 
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  • #2
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
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.
 
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  • #4
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:

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

Where [itex]\theta[/itex] is change of direction in planet's coordinate system, and [itex]v_p[/itex] is planet's orbital velocity.
 
  • #5
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.
 
  • #6
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
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.
 
  • #8
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.
 
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  • #9
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.
 

1. What is the slingshot effect?

The slingshot effect, also known as the gravitational assist or the gravity assist, is a maneuver used by spacecraft to gain speed and change direction by utilizing the gravitational pull of a planet or other celestial body.

2. How is the G force calculated in the slingshot effect?

The G force in the slingshot effect is calculated using the equation F = m * a, where F is the force, m is the mass of the spacecraft, and a is the acceleration. The acceleration is calculated by taking into account the initial velocity of the spacecraft, the gravitational force of the planet, and the angle of the spacecraft's trajectory.

3. What factors affect the G forces experienced in the slingshot effect?

The G forces experienced in the slingshot effect are affected by several factors, including the mass of the spacecraft, the velocity and gravitational force of the planet, and the angle of the spacecraft's trajectory. The speed and direction of the spacecraft before and after the maneuver also play a role in determining the G forces.

4. How can one predict the G forces in a slingshot maneuver?

To predict the G forces in a slingshot maneuver, one can use mathematical equations and simulations that take into account the various factors mentioned previously. These calculations can help determine the maximum G forces that will be experienced by the spacecraft during the maneuver.

5. Are there potential risks associated with high G forces in the slingshot effect?

Yes, there are potential risks associated with high G forces in the slingshot effect. These include structural damage to the spacecraft, potential harm to any living organisms on board, and a decrease in the accuracy of the spacecraft's instruments. It is important for engineers to carefully calculate and monitor the G forces during a slingshot maneuver to minimize these risks.

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