gravity assist to decelerate space ship

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chebby

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Hi,

is there a maximum speed the ship can have in order to use gravity assist to decelerate?

I am thinking of a hypothetical space ship going to nearest star at a significant fraction of speed of light. Let's say it has very little fuel on board. What are the speed requirements if it is to succesfully decelerate around the destination star using gravity assist and position itself near a planet of interest?

thanks
 
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MeteorWayne

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Sure, you can, a little bit, but as for the OP, you will not be able to decelerate enough to slow you down from a near light speed to an orbit.
 
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origin

Guest
SpeedFreek":jwharz5k said:
Can you use a gravity assist to decelerate?
Gravity assist or the 'sling shot' affect uses the motion of the planet for the acceleration - so if you swing around and head in the direction the planet is moving you will accelerate and if you swing around and head in the opposite direction that the planet is moving you will decelerate.
 
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neilsox

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Traveling at 0.2 c, the interstellar space ship arrives at a solar system with two black holes closely orbiting each other. The lower mass black hole might be orbiting it's more massive companion at 0.3 c, so the the space craft can reduce it's speed to approximately zero by means of a slingshot maneuver very close to the lower mass black hole. The accretion discs better be near none, or the space craft will be damaged. The space craft needs to be very sturdy or the tidal forces will tare it apart. The crew needs to be strapped in perpendicular to the tidal force to avoid being spaghettified. With good luck the craft can now travel to a star that orbits both black holes and has a habitable planet.
Edit: There is another problem: The spacecraft is close to the lower mass black hole for only about a microsecond, so the acceleration of the space craft is perhaps a million g = fatal unless the crew members can move a considerable distance inside their space craft, during that microsecond. Neil
 
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derekmcd

Guest
Gravity assists are not measured by a linear acceleration or deceleration from the spot where the gravity assist occurs. The momentum you gain while approaching the planet is lost as you leave (conservation of momentum).

The acceleration is measured in terms of orbital velocity.

If a spacecraft is traveling away from the sun, the sun's gravity will slow the craft down. Using a gravity assist from an outer planet let's you steal some angular momentum from that planet's own orbital velocity and increases your own. The greater your orbital velocity allows for a 'higher' orbit increasing your distance from the sun without using your own power.
 
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chebby

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Thanks all. This is what I understood from reading your replies and wikipedia, correct me if I'm wrong:

1) spaceship's speed relative to body must be larger but close to the escape velocity at the point of approach of the body used to do this maneuver. If it is too small, it will fall on the body or be captured in orbit. If too large, it will just pass by, barely changing its direction. .2c, for example, is much much larger for anything other then a black hole.

2) Assuming #1 is met, and the ship's speed is larger than escape velocity of the star at desired orbit, it is still possible to put the ship into that orbit by gravity assist IF ship speed does not exceed 2U+V where U is orbital speed of the body orbiting around star and V is less than escape velocity from the star. 2U is maximum that can be stolen away from the ship's speed in order to decelerate it. Final veloctiy can be adjusted by playing with angle of approach.

3) Trying to put yourself in orbit around a planet has nothing to do with any of this, it just requires to be in the sphere of influence of the planet with relative speed below its escape velocity. However, #2 can be used as means of accomplishing this.


Neilsox, I really liked your example. At first I didn't understand why you used black holes, but then I realized that because gravity becomes weak proportionally to square root of the radius, no couple of planets or stars can orbit each other at a speed as large as .3c. I looked up escape velocity of our sun, and it's something like .002c at the surface of the sun (which would be a terrible place to turn around.)
 
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neilsox

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Hi derekmcd: Your post appears to be a contradiction. If you still think it is true, please give us an example such as approaching Venus at 20 kilometers per second; closest approach 23 kilometers per seconds with respect to Venus. Then approaching Earth at 18 kilometers per second with respect to Venus. We are still fast for an aerobraking maneuver, but less dangerous than the 20 kilometers per sec (with respect to Venus) before the Venus encounter. My guess is your craft gave momentum to Venus, so that conservation of momentum is preserved. Since the craft changed direction during the slingshot maneuver, apparently Venus also gained or lost angular momentum which must also be conserved.

In the future, 20 kmps may be slow, as it takes a million seconds to travel 20 million kilometers at that average speed. Time is money, so faster is better. Perhaps Venus has a very long day because of too many space craft receiving angular momentum? Neil
 
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MeteorWayne

Guest
neil, could you please tryand rstate that? It looks like you are mixing speed and distance units...I can't really make any sense out of it.
 
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dangineer

Guest
Those are all fine and dandy explanations, but I think some of them may be a little over-complicated (just a little, though). The basics of planetary (or stellar) capture involve the trajectory at the edge of the sphere of influence (SOI) (the region where the influence from other bodies becomes negligable). If the space craft is on a hyperbolic or parabolic trajectory (eccentricity > or = 1) when it enters the star's SOI, then it will pass the star and not be captured in any kind of orbit (note: e=1 [parabolic traj.] represents escape velocity at that particular distance).

Now, iti s true that if there's a planet, you can do a fly-by maneuver near the planet that would reduce the velocity relative to the star enough to put it on an elliptical trajectory (e<1), thus putting it in orbit around the star.

Hope this clarifies things.
 
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neilsox

Guest
Quote: The momentum you gain while approaching the planet is lost ~not exactly except in special cases~ as you leave (conservation of momentum).

If a spacecraft is traveling away from the sun, the sun's gravity will slow the craft down. ~Does this slowing represent a loss of momentum to the Sun?~
Using a gravity assist from an outer planet ~it also can be done with inner planets~ let's you steal some angular momentum from that planet's own orbital velocity and increases your own. ~or the reverse~
~ except for special cases, any close approach to any object in our solar system will either increase or decrease your momentum at least slightly = you got (or lost) some delta v from the object. Your speed and direction with respect to what ever either increased or decreased compared to the same route when the object was somewhere else.
I agree my "such as" has some flaws. I was hoping derekmcd could produce a better example with numbers. Neil~
 
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vogon13

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The Messenger spacecraft is currently using Mercury's gravity to perturb itself into an easier solar orbit to decel into Mercury orbit from.

I am not sure this procedure would be of much use to an interstellar craft approaching a target star with insufficient delta v for capture.

If a craft found itself in this predicament the situation would be dire. And keep in mind, even if an orbit around the target star was achieved, there are many such orbits, and most of them would be ungood. For instance, an orbital period of 13,000 years in a steeply inclined retrograde orientation would be as bad as not making orbit at all. So too an orbit with a period of <30 days if the planet you want to colonize is in an orbit of between 300 and 600 days.
 
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derekmcd

Guest
neilsox":qikrwq4x said:
Quote: The momentum you gain while approaching the planet is lost ~not exactly except in special cases~ as you leave (conservation of momentum).

If a spacecraft is traveling away from the sun, the sun's gravity will slow the craft down. ~Does this slowing represent a loss of momentum to the Sun?~
Using a gravity assist from an outer planet ~it also can be done with inner planets~ let's you steal some angular momentum from that planet's own orbital velocity and increases your own. ~or the reverse~
~ except for special cases, any close approach to any object in our solar system will either increase or decrease your momentum at least slightly = you got (or lost) some delta v from the object. Your speed and direction with respect to what ever either increased or decreased compared to the same route when the object was somewhere else.
I agree my "such as" has some flaws. I was hoping derekmcd could produce a better example with numbers. Neil~
I can barely make out what you are saying. The way you format quotes and responses makes it too much of a pain to respond to.
 
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