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View Full Version : Massive bodies on a collision course - will they miss each other?

DyerWolf
2008-Feb-02, 09:06 PM
In another thread (http://www.bautforum.com/questions-answers/69408-craters-mercury.html) it was pointed out to me that in order for an asteroid to hit a planet it usually must first orbit the planet.

That got me to thinking; hence this question:

Presume two stars are flung from separate galaxies. Based upon the course and speed of each, if they were to continue on a straight line they would impact. (In intergalactic space far from the gravitational interference of any other bodies.)

Would the gravitational attraction of each star change the trajectory of the other sufficient to cause them to miss?

Tobin Dax
2008-Feb-02, 09:16 PM
If they were heading straight at each other? Their gravitational attraction would accelerate them toward each other and they would collide.

If they were going to pass each other somewhat closely, their gravitational influence would at least turn them toward each other a little, and at most bring them close enough to collide, but that depends upon their distance from each other.

grant hutchison
2008-Feb-02, 09:17 PM
Would the gravitational attraction of each star change the trajectory of the other sufficient to cause them to miss?I think not. Gravity is attractive, so its contribution is to make things collide which would otherwise not collide. It's as if the cross-section for collision were larger than either of the bodies involved.

Grant Hutchison

clop
2008-Feb-02, 09:43 PM
The two stars would always be accelerating towards their common centre of mass. If they missed each other on the first pass chances are they'd miss each other on every return trip too, no matter how much they change each other's trajectories. Well that's what I think anyway.

clop

grant hutchison
2008-Feb-02, 10:16 PM
The two stars would always be accelerating towards their common centre of mass. If they missed each other on the first pass chances are they'd miss each other on every return trip too, no matter how much they change each other's trajectories. Well that's what I think anyway.An isolated pair of masses, like the stars in the OP, wouldn't experience any return trips: they'd either collide during the encounter, or continue onwards along their (modified) paths.

Grant Hutchison

clop
2008-Feb-02, 10:29 PM
Oh god, I disagree with Grant, which means I'm almost certainly going to be wrong. Still, I think that two isolated masses floating around in space and exerting a gravitational pull on each other have two choices : 1) their speed is greater than the escape velocity required for one to leave the other, and so they continue on their way and never come back or 2) their speed is below the escape velocity and so they slow down and eventually fall back towards each other again, and pass, and slow down, and so on. I maintain that with no drag they miss each other every time. Or am I missing something here?

clop

neilzero
2008-Feb-02, 11:08 PM
Typically there are 3 or more significant sources of gravity envolved which permits capture into orbit rarely, but sometimes.
If only two gravity sources are significant, then Grant is correct, they either hit or they do not come close again unless an other gravity source or collission becomes significant later. If their paths are close to intersecting simultainiously then their gravity does increase the probability of collission, significantly, But a near miss and sling shot manuver are the usual senario, IMHO = in my humble opinion. Neil

DyerWolf
2008-Feb-02, 11:10 PM
For whatever its worth, I envisioned the trajectories crossing as some form of "x" rather than a head-on collision course.

Jeff Root
2008-Feb-02, 11:58 PM
Oh god, I disagree with Grant, which means I'm almost certainly going to
be wrong.
You are wrong.

Still, I think that two isolated masses floating around in space and
exerting a gravitational pull on each other have two choices :
1) their speed is greater than the escape velocity required for one to
leave the other, and so they continue on their way and never come back
or 2) their speed is below the escape velocity and so they slow down and
eventually fall back towards each other again, and pass, and slow down,
and so on. I maintain that with no drag they miss each other every time.
Or am I missing something here?
Well, you are missing the zeroth possibility, that the stars will collide,
but apart from that, you are wrong: You do not disagree with Grant.
What you say about the stars is compatible with what Grant said.
What gave you the fool idea that you disagreed with him? You just
added a bit that he didn't mention.

Your choice (2) is an elliptical orbit, of course.

-- Jeff, in Minneapolis

grant hutchison
2008-Feb-03, 12:19 AM
Oh god, I disagree with Grant, which means I'm almost certainly going to be wrong. Still, I think that two isolated masses floating around in space and exerting a gravitational pull on each other have two choices : 1) their speed is greater than the escape velocity required for one to leave the other, and so they continue on their way and never come back or 2) their speed is below the escape velocity and so they slow down and eventually fall back towards each other again, and pass, and slow down, and so on. I maintain that with no drag they miss each other every time. Or am I missing something here?I think the only thing that you're missing is the content of the OP, which stipulated that the stars were "flung" towards each other from separate galaxies. That seems to imply that they're starting millions of lightyears apart with some significant velocity towards each other. So your option 1) applies: they're going to have mutual escape velocity during their encounter.
It would be a pretty finely tuned bit of "flinging" that would allow for option 2) :)

Grant Hutchison

clop
2008-Feb-03, 12:25 AM
Heh heh I guess it all boils down to the level of fling. And I don't think it would have to be finely tuned. I imagine the escape velocity of a neutron star is rather high.

clop

grant hutchison
2008-Feb-03, 12:35 AM
Heh heh I guess it all boils down to the level of fling. And I don't think it would have to be finely tuned. I imagine the escape velocity of a neutron star is rather high.At a million light years?
The escape velocity of the sun at a million light years is about 17cm.s-1. If our stars are moving towards each other much faster than that when a million light years apart, then they have mutual escape velocity, and their encounter will be a one-off.
So quite exquisite fine-tuning is required to get them heading towards each other sufficiently slowly that a pair of elliptical orbits will result.

Grant Hutchison

Warren Platts
2008-Feb-03, 02:36 AM
CBDR!!!!!

GHARRRG MATIE!!!

Jens
2008-Feb-03, 04:40 AM
In another thread (http://www.bautforum.com/questions-answers/69408-craters-mercury.html) it was pointed out to me that in order for an asteroid to hit a planet it usually must first orbit the planet.

I just looked at that thread, and although it might be true, that's not what the person wrote. The statement was something like "almost any asteroid..." Which implies that this is not a necessity, merely what almost always happens.

hhEb09'1
2008-Feb-03, 06:11 AM
For whatever its worth, I envisioned the trajectories crossing as some form of "x" rather than a head-on collision course.I thought of that when I read your first question.

It's possible that the course would result in a collision (the X), but because of the difference is sizes, their gravitational attraction could cause them to miss. If everything were completely (perfectly) symmetric, there would be no way to miss, that way.

clop
2008-Feb-03, 06:20 AM
At a million light years?
The escape velocity of the sun at a million light years is about 17cm.s-1. If our stars are moving towards each other much faster than that when a million light years apart, then they have mutual escape velocity, and their encounter will be a one-off.
So quite exquisite fine-tuning is required to get them heading towards each other sufficiently slowly that a pair of elliptical orbits will result.

Grant Hutchison

OK I give in. But still - maybe the parent galaxies will slow them down on their voyage towards each other. I guess it's unlikely.

JohnD
2008-Feb-03, 11:04 AM
And the OP is such a very special case, that they are on exactly opposing paths.
In a million light years, the effect of other gravity sources will deviate either from their original vector. That devaition could seperate or bring together deviating objects, but what ARE the chances of a collision? It is widely said that stars in colliding galaxies do not themselves collide, because Space is Big. You just won't believe how vastly, hugely, mind-bogglingly big - sorry, HHGTTG moment there, but it makes the point.

John

Cougar
2008-Feb-03, 06:42 PM
In another thread (http://www.bautforum.com/questions-answers/69408-craters-mercury.html) it was pointed out to me that in order for an asteroid to hit a planet it usually must first orbit the planet.
Usually must?

Presume two stars are flung from separate galaxies. Based upon the course and speed of each, if they were to continue on a straight line they would impact.
Straight line? They would both follow a "straight line" only if they were originally flung directly at each other, and then of course they would collide. If they were originally sent NOT directly at each other, and the distance between them was decreasing, then their paths would curve toward each other, increasing the likelihood of a collision.

But there are an infinite number of initial conditions (speeds and directions) for the scenario you pose. Some hit and some miss. Some interactions become very complicated....

http://www.xmission.com/~dcc/encounter8.jpg

mugaliens
2008-Feb-03, 06:47 PM
Would the gravitational attraction of each star change the trajectory of the other sufficient to cause them to miss?

No. If we eliminate gravitational attraction, and they would hit, then with gravitational attraction, they will most assuredly hit.

Tobin Dax
2008-Feb-03, 10:03 PM
No. If we eliminate gravitational attraction, and they would hit, then with gravitational attraction, they will most assuredly hit.
By gum, mugs is right. (That was an unintended pun [or palindrome], seriously.)

If they would hit without gravity helping, then they would hit with gravity helping. Changing reference frames to where one star is stationary (at least initinially) makes the other star head right for it, no matter what their relative paths are in a different frame. In that frame, they both accelerate along a straight line right at each other and collide.

hhEb09'1
2008-Feb-04, 02:46 PM
No. If we eliminate gravitational attraction, and they would hit, then with gravitational attraction, they will most assuredly hit.Depends upon the configuration and relative masses I think
By gum, mugs is right. (That was an unintended pun [or palindrome], seriously.) That's xad

If they would hit without gravity helping, then they would hit with gravity helping. Changing reference frames to where one star is stationary (at least initinially) makes the other star head right for it, no matter what their relative paths are in a different frame. In that frame, they both accelerate along a straight line right at each other and collide.Because we're considering the effects of gravity, the reference frame would not be inertial, and then there's no guarantee that the paths remain straight.

Consider a Sun/Earth-like system. If the motion of the whole system is along a direction in the ecliptic that is faster than the orbital motion, then there is some point along the orbit where, if gravity is "turned off" the two would collide. But with gravity on, they orbit.

Ilya
2008-Feb-04, 02:58 PM
In another thread (http://www.bautforum.com/questions-answers/69408-craters-mercury.html) it was pointed out to me that in order for an asteroid to hit a planet it usually must first orbit the planet.

That's just plain false -- see my response on that thread.

DyerWolf
2008-Feb-04, 02:58 PM
If they would hit without gravity helping, then they would hit with gravity helping. Changing reference frames to where one star is stationary (at least initinially) makes the other star head right for it, no matter what their relative paths are in a different frame. In that frame, they both accelerate along a straight line right at each other and collide.

Thanks for the answers.

The majority position seems to be: if we have two stars on trajectories that would intersect without taking into account the star's respective gravitational attraction (or warping of spacetime), adding in the respective warping (or attraction) will change the trajectory, but won't cause them to miss. (and in fact, might guarantee it)

Yet hhEb09'1 suggests they could miss with two objects of different masses.' Couger's post (if I'm reading this correctly) agrees with hhEbo9'1.

So does it change the result if the star's masses are different? (Sun-like and brown dwarf or pulsar?)

grant hutchison
2008-Feb-04, 04:15 PM
Some interactions become very complicated....

http://www.xmission.com/~dcc/encounter8.jpgThat's a surprising image. It looks like there's more going on than an encounter between two point masses. What does it represent?

Grant Hutchison

Neverfly
2008-Feb-04, 04:27 PM
One of my shoestrings looked like that once.

Now I know that I can rest easy and let go of my guilt. It was gravity's fault all this time...

ETA: Prediction for Randi: About half the people that read this post are going to roll their eyes and say to themselves, "...:rolleyes: idiot...."

Jeff Root
2008-Feb-04, 05:08 PM
Depends upon the configuration and relative masses I think...

Because we're considering the effects of gravity, the reference
frame would not be inertial, and then there's no guarantee that
the paths remain straight.

Consider a Sun/Earth-like system. If the motion of the whole system
is along a direction in the ecliptic that is faster than the orbital motion,
then there is some point along the orbit where, if gravity is "turned off"
the two would collide. But with gravity on, they orbit.
I don't understand this. mugaliens and Tobin Dax are right, and
what you say here isn't. The original post specified that two stars
are initially headed straight toward each other, and that they are
far away from the influence of other bodies. In that case, they
will collide. It makes no difference what their sizes are. If they
would collide without mutual gravitational attraction, then they
will collide with mutual gravitational attraction.

If they were originally on trajectories that would make them pass
at close range, then their mutual gravitational attraction might
make them collide, or it might not, with the sizes of the bodies
being an important determining factor.

I think the "Sun/Earth-like system" scenario made no sense.

-- Jeff, in Minneapolis

Cougar
2008-Feb-04, 05:15 PM
That's a surprising image. It looks like there's more going on than an encounter between two point masses. What does it represent?
An encounter between two point masses. :)

The two masses are coming at each other (not directly) from lower right and upper left (see the little arrows?). It is very hard (if not impossible?) to set initial conditions of two objects approaching each other from a large distance to fall into orbit around each other. In this attempt, they do orbit rather chaotically for a while, but then they both get flung out of orbit at high speed.

I suppose, since physical laws are time-symmetric, this little model would work just as well running backwards....

Jeff Root
2008-Feb-04, 05:24 PM
Cougar,

If there are only two point masses in that simulation, then the
simulation is not working. If they do not collide at the first
close encounter, then either they are above their mutual escape
speed and there is only a single encounter, in which they whizz
past each other on hyperbolic trajectories, or they are below
their escape speed and are in elliptical orbits.

-- Jeff, in Minneapolis

hhEb09'1
2008-Feb-04, 05:28 PM
The original post specified that two stars
are initially headed straight toward each other, and that they are
far away from the influence of other bodies.No, the OP did not. In fact, it was even explicitly clarified just a few posts later:
For whatever its worth, I envisioned the trajectories crossing as some form of "x" rather than a head-on collision course.

I think the "Sun/Earth-like system" scenario made no sense.Can you understand me now? :)

Jeff Root
2008-Feb-04, 06:08 PM
The original post specified that two stars are initially headed straight
toward each other, and that they are far away from the influence of
other bodies.
No, the OP did not. In fact, it was even explicitly clarified just a few
posts later:

For whatever its worth, I envisioned the trajectories crossing as
some form of "x" rather than a head-on collision course.
An "x" course IS a head-on collision course!

The original post said:

Presume two stars are flung from separate galaxies. Based upon the
course and speed of each, if they were to continue on a straight line
they would impact. (In intergalactic space far from the gravitational
interference of any other bodies.)
DyerWolf made an error here. The two stars do not have two speeds.
They have only one speed at any given instant: the relative speed
between them.

If you imagine two gravity-less objects on an "x" course, say initially
headed toward the same point in empty space, and equal distances
from that point, but moving at different speeds, then the two bodies
are not headed straight for each other and they are not on courses
that would have them eventually impact.

If you imagine the same two gravity-less objects on the same "x"
course initially headed toward the same point in empty space, and
equal distances from that point, but this time moving at the same
speed, then the two bodies are headed directly toward each other
("x" course notwithstanding!) and they are on courses that will
have them eventually impact.

There is no difference whatever between an "x" course and a head-on
course when you are talking about two bodies in isolation.

Galilean relativity.

I think the "Sun/Earth-like system" scenario made no sense.
Can you understand me now? :)
No!

-- Jeff, in Minneapolis

DyerWolf
2008-Feb-04, 06:09 PM
The majority position seems to be: if we have two stars on straight-line trajectories that would intersect, without taking into account each star's respective gravitational attraction (or warping of spacetime) as they neared one another, accounting for the respective warping (or attraction) will change the trajectory, but won't cause them to miss.

Does the result change if both star's have different masses? (Sun-like vs brown dwarf or pulsar?)

(Typed apparently as Jeff Root typed his last post... Jeff: your response is intriguing - will you break it down a little further for a layman? TIA)

Jeff Root
2008-Feb-04, 06:30 PM
Yes, DyerWolf.

Oh-- I'm a layman, too!

The simplest and easiest thing to do is imagine that only the two
gravity-less bodies exist, and nothing else. Imagine that you are
one of the bodies, watching the other. What do you see?

All you see is the other body, moving closer to you. If you are
on a collision course, it is headed directly toward you. If you are
not on a collision course, it is not headed directly toward you.

I lack a good way to explain what I mean by "directly toward you",
but if you don't agree that it is headed directly toward you, just
continue to try to visualize what you would actually see.

-- Jeff, in Minneapolis

John Mendenhall
2008-Feb-04, 06:36 PM
Usually must?

Straight line? They would both follow a "straight line" only if they were originally flung directly at each other, and then of course they would collide. If they were originally sent NOT directly at each other, and the distance between them was decreasing, then their paths would curve toward each other, increasing the likelihood of a collision.

But there are an infinite number of initial conditions (speeds and directions) for the scenario you pose. Some hit and some miss. Some interactions become very complicated....

http://www.xmission.com/~dcc/encounter8.jpg

That's a surprising image. It looks like there's more going on than an encounter between two point masses. What does it represent?

Grant Hutchison

This happpened to me the last I went fly fishing.

Seriously, the OP question raises the boundary conditions flag. The starting condition of being flung out of separate galaxies on a collision course is fantastically improbable. Unfortunately it is not zero, so we can't throw the question out on impossbile grounds.

Suggestion: If, from the point of view of observers on either object, the other object is always at the same azimuth and right ascension, and the distance between them is decreasing, then they will eventually collide.

grant hutchison
2008-Feb-04, 06:38 PM
That's a surprising image. It looks like there's more going on than an encounter between two point masses. What does it represent?An encounter between two point masses. :)Then I have to agree with Jeff that there seems to be something wrong with the simulation. For the pair to fly apart like that after a few orbits means that they have gained energy from somewhere; for the orbits to precess as illustrated means that something non-Keplerian is happening.

Grant Hutchison

Jeff Root
2008-Feb-04, 06:49 PM
Imagine a milk truck headed east on Arcane Avenue and an Oreo
cookie truck headed north on Slippery Street. They are both a
few seconds away from entering the intersection of the avenue
and the street. Although they are slightly different distances
from the intersection, and travelling at slightly different speeds
relative to the pavement, they are on an "x" course which means
they will both be in the center of the intersection at the same
time, unless one of them changes speed.

Unfortunately, the traffic light is out and Jeff volunteered himself
to direct traffic until it gets fixed. Unfortunate because he is
looking at the pretty girl on the sidewalk instead of the traffic.

Now, let's play back the videotape of what happened. The
video camera is mounted on the milk truck, and is quite unusual:
It is only able to see and record images of trucks, nothing else.
We extract the tape from the milk truck's "black box". What does
the tape show? It shows the Oreo cookie truck headed straight
toward the camera, at constant speed. The truck was at an
angle to the camera the whole time, and its wheels were spinning,
but it headed straight for the camera and never changed speed.

I hafta clean up.

-- Jeff, in Minneapolis

DyerWolf
2008-Feb-04, 06:55 PM
Yes, DyerWolf.

Oh-- I'm a layman, too!

The simplest and easiest thing to do is imagine that only the two
gravity-less bodies exist, and nothing else. Imagine that you are
one of the bodies, watching the other. What do you see?

All you see is the other body, moving closer to you. If you are
on a collision course, it is headed directly toward you. If you are
not on a collision course, it is not headed directly toward you.

I lack a good way to explain what I mean by "directly toward you",
but if you don't agree that it is headed directly toward you, just
continue to try to visualize what you would actually see.

-- Jeff, in Minneapolis

Guess we were getting caught up in the semantics. I presumed "directly toward you" to mean 'in a diametrically opposed direction" (---> <---) (the 'head-on' collision) rather than merely on an intersecting path. Admittedly, objects that meet on a diametrically opposing path by definition follow an intersecting path, but not all objects that meet on intersecting paths follow diametrically opposing paths (hence the 'X').

I can see that as these massive objects approach one another, their respective gravities cause them to depart from a straight line X and meet on paths that describe intersecting parabolic curves - I just didn't know if the change in trajectory would cause them to miss one another.

Jeff Root
2008-Feb-04, 07:09 PM
Guess we were getting caught up in the semantics. I presumed
"directly toward you" to mean 'in a diametrically opposed direction"
(---> <---) (the 'head-on' collision) rather than merely on an
intersecting path. Admittedly, objects that meet on a diametrically
opposing path by definition follow an intersecting path, but not all
objects that meet on intersecting paths follow diametrically
opposing paths (hence the 'X').
No, as I tried to show with the trucks, an 'X' path and "diametrically
opposed paths" are exactly the same, completely identical, with no
difference whatever.

Just try to visualize what you actually see, if all you can see
is the other body. There is no pavement, no traffic lanes in space.

-- Jeff, in Minneapolis

grant hutchison
2008-Feb-04, 07:09 PM
Admittedly, objects that meet on a diametrically opposing path by definition follow an intersecting path, but not all objects that meet on intersecting paths follow diametrically opposing paths (hence the 'X').It's just your choice of the observer's velocity that makes the difference between these scenarios, though.
Imagine a pair of objects are on diametrically opposed paths from your point of view, and you're standing at the point where they whisk past each other, barely missing.
Now, I am flying at right angles to their paths, at some velocity relative to you, such that I pass over your head just as the objects pass each other. I'll see them travelling in that 'X', in my reference frame, since they'll appear to converge on each other from up ahead of me, cross as I pass over them, and then diverge behind me.

Grant Hutchison

DyerWolf
2008-Feb-04, 07:58 PM
Egads, I think I understand now... Although it hurts my brain a bit to think like this.

Frames of reference. Ouch.

...:think:

...:think::doh:

...okay, so if this is the case - there is no 'bending' of their trajectories at all; the perceived warping of the previously straight-line path into a parabolic curve is relevant only to an outside observer - but is irrelevant to whether the two planets will impact each other.

Yes?

cmsavage
2008-Feb-04, 08:00 PM
The trajectories in a 2-body gravitational system are either two ellipses (bound system) or two hyperbolas (unbound system) with focus points at the center of mass. The Wikipedia article (http://en.wikipedia.org/wiki/Two-body_problem) on the 2-body problem gives part of the derivation of these orbits and two animations. The first animation shows two bound, equal mass objects, which have elliptical orbits. Note how the objects are always on opposite sides of the center of mass, so that, while the orbits cross at two points, they are never at those points at the same time. For unequal masses, the larger mass would have a smaller ellipse, similar to the second animation (which shows circular orbits of two bound, unequal mass objects). Both objects always have the same orbital period, even if the masses are unequal, and the orbits never change.

For unbound objects, like the two stars of the OP, the orbits are hyperbolas. The orbits would look very similar to what is in the center portion of the first animation (basically, take the ellipses to be very big). The two objects will come in from very far away, reach their point of closest approach, then move off into the distance. They come close only once and do not enter into an orbit: if they came in unbounded (from infinity, more or less), they leave unbounded (to infinity).

If your two stars travelled in straight lines without gravity, the distance between them (more appropriately, their centers) at their closest point is called the impact parameter. The orbital solutions with gravity show that the point of closest approach is always closer than the impact parameter. So if your stars were going to collide without gravity (the sum of their radii is smaller than the impact parameter), then they will definitely collide with gravity.

This only applies to the 2-body problem (also assuming point masses, although that does not matter much). If you have 3 or more bodies, then the behavior gets very complicated. And when you are talking about galaxies, you are talking about a 100,000,000,000-body problem or worse. If you assume two stars were thrown out of two nearby galaxies (100,000 light years apart) at typical velocities (1,000 km/s) directly at each other (center to center), they would have to have their trajectories deflected by less than 10^-11 degrees over the 10 million years they take to reach each other in order to still hit. Even well outside galaxies, the gravitational pull of the closest galaxies and galaxy clusters will deflect those stars by much larger angles than that.

Cougar:
The elliptical/hyperbolic orbits are exact analytical solutions (and are the only solutions), so your image could not possibly be correct for two point masses (as Jeff noted). Did you use a numerical orbit simulator? That looks like the results of a simulator with time steps that are way too large.

grant hutchison
2008-Feb-04, 08:10 PM
Egads, I think I understand now... Although it hurts my brain a bit to think like this.Not having to think about the milk and cookies stuff makes it easier, though, doesn't it? ;)

...okay, so if this is the case - there is no 'bending' of their trajectories at all; the perceived warping of the previously straight-line path into a parabolic curve is relevant only to an outside observer - but is irrelevant to whether the two planets will impact each other.The paths will appear to bend to some observers (those who see the 'X'), but not to those who see the two stars heading straight at each other along diametrically opposed paths: they'll simply see the stars accelerate towards each other in straight lines.

Grant Hutchison

Cougar
2008-Feb-04, 09:43 PM
Cougar:
The elliptical/hyperbolic orbits are exact analytical solutions (and are the only solutions), so your image could not possibly be correct for two point masses (as Jeff noted). Did you use a numerical orbit simulator? That looks like the results of a simulator with time steps that are way too large.
I expect you are right. Thanks for the correction. But just to complicate matters, what if the masses in my simulation are neutron stars, and therefore the precession of their orbits is due to the strong relativistic effects? Wouldn't this then be an accurate model for the given initial conditions?

grant hutchison
2008-Feb-04, 10:07 PM
But just to complicate matters, what if the masses in my simulation are neutron stars, and therefore the precession of their orbits is due to the strong relativistic effects? Wouldn't this then be an accurate model for the given initial conditions?I think the precession would be more regular than is depicted (thinking of the steady variation of Mercury's perihelion under GR); and there's still the matter of the apparently varying energy of the system, especially that final mutual escape.

Grant Hutchison

cmsavage
2008-Feb-04, 10:21 PM
I expect you are right. Thanks for the correction. But just to complicate matters, what if the masses in my simulation are neutron stars, and therefore the precession of their orbits is due to the strong relativistic effects? Wouldn't this then be an accurate model for the given initial conditions?
Not very likely. It is not just precession because the size and shape of the orbits as well as the orientation change in your figure. If the relativistic effects are strong enough to radically change the orbit from an initially hyperbolic trajectory to elliptical one, or from one elliptical orbit to a very different one, then a significant amount of energy is being lost somewhere (mainly gravitational waves). The objects would become more tightly bound, never less. The fact that they shoot back out at the end means they somehow got a lot of energy back, which is uncharacteristic of relativity. Around black holes, you can have a "single" pass of an object that goes around the BH multiple times, but that type of orbit has only one minimum: it spirals in, reaches a minimum, then spirals back out (not in and out multiple times, then escaping off to infinity). Your example shows many minima. Basically, for two bodies under Newtonian gravity or General Relativity, you can have only one or an infinite number of closest approaches.

The chaotic behavior is not a result of relativity, but looks very much like what can occur in 3 or 4 body systems or with a numerical simulation taking steps that are too large.

tony873004
2008-Feb-04, 10:26 PM
There is something wrong with that simulation. I would guess that a massive 3rd body exists. That trajectory looks like something orbiting Jupiter from a distance, where the Sun's gravity continuously perturbs the orbit.

If it were two point masses, you could only have one out of 3 outcomes:

1. Collision (only if the objects had some size to them)
2. Hyperbolic trajectory - one and only one close approach
3. Elliptical orbit - which would have to be the starting condition of the system.

Cougar
2008-Feb-05, 01:35 AM
There is something wrong with that simulation. I would guess that a massive 3rd body exists. That trajectory looks like something orbiting Jupiter from a distance, where the Sun's gravity continuously perturbs the orbit.
No, cmsavage is right. It's just a large time step in the numerical program. It's downright unfair that Nature does integration without even thinking about it.

cmsavage
2008-Feb-05, 04:29 AM
....It's downright unfair that Nature does integration without even thinking about it.
That damn Nature, always showing us up. :lol:

Neverfly
2008-Feb-05, 04:38 AM
Originally Posted by Cougar
....It's downright unfair that Nature does integration without even thinking about it.That damn Nature, always showing us up. :lol:
I'm sure modern computers say that about us with our brains.

ETA: or does that go back to Nature?:doh: It's a vicious cycle....

hhEb09'1
2008-Feb-05, 05:17 AM
No!neither can I, musta been that early wake up from hibernation

I think I dropped a vector on the back of the envelope :)