1. Subscribersonhouse
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    17 Aug '10 02:55
    Originally posted by zeeblebot
    equilateral triangle = 60 degrees.

    http://en.wikipedia.org/wiki/Lagrangian_point#L4_and_L5

    The L4 and L5 points lie at the third corners of the two equilateral triangles in the plane of orbit whose common base is the line between the centers of the two masses, such that the point lies behind (L5) or ahead of (L4) the smaller mass with regard to its o ...[text shortened]... asteroids are (largely) named after characters from the respective sides of the Trojan War.
    I think I am correct in my assumption there is a maximum mass of an object at L4 and L5, That is to say, you couldn't put Jupiter in an L4 from Neptune. I don't think that would be stable.
  2. Cape Town
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    17 Aug '10 06:021 edit
    Originally posted by sonhouse
    I didn't say the forces canceled out, they couldn't because the angle is not right, the cancelation point would be on a line between the to objects dead center if they had equal mass and closer to the smaller of the two masses. I know it's a guiding force, the natural perturbation forces being smaller than the force keeping on that orbital path.
    My point was that your post never mentioned that the stars were orbiting each other, and that the Lagrangian object was orbiting their center of mass. That orbital motion is essential to understanding the situation.
    If the two stars were magically suspended in space without orbiting each other, there would be only one Lagrangian.
  3. Subscribersonhouse
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    17 Aug '10 13:30
    Originally posted by twhitehead
    My point was that your post never mentioned that the stars were orbiting each other, and that the Lagrangian object was orbiting their center of mass. That orbital motion is essential to understanding the situation.
    If the two stars were magically suspended in space without orbiting each other, there would be only one Lagrangian.
    I think I used the term orbit and such several times. I didn't know I needed to specify they were not magically standing still. I would think the underlying conditions that make for lagrangians would still be present but static at L4 and L5 however, if the objects were stopped in their tracks.

    The gravitational interactions would not go away if they were still, it would still hold something at those points, suppose the system were still relative to each other but the two were moving at some velocity as a unit, something at L4 or 5 I think would track along with the main masses. IMO anyway.
  4. Cape Town
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    17 Aug '10 14:05
    Originally posted by sonhouse
    I think I used the term orbit and such several times. I didn't know I needed to specify they were not magically standing still.
    My point is you didn't use the concept in your explanation. In other words your description should hold true even if they are not in orbit.

    I would think the underlying conditions that make for lagrangians would still be present but static at L4 and L5 however, if the objects were stopped in their tracks.
    And I am claiming that those conditions do not hold unless the system is in orbit.

    The gravitational interactions would not go away if they were still, it would still hold something at those points, suppose the system were still relative to each other but the two were moving at some velocity as a unit, something at L4 or 5 I think would track along with the main masses. IMO anyway.
    Surely it is obvious that that is false? An object placed at L4 or L5 would immediately fall towards the barycenter of the system (or close to it).
  5. Subscribersonhouse
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    17 Aug '10 19:46
    Originally posted by twhitehead
    My point is you didn't use the concept in your explanation. In other words your description should hold true even if they are not in orbit.

    [b]I would think the underlying conditions that make for lagrangians would still be present but static at L4 and L5 however, if the objects were stopped in their tracks.

    And I am claiming that those conditions ...[text shortened]... laced at L4 or L5 would immediately fall towards the barycenter of the system (or close to it).[/b]
    Godda admit you have a point there. Not sure if it would automatically collide with the main body, it might end up in an elliptical orbit. That would depend on whether there were forces acting on the smaller body on the way in, such as sunlight, solar storms, etc. Forgot about that part, gravity acting on it๐Ÿ™‚
  6. Cape Town
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    18 Aug '10 05:03
    Originally posted by sonhouse
    Godda admit you have a point there. Not sure if it would automatically collide with the main body, it might end up in an elliptical orbit. That would depend on whether there were forces acting on the smaller body on the way in, such as sunlight, solar storms, etc. Forgot about that part, gravity acting on it๐Ÿ™‚
    I'll use the sun - earth system for labels:
    My understanding is that in and orbiting system, a satelite placed at in the same orbit as earth would need to be faster or slower (I don't know which) than earth to maintain a the same orbital path as earth. However, at L4 and L5, the earth and sun exert small forces on the satellite which either slow it down slightly or speed it up thus keeping it at a constant speed relative to earth.
  7. Subscribersonhouse
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    18 Aug '10 12:54
    Originally posted by twhitehead
    I'll use the sun - earth system for labels:
    My understanding is that in and orbiting system, a satelite placed at in the same orbit as earth would need to be faster or slower (I don't know which) than earth to maintain a the same orbital path as earth. However, at L4 and L5, the earth and sun exert small forces on the satellite which either slow it down slightly or speed it up thus keeping it at a constant speed relative to earth.
    I wonder if anyone worked out the forces involved in multiple orbiters, like say, 4 equal mass planets in the same orbit but 90 degrees apart. Or 2,3,4,5, etc. I just looked at the whole diagram of equilateral triangles which would be 6 triangles all centered on the sun.

    With 6 equally spaced points you could have 6 main bodies, say Earth mass planets at the half way points equally spaced around the orbital path and then 6 small bodies at the resultant multiple L4's and L5's. Has anyone done a simulation for stability with that configuration?

    You could picture some super advanced civilization maybe coming across this configuration of planets and planetoids as making for a compact solar system with 6 major planets allowing for a very large population and 6 asteroids or small planets that would give a permanent mining facility for metals and such.

    Sounds like a good background plot for a sci-fi story, eh.
  8. Cape Town
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    18 Aug '10 14:55
    Originally posted by sonhouse
    I wonder if anyone worked out the forces involved in multiple orbiters, like say, 4 equal mass planets in the same orbit but 90 degrees apart. Or 2,3,4,5, etc. I just looked at the whole diagram of equilateral triangles which would be 6 triangles all centered on the sun.
    Its an interesting concept. Lagrangians assume the the mass at the Lagrangian is negligible compared to the other two. I don't think a large body is stable at that point, though I don't know why not. If it was, I think we would have more double planet systems.
  9. Subscribersonhouse
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    18 Aug '10 19:573 edits
    Originally posted by twhitehead
    Its an interesting concept. Lagrangians assume the the mass at the Lagrangian is negligible compared to the other two. I don't think a large body is stable at that point, though I don't know why not. If it was, I think we would have more double planet systems.
    What about 6 planets 60 degrees apart, equal masses, all Earth mass, wonder if they could have a stable orbit. They would all be at each other's L4 and 5. It looks intuitively like each planet there would have a double force, one backwards in its orbit the other forwards in its orbit. It might even be stable!

    Another variation: Three planets 120 degrees apart with 3 asteroids at L4,5. In that configuration, each asteroid would be simultaneously at L4 and L5. That might increase the maximum mass of the L objects and still be stable orbit wise.

    Googled it and found this:

    http://www.bautforum.com/archive/index.php/t-14254.html

    Three equal mass planets co-orbiting 120 degrees apart, this article claims to be stable, called 'Kemplerer Rosette' orbits. Seems I am not terribly original here.....

    By extension then, three smaller objects would be stable at the respective L4/5 points.
  10. Cape Town
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    19 Aug '10 08:54
    Originally posted by sonhouse
    Three equal mass planets co-orbiting 120 degrees apart, this article claims to be stable, called 'Kemplerer Rosette' orbits. Seems I am not terribly original here.....
    Much more exciting is co-orbital moons (that I found out about from your link):

    http://en.wikipedia.org/wiki/Co-orbital_moon

    Two moons that keep swapping orbits. Really cool.


    The problem with your 'Kemplerer Rosette' orbits is that for two bodies to follow the same orbit without any interaction, they must be of exactly the same mass. So unless an interactions such as the Lagrange points system is keeping a body in an unusual orbit, we need several planets that are exactly the same mass to somehow get into the same orbit.
    This is probably so unlikely that it never happens.

    It might however be something intelligent creatures like humans could arrange. For example if we had say 4 death stars and we wanted them orbiting earth, that would probably be the best arrangement as it is fairly stable.
  11. Subscribersonhouse
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    20 Aug '10 01:10
    Originally posted by twhitehead
    Much more exciting is co-orbital moons (that I found out about from your link):

    http://en.wikipedia.org/wiki/Co-orbital_moon

    Two moons that keep swapping orbits. Really cool.


    The problem with your 'Kemplerer Rosette' orbits is that for two bodies to follow the same orbit without any interaction, they must be of exactly the same mass. So unless ...[text shortened]... anted them orbiting earth, that would probably be the best arrangement as it is fairly stable.
    So they would be stable if there were 2,3,4,5 etc., as long as they were equispaced, and equimass? I guess if so, then by extension, Larry Niven's Ring World would be stable also, the ultimate extension of that idea, one solid ring of mass all the way round the sun. If that is stable as an orbiting body, would it also be stable if it were not rotating but still as far as circular motion goes? Wouldn't such a system be stable because it cannot get any closer to its parent star because it would be under somewhat compressive forces which would be supported by the strength of the ring material?
  12. Cape Town
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    20 Aug '10 05:29
    Originally posted by sonhouse
    Wouldn't such a system be stable because it cannot get any closer to its parent star because it would be under somewhat compressive forces which would be supported by the strength of the ring material?
    If the material is strong enough.
    But I suspect that if it wasn't rotating at the correct speed, there may be issues if it gets slightly out of place. It might start to wobble or something like that.
  13. Subscribersonhouse
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    20 Aug '10 11:39
    Originally posted by twhitehead
    If the material is strong enough.
    But I suspect that if it wasn't rotating at the correct speed, there may be issues if it gets slightly out of place. It might start to wobble or something like that.
    It's an interesting concept though. If it is not spinning there would be no wobble but if it was spinning, I wonder what the right RPM, RPY?, would be to balance the inward force of gravity with the outward centripital forces?

    If there was sufficient mass you could have with a no spin situation, gravity on both the inner and outer surfaces, doubling the effective usable surface area, gravity needed by humans to keep up muscles and such.

    If the mass were not enough for self gravity, then spin would be able to make for artificial gravity, but only on the inner surface. The outer surface would have to have everything battened down or stuff would fly away.

    So at a distance of 1 AU, what would the gravity be on the outer surface if it was not spinning, gravity due to the sun? And what spin rate would result in 1 gravity on the inner surface if it had to spin?
  14. Cape Town
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    20 Aug '10 12:08
    Originally posted by sonhouse
    It's an interesting concept though. If it is not spinning there would be no wobble but if it was spinning, I wonder what the right RPM, RPY?, would be to balance the inward force of gravity with the outward centripital forces?

    If there was sufficient mass you could have with a no spin situation, gravity on both the inner and outer surfaces, doubling the ...[text shortened]... o the sun? And what spin rate would result in 1 gravity on the inner surface if it had to spin?
    I think the idea of creating gravity using the spin would require such an enormously high spin rate that it would not really be viable.
    The spin rate required to create a stable ring should be similar to the normal orbit of a planet whose diameter is slightly larger than the rings diameter (diameter of the rim, not of the whole ring).

    I believe that a non-rotating ring would be highly unstable ie if the sun is not exactly in the center, the ring would tend to drift until it touches the sun.
  15. Subscribersonhouse
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    21 Aug '10 02:062 edits
    Originally posted by twhitehead
    I think the idea of creating gravity using the spin would require such an enormously high spin rate that it would not really be viable.
    The spin rate required to create a stable ring should be similar to the normal orbit of a planet whose diameter is slightly larger than the rings diameter (diameter of the rim, not of the whole ring).

    I believe that ...[text shortened]... if the sun is not exactly in the center, the ring would tend to drift until it touches the sun.
    Remember we are talking about a Niven ring, or is it Dyson? Not sure, I know Niven wrote the ring world series. And we are talking about a ring with a radius of 1 au, say 100 million miles, 160 million km. Did you do the math for that size?

    I'll see if I can figure it out. Ok, with A=V^2/R, we'll make R+160 E6 Km or 160 E9 meters (1.6 E11 meters) and a velocity of 31,850 meters per second (that is about a one year orbit. It's interesting the distance from the sun has no bearing on what velocity the ring goes at.

    So 31850^2=~1E9/1.6E11 =~.006 meters/second^2, about 1/1600th of a G. Not much. So we speed up. So we need about 1600 times the force, Square root of 1600, 40, so 40 times that 31850 m/s gives about 9.8 M/S^2 or one G of centripetal
    force on the inside of the ring. That is about 10 days per revolution, roughly.

    So if the ring spins with a velocity of 1200 km/sec you get about one G on the inside.
    That works out to about 1.3 degrees or revolution per minute.
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