Notices
Results 1 to 36 of 36

Thread: Orbit velocity, what determines it?

  1. #1 Orbit velocity, what determines it? 
    Forum Freshman
    Join Date
    Sep 2008
    Posts
    24
    Is there a law which states that all orbital velocities (that is, the speed at which a planet travels around a star) are the product of some combination of forces? I do not mean simple inertial forces, such as an instance where a rogue astroid is caught in a planets gravitational pull and developes a stable orbit. I mean is the orbital velocity determined by some sort of relationship which exists between the planet and the star?


    Reply With Quote  
     

  2.  
     

  3. #2  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    Orbital velocity depends on only two things, IIRC: mass of the star and the distance from it. I don't remember the exact formula though.

    Also, the asteroid-planet system is exactly the same as the planet-sun system. There's no difference.


    Reply With Quote  
     

  4. #3  
    Forum Freshman
    Join Date
    Sep 2008
    Posts
    24
    Do inertial forces have nothing to do with it? Such as in the case of the astroid.. if it first found our solar system after being thrust at speeds much higher than its natural orbital velocity, would it not enter the orbit at a slightly higher velocity? Or would it simply not be stable for orbit and pass by?

    Also, I know this is a bit off topic, but would anyone happen to know if time dialation is relative to gravitational field strength? That is, does a gravitational field with half the strength of another also experience the passage of an hour in half the time?
    Reply With Quote  
     

  5. #4  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,070
    If the object is traveling slower than



    Where G is the universal gravitational constant
    M is the mass of the capturing body
    and r is the center to center distance between the objects

    There is a chance of capture.

    Whether there is a capture or not depends on some other things.

    For instance, if the capturing body orbits another body (such as a planet that orbits a Sun) , then the new orbit has to be within that body's Hill sphere. Outside of this, the third body's gravity will prevent an orbit.

    What type of orbit it is captured into depends on its actual speed and trajectory when captured. If, (and this is very unlikely. ) it was traveling exactly at a speed of



    at an angle of 90 to the line joining it to the capturing body, it will go into a circular orbit. At any other speed and trajectory, it will go into an elliptical orbit of some eccentricity.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  6. #5  
    Forum Freshman
    Join Date
    Sep 2008
    Posts
    24
    Alright Janus, that clarified some things very nicely for me Thanks!

    I raise these questions because while fiddling with some numbers a few weeks ago I came up with this formula:

    = [(Orbital Period)/((1)/(Sum of masses of two bodies)/(distance between^2))]/[Orbital Velocity]

    It really interested me because I approached it with the assumption that gravitational dialation is relative to time dialation. I was trying to identify a constant speed at which a body must travel around a Co(orbital)G. This formula gives near-exact same answer for any of the planets and our sun. Does this formula simply make more complicated the formulas which you (Janus) have just shown me? Or is there something more at work here?
    Reply With Quote  
     

  7. #6  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,070
    Gravitational time dilation follows the formula:



    So, No, there is no linear relationship between field strength and time dilation. (and if by field strength you mean local acceleration due to gravity, there is no direct relationship at all.)
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  8. #7  
    Forum Freshman
    Join Date
    Sep 2008
    Posts
    24
    Any rough idea on what the difference in time dialation would be on a scale say, between mercury and neptune? Anywhere near the magnitude of difference between field strengths?

    lol and by field strength I DO mean local acceleration due to gravity! haha
    Reply With Quote  
     

  9. #8  
    Suspended
    Join Date
    Apr 2008
    Posts
    2,178
    Quote Originally Posted by MagiMaster
    Orbital velocity depends on only two things, IIRC: mass of the star and the distance from it. I don't remember the exact formula though.

    Also, the asteroid-planet system is exactly the same as the planet-sun system. There's no difference.
    Wouldn't it also depend upon the elliptical orbit that some planets take around the sun as well?

    Sincerely,


    William McCormick
    Reply With Quote  
     

  10. #9  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    Actually, that's quite true, but since an elliptical orbit means that the distance is changing, the same relationship implies that the speed is changing too. Someone correct me though if it's not that simple.
    Reply With Quote  
     

  11. #10  
    Forum Freshman
    Join Date
    Sep 2008
    Posts
    24
    That sounds right, angular momentum would take over at that point though so that distance and velocity become inverted forces of one another to keep stability.
    Reply With Quote  
     

  12. #11  
    Suspended
    Join Date
    Apr 2008
    Posts
    2,178
    Quote Originally Posted by MagiMaster
    Actually, that's quite true, but since an elliptical orbit means that the distance is changing, the same relationship implies that the speed is changing too. Someone correct me though if it's not that simple.

    It will not be the same thing as a circular orbit. Because you are changing from a tangent orbit, to an elliptical orbit, that puts the path of the object on a non-tangent path.

    You are going to get the positive acceleration along the long sides of the ellipse heading to the minor axis. And it will slow at the ends of the approaching the major axis.

    I am sure you could create huge time differences in orbit completion if you change from a tangent path. I believe it is something totally different then a tangent orbit.



    Sincerely,


    William McCormick
    Reply With Quote  
     

  13. #12  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    Note though that I did not say that the orbital period is dependent on only mass and distance. Orbital rate depends on mass and distance, and distance can vary during the orbit. Orbital period is the integral of the distance divided by the rate over the whole orbit, which means it does depend on the eccentricity.
    Reply With Quote  
     

  14. #13  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,070
    Quote Originally Posted by MagiMaster
    Note though that I did not say that the orbital period is dependent on only mass and distance. Orbital rate depends on mass and distance, and distance can vary during the orbit. Orbital period is the integral of the distance divided by the rate over the whole orbit, which means it does depend on the eccentricity.
    Orbital period depends on the semi-major axis (or mean orbital distance) and mass or:



    Where a is the semi-major axis. It does not matter what the eccentricity of the orbit is as long as the the semi-major axis remains unchanged. A circular orbit with with a given radius will have the same period as an elliptical orbit with an equal semi-major axis.

    If however a different eccentricity leads to a different semi-major axis, then you will get a different period.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  15. #14  
    Suspended
    Join Date
    Apr 2008
    Posts
    2,178
    Quote Originally Posted by Janus
    Quote Originally Posted by MagiMaster
    Note though that I did not say that the orbital period is dependent on only mass and distance. Orbital rate depends on mass and distance, and distance can vary during the orbit. Orbital period is the integral of the distance divided by the rate over the whole orbit, which means it does depend on the eccentricity.
    Orbital period depends on the semi-major axis (or mean orbital distance) and mass or:



    Where a is the semi-major axis. It does not matter what the eccentricity of the orbit is as long as the the semi-major axis remains unchanged. A circular orbit with with a given radius will have the same period as an elliptical orbit with an equal semi-major axis.

    If however a different eccentricity leads to a different semi-major axis, then you will get a different period.

    I am curious how that formula calculates the varying velocity of the object in an elliptical path.

    In other words the object is going to be changing speeds, in what I call an unstable orbit. An elliptical path is not a circular path. There are other factors at work. As soon as you leave a tangent path, you introduce many other factors. And I do not see them just working into the perfect orbit formula.

    What are all the variables in that formula. Could you list them out?

    I see where an elliptical orbit might create the possibility for the sun to be effected by the orbit. Changing the whole game.

    Just like if the moons orbit was effected to become more elliptical, it could cause strange effects here on earth with tides. And to me inevitably change the orbit time of the moon. Compared to a more circular orbit.

    An eccentric orbit is very different then a circular orbit. People that work with spinning things would probably agree with me. I would think that an elliptical orbit would take less time then a circular orbit. Because of the long straight path the object takes.


    At what midpoint axis do you mean? The circle in this first picture cuts the ellipse at the midpoint of one of the ellipses, quarter segments midpoint.

    Just so you know, the distance around is still greater in the circle then the ellipse. In this first picture.




    In this picture the circler intersects the ellipse at the bisection of the right angle formed by the major and minor axis.

    In this second picture the ellipse circumference is much longer then the circle.





    Sincerely,


    William McCormick
    Reply With Quote  
     

  16. #15  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    Ah, I didn't know it reduced so well.

    William, the two variables are a, the semi-major axis, and M, the mass of the star. G and are constants. In this case, the formula doesn't deal with the speed during different parts of the orbit; it's just the total.
    Reply With Quote  
     

  17. #16  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,070
    Quote Originally Posted by MagiMaster
    Ah, I didn't know it reduced so well.

    William, the two variables are a, the semi-major axis, and M, the mass of the star. G and are constants. In this case, the formula doesn't deal with the speed during different parts of the orbit; it's just the total.
    If you want, you can find the speed at different radial distances(r) of the orbit by



    Note that for the instance of a circular orbit a=r for all points of the orbit and the equation reduces to that for a circular orbit given above.

    If you want to know the velocity at some number of degrees from perigee, you can use



    to find r for any angle theta and use the value in the equation above to find velocity at that angle.

    where e is the eccentricity,

    For perigee and apogee this equation reduces respectively to



    and



    because

    cos (0) = 1 and cos(180)=-1

    Substituting into the velocity equation we get



    and


    Giving the perigee and apogee velocity for an orbit with known semi-major axis and eccentricity.

    Example: Given Earth's mean orbital distance of 1.496e8 km, its orbital eccentricity of 0.01671, and a mass for the Sun of 1.99e30 kg we get a perihelion velocity of 31.86 km/sec and a aphelion velocity of 29.30 km/sec
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  18. #17  
    Forum Freshman
    Join Date
    Sep 2008
    Location
    Adelaide South Australia
    Posts
    8
    i would say our star(sun) does.
    Reply With Quote  
     

  19. #18  
    Suspended
    Join Date
    Apr 2008
    Posts
    2,178
    Quote Originally Posted by MagiMaster
    Ah, I didn't know it reduced so well.

    William, the two variables are a, the semi-major axis, and M, the mass of the star. G and are constants. In this case, the formula doesn't deal with the speed during different parts of the orbit; it's just the total.
    Then you are saying that the path of the ellipse could be about 3/4's as long as the circular path. That would mean that the elliptical path would take longer then the circular path, if they were the same length perimeter and circumference?

    It just does not seem to make sense to me. To me the elliptical path if equal to the circular path would take less time. But according to that formula you posted. The formula is claiming that the circular path is the same even though the circular path is almost 25 percent longer in this case then the ellipse I posted, with picture.





    Sincerely,


    William McCormick
    Reply With Quote  
     

  20. #19  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    First, it wasn't the formula I posted. I pointing out what had already been said. More importantly though:
    Quote Originally Posted by Wikipedia
    Note that for all ellipses with a given semi-major axis, the orbital period is the same, regardless of eccentricity.
    So yes, the ellipse and the circle would take the same time. This is certainly counterintuitive, but that doesn't mean it's not true. This is one case where doing the math would help illustrate what's really going on.
    Reply With Quote  
     

  21. #20  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,070
    Quote Originally Posted by MagiMaster
    First, it wasn't the formula I posted. I pointing out what had already been said. More importantly though:
    Quote Originally Posted by Wikipedia
    Note that for all ellipses with a given semi-major axis, the orbital period is the same, regardless of eccentricity.
    So yes, the ellipse and the circle would take the same time. This is certainly counterintuitive, but that doesn't mean it's not true. This is one case where doing the math would help illustrate what's really going on.

    Here's an animation showing both the circular and elliptical orbits with the same semi-major axises.

    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  22. #21  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    Nice picture. That does make things clearer. I had forgotten that the sun should be at one of the foci of the ellipse instead of the center.
    Reply With Quote  
     

  23. #22  
    Suspended
    Join Date
    Apr 2008
    Posts
    2,178
    Quote Originally Posted by Janus
    Quote Originally Posted by MagiMaster
    First, it wasn't the formula I posted. I pointing out what had already been said. More importantly though:
    Quote Originally Posted by Wikipedia
    Note that for all ellipses with a given semi-major axis, the orbital period is the same, regardless of eccentricity.
    So yes, the ellipse and the circle would take the same time. This is certainly counterintuitive, but that doesn't mean it's not true. This is one case where doing the math would help illustrate what's really going on.

    Here's an animation showing both the circular and elliptical orbits with the same semi-major axises.


    I would have to see the model before I buy that one. Do they have a working model of this somewhere. With all multi planetary variables taken out?


    Sincerely,


    William McCormick
    Reply With Quote  
     

  24. #23  
    Forum Freshman
    Join Date
    Sep 2008
    Posts
    24
    Yeah, it looks good.. but it seems like they wouldn't be able to maintain that stability. I would think that the gravitational force of the two planets on one another would pull them constantly together, inevitably forming a single orbit where they would collide.
    Reply With Quote  
     

  25. #24  
    Moderator Moderator Dishmaster's Avatar
    Join Date
    Apr 2008
    Location
    Heidelberg, Germany
    Posts
    1,624
    Quote Originally Posted by camtaylor17
    Yeah, it looks good.. but it seems like they wouldn't be able to maintain that stability. I would think that the gravitational force of the two planets on one another would pull them constantly together, inevitably forming a single orbit where they would collide.
    This model is not meant that you have both orbits at the same time. It only gives you an idea about the differences and similarities of both cases within one single model. In particular, you notice that the orbital periods are the same. By the way, Johannes Kepler was the first to describe planetary orbits as ellipses.
    Reply With Quote  
     

  26. #25  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,070
    Quote Originally Posted by camtaylor17
    Yeah, it looks good.. but it seems like they wouldn't be able to maintain that stability. I would think that the gravitational force of the two planets on one another would pull them constantly together, inevitably forming a single orbit where they would collide.
    As already pointed out, this image was only meant as a comparsion between the two orbits.

    That being said, if you were to start with such a two moon situation, a lot depends on the mass ratios between the bodies involved. Merging into the same orbit under any case is not a likely outcome. For instance, one effect will be for the circular orbit moon to cause the other moon's orbit to precess. As this happens, the moons will drift out of sync, complicating the system even more.
    The upshot is that a close miss is more likely to occur than a collision. And such a close miss would slinghot the moons into wildy different orbits, and might even eject one of the moons from the system entirely.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  27. #26  
    Forum Senior Booms's Avatar
    Join Date
    Sep 2008
    Location
    The perceptual schematic known as earth
    Posts
    361
    Simply put Orbital Velocity is exactly the same as normal velocity and is based upon distance from the sun, mass of the planet and sun, and gravity from the sun and any other nearby mass however small or huge





    however I think I remeber reading somewhere that Orbital Velocity is one of those 'greater than the sum of it's parts' freaks of nature type thing, that basically mean, if you swing a bucket round your head you need the rope to stop it flying off right? and if you spin it too slowly it falls on your head, this mean that following the basics our planet should be either flying away from the sun, or falling into it, but it isn't same goes for electrons around atom nuclei, the only way for things to be the way things are is if there is a rope, but there isn't.. weird huh?


    p.s sorry if that final paragraph is wrong, I just think I remeber reading it somewhere
    It's not how many questions you ask, but the answers you get - Booms

    This is the Acadamy of Science! we don't need to 'prove' anything!
    Reply With Quote  
     

  28. #27  
    Moderator Moderator Janus's Avatar
    Join Date
    Jun 2007
    Posts
    2,070
    I decided to run a couple of scenerios on Gravsim.

    I used two satellites of equal mass orbiting a Sun sized mass with the circuar orbit having a radius of 1 AU and the second orbit having an eccentricity of 0.5.

    On the first run, I gave the satellites 1/81 the mass of the Sun. (about the Earth/moon ratio. )
    The satellites quickly fell out of sync and eventually one satellite was ejected from the system.

    I then reduced the mass of the satellites by a factor of ten.

    The orbits began to change, but only by precession and eccentricity. the satellites remained in a 1 to 1 resonance. The orbits kept trading off between being more circular or eliptical but always maintained the same period and the satellites never got close to a collision. So far I've run the sim for over 1300 orbits, and there has been no sign of the satellites leaving this resonance.
    "Men are apt to mistake the strength of their feelings for the strength of their argument.
    The heated mind resents the chill touch & relentless scrutiny of logic"-W.E. Gladstone


    Edit/Delete Message
    Reply With Quote  
     

  29. #28  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    I can't remember, but has the general three body problem been solved? If it has, it might be interesting to look up the solution in this case. If it hasn't you might could work out something new here.
    Reply With Quote  
     

  30. #29  
    Moderator Moderator Dishmaster's Avatar
    Join Date
    Apr 2008
    Location
    Heidelberg, Germany
    Posts
    1,624
    Quote Originally Posted by MagiMaster
    I can't remember, but has the general three body problem been solved? If it has, it might be interesting to look up the solution in this case. If it hasn't you might could work out something new here.
    It can only be solved numerically, not analytically, unless approximations are introduced.
    Reply With Quote  
     

  31. #30  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    I'm pretty sure they've analytically solved at least some special cases. If this isn't already one of them, it might be worth looking into.
    Reply With Quote  
     

  32. #31  
    Moderator Moderator Dishmaster's Avatar
    Join Date
    Apr 2008
    Location
    Heidelberg, Germany
    Posts
    1,624
    Quote Originally Posted by MagiMaster
    I'm pretty sure they've analytically solved at least some special cases. If this isn't already one of them, it might be worth looking into.
    Indeed, there are solutions for special cases. But not for the general one. If you put some constraints to the problem (i.e. equal mass), you can find solutions like described in these references:

    http://www.ams.org/notices/200105/fea-montgomery.pdf
    http://www.ams.org/featurecolumn/archive/orbits1.html
    http://www.emis.de/journals/Annals/152_3/chencine.pdf
    Reply With Quote  
     

  33. #32  
    Forum Radioactive Isotope MagiMaster's Avatar
    Join Date
    Jul 2006
    Posts
    3,440
    Thanks for the links. There's some interesting stuff there.
    Reply With Quote  
     

  34. #33  
    Forum Freshman
    Join Date
    Sep 2008
    Posts
    24
    Quote Originally Posted by Dishmaster
    It only gives you an idea about the differences and similarities of both cases within one single model. In particular, you notice that the orbital periods are the same.
    But that is only because the masses of the planets which are orbiting in this model are assumed to be the same as one another, correct? Would a slight deviation in mass not affect the orbital period directly?
    Reply With Quote  
     

  35. #34 Re: Orbit velocity, what determines it? 
    . DrRocket's Avatar
    Join Date
    Aug 2008
    Posts
    5,486
    Quote Originally Posted by camtaylor17
    Is there a law which states that all orbital velocities (that is, the speed at which a planet travels around a star) are the product of some combination of forces? I do not mean simple inertial forces, such as an instance where a rogue astroid is caught in a planets gravitational pull and developes a stable orbit. I mean is the orbital velocity determined by some sort of relationship which exists between the planet and the star?
    You are looking for the vis viva equation of orbital mechanics.

    http://en.wikipedia.org/wiki/Vis-viva_equation
    Reply With Quote  
     

  36. #35  
    Moderator Moderator Dishmaster's Avatar
    Join Date
    Apr 2008
    Location
    Heidelberg, Germany
    Posts
    1,624
    Quote Originally Posted by camtaylor17
    Quote Originally Posted by Dishmaster
    It only gives you an idea about the differences and similarities of both cases within one single model. In particular, you notice that the orbital periods are the same.
    But that is only because the masses of the planets which are orbiting in this model are assumed to be the same as one another, correct? Would a slight deviation in mass not affect the orbital period directly?
    No. As long as the planet's mass is much larger than the mass of the central body (the sun), the mass of the orbiting body is independent of the orbital period. It is the distance that determines the orbital speed. You might want to look up Kepler's laws. Note that even Jupiter, the heaviest planet in the solar system has only 1/1000 of the solar mass.
    Reply With Quote  
     

  37. #36 Re: Orbit velocity, what determines it? 
    Moderator Moderator Dishmaster's Avatar
    Join Date
    Apr 2008
    Location
    Heidelberg, Germany
    Posts
    1,624
    Quote Originally Posted by DrRocket
    Quote Originally Posted by camtaylor17
    Is there a law which states that all orbital velocities (that is, the speed at which a planet travels around a star) are the product of some combination of forces? I do not mean simple inertial forces, such as an instance where a rogue astroid is caught in a planets gravitational pull and developes a stable orbit. I mean is the orbital velocity determined by some sort of relationship which exists between the planet and the star?
    You are looking for the vis viva equation of orbital mechanics.

    http://en.wikipedia.org/wiki/Vis-viva_equation
    Indeed, this follows from a fundamental principle in physics, the conservation of energy. Here it is the sum of kinetic and potential (gravitational) energy. Another similar law that is often used is the Virial theorem. In its simplest form it says that for bound orbits (circular, elliptic) the value of the kinetic energy is half of the potential energy. This means:

    This demonstrates that for planetary orbits (bound, i.e. elliptical) the orbital speed only depends on the mass of the central body (the sun) and the distance. This cannot be used for comets or unbound spacecrafts, because they have usually parabolic or hyperbolic orbits.
    Reply With Quote  
     

Bookmarks
Bookmarks
Posting Permissions
  • You may not post new threads
  • You may not post replies
  • You may not post attachments
  • You may not edit your posts
  •