Calculating a planet's rotation

Discussion in 'Highly Technical' started by Rakehellion, Jan 16, 2013.

  1. Rakehellion

    Rakehellion Lifer

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    [​IMG]

    I'm calculating gravitic influence as pow(distance, -2) * mass1/mass2.

    The side of the planet farthest from its neighbor would be less affected by gravity, thus the planet should rotate as it orbits. How would you find the exact amount?
     
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  3. TuxDave

    TuxDave Lifer

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    First point... great drawing. :)

    Next point, I didn't think there was a requirement that all planets must rotate in steady state when orbiting another mass. Take the moon and earth for example. So I guess if you could clarify the "thus the planet should rotate" part. I'm rusty on physics so I could be completely off base.
     
  4. Sunny129

    Sunny129 Diamond Member

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    fixed. the fact that the gravitational force exerted by a body on another body's far side is less than the gravitational force exerted on that same body's near side does not imply rotation. if anything, such a gravitational interaction would tend to slow the rotation of a body, provided it is close enough to its gravitating neighbor. taking it one step further, if the bodies are similar in mass, then the rotation of both bodies will tend to slow over time (again, provided they're close enough in proximity), eventually resulting in a tidal lock.

    but if you think about it, the moon does rotate...its rotational period just happens to be the same as its orbital period. because its rotation period matches its orbital period, one side of the moon always faces earth, and the other side always faces away from earth.
     
  5. TuxDave

    TuxDave Lifer

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    Woops, that's right.
     
  6. Pia

    Pia Golden Member

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    That kind of damping is due to deformation of the bodies, so the normal rigid physics model OP is using would not model it. Just pointing this out to prevent confusion.
     
  7. DrPizza

    DrPizza Administrator Elite Member Goat Whisperer
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    Just to be technical, (given the forum), inertia isn't a vector. You're confusing it with momentum. Inertia is equivalent to mass.

    Wikipedia's entry for tidal locking includes a formula to estimate how much time would elapse before two bodies became tidally locked.
     
  8. Rakehellion

    Rakehellion Lifer

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    And if the body escapes orbit, it should continue rotating. Or if the body wasn't rotating before it met its neighbor, it will begin.

    I'm needing to calculate these values on the computer, so my numbers are appropriately naive and parsimonious.

    [​IMG]

    Is there a lazy way to plug Θ, mass1, mass2, diameter1, diameter2, and distance into a function to approximate the angle of rotation?
     
  9. Pia

    Pia Golden Member

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    ???
    Gravity is a force. Inertia is not a force. "Gravity + inertia" is nonsense, and any angle dependent on it is likewise nonsense.
    There is no way, whether easy or hard. If you are working with a rigid physics model, the bodies do not change the rotation of each other and will just keep rotating at a constant rate. And if it's supposed to be a non-rigid model, then you haven't given any parameters at all describing the non-rigid part.
     
  10. Rakehellion

    Rakehellion Lifer

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    Kinetic energy of the planet's current motion. I'm not really trying to calculate force, I'm trying to calculate movement in response to force.

    How is that possible when the force exerted on them is not constant?
     
    #9 Rakehellion, Jan 19, 2013
    Last edited: Jan 19, 2013
  11. KillerCharlie

    KillerCharlie Diamond Member

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    That's not a simple technicality!

    OP, you need to read a physics book. Your illustration is so far from making any sense that you might as well plot apples vs. oranges.
     
    #10 KillerCharlie, Jan 20, 2013
    Last edited: Jan 20, 2013
  12. mpo

    mpo Senior member

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    Uranus has an axial tilt of 97 degrees. The planet rotates on its 'side' so the north pole will experience 42 years of sunlight, followed by 42 years of darkness.

    Venus' axial tilt is 177 degrees. It rotates in the opposite direction compared to most of the other planets.

    How can you account for those two planets in your model?
     
  13. sufs

    sufs Junior Member

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    Afaik a planets rotation exists because of conservation of angular momentum. If I remember correctly applied force will either add to this or subtract from it. As somebody pointed out earlier the effect you are trying to describe will subtract from the angular momentum, and thus slowing the rotation.

    Trying to clarify what I think you have problems understand I can say this, linear motion (which is usually easy to understand) has analogy to angular motion like this:

    Mass -> Interia
    Force -> Torque

    Mass and inertia are not vectors, force and torque is.
     
  14. Rakehellion

    Rakehellion Lifer

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    I forgot to mention that I'm doing this in real time on a computer, so what I need is a very close approximation. I can't model every particle and I probably can't do non-rigid bodies.

    Despite my mixing of terminology, several people seemed to understand what I meant just fine. I'm posting here apparently because I'm not familiar with the physics involved. I need help, not putdowns.

    I don't. The idea is that I'll create my own solar system wherein orbit, rotation, and axial tilt will render themselves organically.

    I meant velocity sustained by inertia. Can't get too verbose in iPhone sketches.


    -Planet A moves in a path past Planet B
    -The far side of Planet A will be less affected by gravity, thus continue moving along the path
    -The near side of the planet will be more compelled to follow gravity
    -The near and far sides of the planet are moving in different directions, thus causing it to rotate

    Right? I need to know by how much. I don't know anything about the calculus involved and it would help immensely if someone could point me in the right direction.
     
  15. KillerCharlie

    KillerCharlie Diamond Member

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    I am not trying to put you down, I am trying to point you in the right direction. I work as an aerospace engineer, have taken multiple orbital mechanics courses, and tutored college physics. I can tell that you are struggling with some very basic concepts, like the difference between force and mass. You simply cannot try to jump to an understanding of more advanced concepts without understanding the basics.

    First of all, if you make a sketch with vectors on it, they have to be indicating the same thing. Plotting inertia and a force on the same plot is not useful. Inertia is not even a vector, it's essentially a measurement of mass like Dr. Pizza said.

    You're also trying to get a rotation out of a point mass - you're drawing forces that are acting at that point and there's no torque, thus no source of angular momentum. The only force acting in your sketch and modelling simplification is gravity. That's it. From that force, and a few starting values, you can easily compute the orbital speed of one body around another, but not its rotational velocity at all.

    In fact, It is 100% impossible to predict the rotational speed of a planetary body short of directly measuring it.

    In your model you're assuming constant density, uniform planets. A spherical, uniform planet behaves exactly like an infinitely small point mass (when you integrate the gravitational forces over its volume). Sure, the far side of the planet has less gravitational force, but the close side has more, balancing it out. No rotation is created.

    If you take a massive leap in the fidelity of you're modelling, it is possible to model the non-spherical, non-constant density planet. Because the planet isn't uniform, a small torque could be imparted on a planet, changing its angular velocity - but the angular momentum of the system is constant without outside force. It is possible to compute this torque with detailed data, but it won't tell you anything about the current rotational speed of the planet. Tidal forces and such can damp the rotation of a body, but the total angular momentum of the system is still conserved.

    The rotational speed of a planet about its axis is mostly a function of how the planet was formed (how the space dust came together) and discrete events like asteroid impacts. There's nothing that says the planets even have to spin in the same direction - some do not (retrograde motion). The spin of a planet is mostly unrelated to its orbit.

    And the "computer program" you're using - I can tell you as a practicing engineer that unless you understand exactly what a computer program is doing, it will do you more harm than good 100% of the time. You need to understand the physics and math behind the problem, starting with the basics. There is no shortcut.
     
    #14 KillerCharlie, Jan 20, 2013
    Last edited: Jan 20, 2013
  16. Rakehellion

    Rakehellion Lifer

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    Why would it behave as a point? Wouldn't uneven forces make things less balanced, not more?

    A cannon ball is fired from a cannon, hits the ground, and begins to roll in the direction of travel. What's wrong with this analogy?

    That sounds like an exaggeration.
     
    #15 Rakehellion, Jan 21, 2013
    Last edited: Jan 21, 2013
  17. Pia

    Pia Golden Member

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    The friction between the cannon ball and the ground is only on one side of the ball, thus there's non-zero torque. In the case of rigid planets and gravity, if you consider the line that goes through the centers of the planets, you can mirror every point of a planet about that line to find another point whose torque from gravity cancels out the torque from the first point. Thus, the rotation cannot change.
     
  18. KillerCharlie

    KillerCharlie Diamond Member

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    It is not possible to calculate the rotation of a planet without measuring it directly. What other proof do you need?

    I can pull out my textbooks and show you the derivation if you really want to see it. However, that depends on understanding calculus and physics, so it'll just be nonsense to you.
     
  19. silverpig

    silverpig Lifer

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    The fact is the rotational period of a planet will change throughout time too. It can't just be predicted.

    Some planets spin quickly, some are tidally locked, some spin "backwards", some spin on their side.
     
  20. serpretetsky

    serpretetsky Senior member

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    I like this question. Let's try this perspective.

    Any time an objects changes rotational velocity, there MUST be an acting torque on it.

    If you're not familiar with torque, torque is a rotational force. Torque can be measured by multiplying the force being applied onto a lever and the length of the level (so a force applied to a long lever gives more torque than the same force applied to a shorter lever).

    If an object experiences no torque, there will be no difference in its rotational velocity. That means that whatever rotational speed the object had before some zero-torque event is the exact same rotational speed it will have after that event.

    Let's see if there are any sources of torque during the orbit of a spherical celestial object about a large source of gravity.

    We will be using one assumption that I am not proving. For solid spherical objects that give of a field that is inverse-square, the entire sphere can be modeled as a point. This means, if for example we were going to see if the Sun should impart any rotation onto the earth, we can model the sun as a single point. If this assumption isn't satisfactory, please reply and i'll try to prove it as well (be warned, it probably involves calculus).

    edit: Actually, because we have so much symmetry, i think it's possible to due this entire proof without assuming a point gravitational source.

    Alright, so here is the setup:
    [​IMG]

    We have large source of gravity that we will be treating like a point source, and it imparts a gravitation field onto a smaller celestial sphere. Now, I am assuming this sphere has a symmetrical density about its radius.

    Let's choose an arbitrary, but small volume, somewhere in our celestial sphere. This volume is shown red in the image. For my example, it's irrelevant where you choose this volume. If you choose closer to the gravity source, it will experience a stronger force, if you choose farther from the gravity source it will have a weaker force. It doesn't matter, it's irrelevant to this proof.

    Now, if we calculate the force on this volume, we find that volume is some distance from the center of the celestial sphere, and therefore any force on this volume will impart a torque onto the celestial sphere.

    But hold on there! We have a dotted line from the center of our gravity source through the center of our celestial object. For ANY arbitrary volume that we pick on our celestial sphere, we can find another equal volume symmetrical opposed across the dotted line. (the volume pairs are such that a line drawn between them will intersect our dotted line and create a right angle)

    Because we chose a dotted line that goes from the center of our gravity source to the center of our celestial sphere, any pair of equal volumes radially symmetrical about this dotted line will have equal density and be equidistant from our gravitational source. This means an equal force due to gravity will be acting on them.

    And because we also choose these two volumes to be on opposite ends, then any torques they have will cancel.

    Since there force due to gravity is equal, and since they are equalidistant from the center of our celestial sphere, their torques completely cancel.

    For ANY arbitrary volume that you choose on our celestial sphere, the torque it imparts onto the sphere is completely canceled by it's opposite pair. If you sum up all the pairs you can create the entire volume of the celestial sphere, and the sum of torques will still be zero. Therefore there is zero net torque on our celestial sphere.


    therefore, whatever rotation velocity the celestial sphere had when it entered the gravitation field, is the same rotational velocity it will have throughout orbit, and will be the same rotation velocity it will have if it ever leaves, unless some other outside torque acts on it.
     
    #19 serpretetsky, Jan 23, 2013
    Last edited: Jan 23, 2013
  21. Pia

    Pia Golden Member

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    I kinda said that already, but with one sentence. :)
     
  22. silverpig

    silverpig Lifer

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    That only works for perfect spheres that are perfectly rigid... which planets are not. Tidal forces will transfer energy until the planet is in tidal lock.
     
  23. Sunny129

    Sunny129 Diamond Member

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    yes, but nothing drives the point home quite like an illustration, and as someone who is fairly novice in the physics department, i think this may be exactly what the OP needs to see in order to understand why gravitation does not induce rotation of the body being acted upon. we'll see how he responds...
     
    #22 Sunny129, Jan 23, 2013
    Last edited: Jan 23, 2013
  24. serpretetsky

    serpretetsky Senior member

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    Excuse me, i thought we were simply talking about rigid physics since the OP already admitted can't do non rigid physics.
     
  25. silverpig

    silverpig Lifer

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    Right you are then.
     
  26. Rakehellion

    Rakehellion Lifer

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    I said I might be able to, but processor power is very limited. I want to simplify the model as much as possible.

    I see that my original notion was way off and I'm looking into tidal locking.