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Orbits of planets/Angular momentum

  1. Mar 9, 2010 #1
    1. The problem statement, all variables and given/known data

    Show that the angular momentum in circular orbit around a mass
    M can be written as functions of just the masses, the
    orbit radius, and G.


    2. Relevant equations

    L = r x p = r x mv
    L = Iw


    3. The attempt at a solution

    I had no trouble showing that the total energy can be written with these variables, but the angular momentum is proving much tougher. I know this has to do with Keplars Law and maybe this will help:
    Newton's law of universal gravitation states that the force is proportional to the inverse square of the distance; a_r = -GM/r^2
     
  2. jcsd
  3. Mar 9, 2010 #2

    collinsmark

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    So far so good. Now think about how centripetal acceleration might fit into the picture too.
     
  4. Mar 9, 2010 #3

    ideasrule

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    You also know that the acceleration is equal to the centripetal acceleration, v^2/r.
     
  5. Mar 9, 2010 #4
    Centripital acceleration = force of gravity! (GMm/R^2)

    bing bong boom do some calculations and convert the angular velocity from centripital to velocity through v=wr and then I should get the angular momentum. One question, so
    L=mvr and I found what v was and I already have r but to which mass does that m in the equation refer to??
     
  6. Mar 9, 2010 #5

    ideasrule

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    No, centripetal acceleration = ACCELERATION of gravity. Remember that F=ma, so a force can never equal an acceleration.
     
  7. Mar 9, 2010 #6
    oops sorry, I wrote out my thoughts too quickly, thusly they made no sense

    Here was my thought process;

    Fgrav = GMm/R^2
    Centripital Force = mw^2R so I decided to set these two equal to eachother and find the angular velocity which ended up being this:
    w = the square root of (GMm/mR^3)
    then keeping in mind that v=wR I get
    v = R * the square root of (GMm/mR^3)
    and because we know L = mvR I just plugged in the v I just got -->
    L = m * (the square root of (GMm/mR^3) * R) * R

    I really hope that's right because it makes sense to me. My only question was whether the beginning m was the orbiting object or the stationary one??
     
  8. Mar 9, 2010 #7

    ideasrule

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    Seems right, but why didn't you cancel out the m's in GMm/mR^3?

    "m" is the orbiting object's mass because centripetal (=towards center) force applies to the object that's moving in a circle.
     
  9. Mar 9, 2010 #8

    Angular velocity is given as [itex]\omega[/itex], where [itex]\theta[/itex] is the angle with some value, and [itex]\Delta[/itex] is the change operator.

    They have the given relationships:

    [itex]\omega = \frac{\Delta \omega}{\Delta t}[/itex]

    [itex]\alpha = \frac{\Delta \omega}{\Delta t}[/itex]

    which reads that the angular acceleration is the change in angular velocity over time. Since you have the condition of [itex]L=I \omega[/itex], this means that we may as well describe this in terms of a linear momentum, without any perturbations.

    It can be calculated that the velocity of the circular motion by the formula:

    [itex]V_{constant} = \frac{2 \pi R^2}{t}[/itex]
     
  10. Mar 9, 2010 #9
    In a similar vein, I was trying to figure out what would happen to the length of our days if the moons distance was 1.5 times its current value from earth. I found an equation online that said that P = A^1.5 where P is the orbital period and A is the average distance. I don't understand how they derived or got that equation? any ideas?
     
  11. Mar 9, 2010 #10

    collinsmark

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    I wouldn't say that it's equal, but the period is proportional to distance raised to the 1.5 power, yes.

    It can be derived with the same equations discussed in this thread. Just recognize that the period is time taken for the object to go all the way around the circle. In other words, the time taken for the object to traverse a distance of [tex] 2 \pi R [/tex]. Noting that speed times time equals distance,

    [tex] vP = 2 \pi R [/tex]

    or

    [tex] P = \frac{2 \pi R}{v} [/tex]

    With that, combined with some other equations discussed in this thread, you should be able to calculate where the R1.5 proportionality comes from.
     
    Last edited: Mar 9, 2010
  12. Mar 10, 2010 #11
    I have just noticed i've used [itex]R^2[/itex] in my last post. This post has correctly stated the equation:

    [tex] P = \frac{2 \pi R}{v} [/tex] without any sqaure term - sorry about that. Glad no one noticed... i think ;)
     
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