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3 explanations of comet velocity. OK?

  1. Dec 6, 2007 #1
    [SOLVED] 3 explanations of comet velocity. OK?

    1. The problem statement, all variables and given/known data

    This is from Advanced Physics by Adams and Allday, spread 3.31.

    Explain why comets in very eccentric orbits move very slowly far from the Sun and very fast close to it. Give three explanations, one in terms of energy, one in terms of angular momentum, and one in terms of forces acting on the comet around its orbit.

    2. Relevant equations

    Rotational kinetic energy: [itex]R.K.E. = 0.5 I {\omega}^{2}[/itex]

    Moment of inertia of point mass: [itex]I = m r^{2}[/itex]

    Angular momentum: [itex]L = I \omega[/itex]

    Centripetal acceleration: [itex]a = r {\omega}^{2}[/itex]

    Gravitational force: [itex]F = G m_1 m_2 / r^{2}[/itex]

    3. The attempt at a solution

    I would be confident this was right if I could understand and accept the implication.

    In each case I answered the question by showing that [itex]r^{a} {\omega}^{b}[/itex] is constant so [itex]\omega[/itex] must become small as r becomes big and vice versa.

    For energy, conservation of energy yields [itex]0.5 m r^{2} {\omega}^{2}[/itex] is constant so [itex]r^{2} {\omega}^{2}[/itex] is constant.

    For angular momentum, conservation of angular momentum yields [itex] m r^{2} {\omega}[/itex] is constant so [itex]r^{2} \omega[/itex] is constant.

    For forces, centripetal acceleration and gravity yield [itex]G m_1 m_2 / r^{2} = m r {\omega}^{2}[/itex] so [itex]r^3 {\omega}^{2}[/itex] is constant.

    Is it correct that [itex]r^{2} {\omega}^{2}[/itex], [itex]r^{2} {\omega}[/itex], and [itex]r^3 {\omega}^{2}[/itex] are all constant? It feels unlikely -- very constrained. If it is true then what is its significance, physically? Am I being naieve -- are these simply properties of an ellipse?
     
  2. jcsd
  3. Dec 6, 2007 #2
    Well it's clear what the significance is from the equations. You're saying their constant, so you're saying the things on the other side of the equal sign are constant. Are they? For the first two you assumed conservation of energy and momentum, which is correct, for the second, you have r^3w^2=masses and G, do the masses or the gravitational constant ever change? Nope
     
  4. Dec 7, 2007 #3

    Shooting Star

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    Homework Helper

    Hi catkin,

    Let's see what the energy concept tells us. The total energy is constant and is the sum of the PE and KE. Now, the PE is low when near to the sun, being the least at the perihelion, and so the KE is high, and thus the speed is the high. (Why is the PE low close to the sun?)

    Can you give similar qualitative arguments for the other cases?
     
  5. Dec 10, 2007 #4
    Sorry for the delay -- other priorities. Here goes ...

    PE is low(est) close to the sun because distance from Sun ("height") is smallest.

    Angular momentum is constant. Moment of Inertia is greatest when furthest from the Sun and smallest when closest to the Sun, hence angular velocity is smallest when furthest from the Sun and greatest when closest to the Sun.

    Gravity provides the centripetal acceleration and is greatest nearest the Sun, smallest furthest from the Sun but centripetal acceleration is proportional to the distance from the Sun hence the angular velocity is smallest when furthest from the Sun and greatest when closest to the Sun.

    How's that? It hasn't yieded the Eureka moment yet, regards all 3 terms being constant.
     
  6. Dec 10, 2007 #5

    Shooting Star

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    Homework Helper

    Quite good reasoning. Let's add just a bit of math here.

    Using the energy concept, the PE is -GMm/r. As r decreases, GMm/r increases, and so -GMm/r decreases.

    Gravity does not provide the whole of the centripetal force, except in circular orbits. That is to say, the normal to the path of the comet is not directed toward the sun, except at the closest and the farthest points. But nearer to the sun, the centripetal force is almost in the direction toward the sun, and so mv^2/r is high, and r is low, implying v^2 is high.

    But all the three "constants" you had mentioned cannot be constants. If they are, then both omega and r are constants, resulting in circular motion.
     
  7. Dec 11, 2007 #6
    Thanks, Shooting Star, that makes perfect sense. For uniform circular motion any f(r,ω) will be constant because r and ω are constant.

    The question is the second of a handful of four "practice questions" after a section on circular motion; we haven't covered elliptical orbits yet. Hence my attempt, using circular concepts to answer an elliptical problem.

    Don't ask -- I don't understand why the authors ask a question beyond the topic. They often do it and it doesn't help when you're learning a new topic. Useful when it comes to synoptics and we're trying to pull it all together but not early on and definitely not as the second practice question!
     
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