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Kepler's Laws - implications of changed theory

  1. Mar 3, 2006 #1
    I've been presented with the following problem:

    [itex]Problem[/itex]: As a thought experiment it's presumed that the attracting force of gravitation instead of being inversely proportional to [itex]r^2[/itex], is inversely proportional to [itex]r^3[/itex]. How does this theory change Kepler's 3 laws?

    (For one of the laws it isn't easy to determine, so a "Can't easily be determined" is an acceptable answer.)

    a) How is Kepler's 1st law changed?
    b) How is Kepler's 2nd law changed?
    c) How is Kepler's 3rd law for cirkular planetary orbits changed?

    ----

    I'm merely looking for a few hints. I assume it to be implicit that no advanced equation juggling will be needed (the topic is, accordingly, treated like this in the textbook too).

    a) Kepler's 1st law.
    -----
    "Each planet moves in an elliptical orbit, with the sun at one focus of the ellipse."
    -----
    Applying the theory will make a given planet accelerate less towards the sun, thus changing its orbit. Its period of orbit will drop. It might settle itself in an orbit with larger axes, or continue spiraling outwards - eventually escaping the orbit - since the escape speed will be smaller (due to its relation with [itex]G[/itex]).


    b) Kepler's 2nd Law
    -----
    "A line from the sun to a given planet sweeps out equal areas in equal times."
    -----
    This law doesn't, like the 1st and 3rd, in reality, work only for a [itex]1/r^2[/itex] force. That might hint to the conclusion that this is the law whose change isn't easy to determine. Irrespectively - it seems a bit hard. Unless some analysis of the [itex]dA/dt[/itex] and angular momentum, if e.g. the angular momentum [itex]L[/itex] isn't constant anymore.


    c) Kepler's 3rd Law
    -----
    "The periods of the planets are proportional to the 3/2 powers of the major axis lengths of their orbits."
    -----
    Newton's 2nd Law gives, for the radial speed, [itex]v=\sqrt{(G*m_E)/r}[/itex], which with the theory must be changed to [itex]v=\sqrt{(G*m_E)/r^2}[/itex], thus making the period [itex]T[/itex] larger - so that it's proportional to (3/2+1) = 5/2 powers of the major axis length: [itex]T=(2*\pi*r^{3/2})/\sqrt{G*m_e}[/itex], because of the relation [itex]T=(2*/pi*r)/v[/itex].

    ----------
    How much of this is on the right track?
     
    Last edited: Mar 3, 2006
  2. jcsd
  3. Mar 3, 2006 #2
    1. An additional possibility is that it will orbit, but the orbit won't be closed. The orbit would look like something you get off a spirograph, if you're old enough to know what one of those is! (I don't think they make them anymore, do they?)

    2. Regardless of the power of the force law, the angular momentum is still conserved.

    3. I glanced over this and everything looks okay. Again, in reference to my comment on 1, the orbit may not be closed so talking about the period of motion may not make sense.

    Hope it helps, rather than confuses! :wink:

    -Dan
     
  4. Mar 3, 2006 #3
    Guesses at answers

    Thanks for the quick reply!

    I'm prone to thinking that the answer in 1) is a spiralling orbit, which would make 2) the hard-to-determine-one (since the type of orbit would be hard to determine). However, if the orbit isn't closed, it might be 3) that's hard to determine, since it only deals with a period.

    So either:

    Answer I -
    1. The orbit's some kind of spiral, and thus not a closed orbit
    2. This is hard to determine, because it's hard to determine how [itex]dA/dt changes with changes in radius[/itex]
    3. ?

    Answer II -
    1. The orbit's closed, but larger than before
    2. Hard to determine.
    3. The period is now related to 5/2 powers of the radius (instead of 3/2).

    It seems, though, that nr. 2 in Answer II wouldn't be hard to determine (maybe nr. 1 will be instead).
     
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