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Kepler's third law of planetary motion

  1. Sep 29, 2015 #1
    1. The problem statement, all variables and given/known data
    Using the equation

    T = kr^p
    I drew a loglog graph of orbital time period against orbit radius for the planets mars to saturn.
    lnT = plnr + lnk

    My value for gradient was 5.25/0.48 = 10.94
    Meaning p = 10.94
    How do I compare this to the actual value? What is it?


    2. Relevant equations


    3. The attempt at a solution
     
  2. jcsd
  3. Sep 29, 2015 #2
    The question, as stated, does not make much sense. Is your system for Mars orbiting the Sun and Saturn orbiting the Sun? Or for moon's orbiting those planets? More information would be useful.
     
  4. Sep 29, 2015 #3
    Also, you should have some intuition about what p should be; find Kepler's third law, whether online or in a textbook, and look at the functional relationship between T and r. Most sites/texts are going to state the law with powers on both variables. If you solve for T, what should the power on r be?
     
  5. Sep 29, 2015 #4

    phyzguy

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    I suspect you made a mistake. Can you share the actual values you used for r, T, ln T, and ln r? Given Kepler's 3rd law, what should the value of p be?

    P.S. By the way, It is Kepler, not Keplar.
     
  6. Sep 29, 2015 #5

    gneill

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    Staff: Mentor

    You could start by stating Kepler's 3rd law as a Relevant Equation, then attempting a solution using it...
     
  7. Sep 29, 2015 #6
    gneill, I don't know what you mean by relevant equation.

    phyzguy, my values are follows:
    All radiuses stated are x10^10
    R:
    Mercury - 5.785
    V - 10.81
    E - 14.95
    M - 22.78
    J - 77.76
    S - 142.58

    T: (in earth days)
    M - 87.97
    V - 224.7
    E - 365.3
    M - 687.1
    J - 4333
    S - 10 760

    lnR (In the respective order):
    24.78
    25.41
    25.73
    27.38
    27.99

    lnT:

    4.477
    5.415
    5.901
    6.532
    8.374
    9.284

    I'm also going to look at Kepler's 3 planetary law a bit more and see if I can make sense of my results. In my textbook it is written Keplar in my defense.
    I will also add that I plotted all of my points carefully, using all of the significant figures where possible and used the biggest gradient possible.

    gTurner, I wasn't given a lot of information about these values. The table was put on the board and we were told to graph it.

    Thanks for replying :)
     
  8. Sep 29, 2015 #7

    phyzguy

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    So explain to me how you got a slope of 10.94 out of these numbers. From Mercury to Saturn, the change in ln T is about 4.8 and the change in ln R is about 3.2. How does this give a slope of 10.94??
     
  9. Sep 29, 2015 #8

    gneill

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    Staff: Mentor

    It's part of the post formatting template (See your first post in this thread):

     
  10. Sep 29, 2015 #9
    Ohh, I probably read my graph wrong because I used compressed scales. I'm a total fool sometimes, sorry.
    Those numbers should give 3/2 which is right because
    T^2 = a^3
    so T = a^3/2
    yes?
     
  11. Sep 29, 2015 #10

    gneill

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    Staff: Mentor

    That looks good.
     
  12. Sep 29, 2015 #11
    Oh, I understand. We haven't actually been through that equation in my lesson yet so I wasn't really sure about that. In my textbook it just said 'Keplar's third law".
     
  13. Sep 29, 2015 #12
    Can I just ask another thing, my teacher said the x intercept should give the radius of the sun but my value was out. Because I was just reading where the line crossed the axis I don't think I got that bit wrong.
    My intercept was 24.46
    So I did e^24.46
    4.196 x 10 ^ 10 m / 41.96 x 10^6 km
    Any ideas about that?
     
  14. Sep 29, 2015 #13

    gneill

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    It seems a strange idea to me. I don't see how Kepler's 3rd Law could account for the density of the Sun, which would depend (approximately) upon its mass and volume and hence its radius. Ultimately the periods of the planets depends upon the Sun's mass and their distances (See: Newton's derivation of Kepler's 3rd Law).

    Besides, how do you choose where the x-axis is on a log-log plot? There's no "y = 0" line.
     
  15. Sep 29, 2015 #14

    phyzguy

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    The intercept will give you the mass of the Sun, not the radius. But it will not give it to you directly, you will need to do some math. Try writing out the derivation of Kepler's 3rd law, taking the log, and calculating what the intercept should be.
     
  16. Sep 30, 2015 #15
    It makes even less sense to me. My teacher hasn't even talked about Kepler yet but we've started doing Gravity so he'll probably bring the two topics together and explain.

    I'm confused about you saying there's no y = 0 line?
    I think perhaps you think I meant a graph like this one

    https://www.msu.edu/course/fsc/441/logaxis.gif

    I just did a normal y = mx + c graph but using the logs of the numbers. My graph definitely has a y = 0 line :biggrin:
    I thought about what my teacher said in terms of the equation and worked out that if the x intercept gave the radius of the sun it would mean

    lnT = 0 and plnR + lnK = 0
    Meaning T = 1
    plnR = - lnK
    R ^ p = - k
    R ^ 3/2 = - k
    You should have to know what k is to get the radius of the sun? Have I interpreted that correctly? Looking at my graph, the y intercept is below the x axis so the negatives cancel.

    I'm sorry, I somehow quoted you three times!
     
  17. Sep 30, 2015 #16

    gneill

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    Staff: Mentor

    Still doesn't make sense to me. If T = 1, then 1 what? Year? Day? Hour? The result would appear to depend on the choice of units. And there's nothing that I can see that pins the R in the equation to the body of the Sun.

    Suppose the Sun were replaced by a black hole of the same mass. The planets would all still feel the same gravity and Kepler's laws would continue to work just fine and give the same results. Yet the size of the black hole would be much, much smaller than the Sun was.

    Perhaps the teacher was looking at a way to isolate the gravitational parameter k of the Sun?
     
  18. Sep 30, 2015 #17
    Ah, my units for T are in terms of day. I need to ask him to explain what he meant better I think.
     
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