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Kepler's Laws-What am I doing wrong here?

  1. Jan 5, 2007 #1
    Kepler's Laws--What am I doing wrong here?

    1. The problem statement, all variables and given/known data
    Io, a satellite of Jupiter, has an orbital period of 1.77 days and an orbital radius of 4.22 x 10^5km. From these data, determine the mass of Jupiter.


    2. Relevant equations
    T^2=(4(pi)^2/GM)r^3


    3. The attempt at a solution
    I converted the radius from km to meters and substituted all of the numbers to their variables. However, I still came up with the wrong answer. Please help me check my work, thanks!

    Divided r^3 from both sides:
    (1.77)^2/(4.22x10^8m)^3= 4(pi)^2/(6.673 x 10^-11)M

    Multiplied universal gravitation constant with M(mass) on both sides:
    (4.1688x10^-26)(6.673x10^-11)M=4(pi)^2

    solved for M: M=4(pi)^2/ (2.78184 x 10^-36)

    I came up with: 1.4192 x 10^37

    The correct answer is: 1.90 x 10^27 kg

    What am I doing wrong? Thanks!!
     
  2. jcsd
  3. Jan 5, 2007 #2

    chroot

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    Don't plug in numbers until you're done with the algebra.

    Solve for M:

    [itex]T^2 = \frac{4 \pi^2 r^3}{G M}[/itex]

    [itex]M = \frac{4 \pi^2 r^3}{G T^2}[/itex]

    Convert your units into mks:

    [itex]r = 4.22 \cdot 10^8\,m[/itex]

    [itex]T = 152928\,s[/itex]

    Plug into the equation, and you're done. You appear to have left your T in units of days, which is inconsistent with the SI mks system.

    - Warren
     
    Last edited: Jan 5, 2007
  4. Jan 5, 2007 #3
    Ah I see.. So is every single variable in Physics in the SI unit? Thanks a lot!
     
  5. Jan 5, 2007 #4

    chroot

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    You're free to use any unit system you want, but you must be consistent. If you want to use G = 6.673 x 10^-11, you must understand its units:

    [itex]G = 6.67300 \cdot 10^{-11} \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{s}^{-2}[/itex]

    That means that everything else in your equation must in terms of meters, kilograms, and seconds: the standard mks SI system.

    If you really wanted to use days as your fundamental unit of time, you could, but G would be:

    [itex]G = 0.498136781 \,\textrm{m}^{3} \,\textrm{kg}^{-1} \,\textrm{day}^{-2}[/itex]

    - Warren
     
  6. Jan 5, 2007 #5
    Wow, thanks for all the superfluous replies! Okay, I understand it clearly now. Thanks to you! :smile:
     
  7. Jan 5, 2007 #6

    chroot

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    No problem. Try plugging in your values for r in meters and T in days, using the "version" of G that I provided. :wink:

    - Warren
     
  8. Jan 5, 2007 #7

    berkeman

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    Your adjective "superfluous" is incorrectly applied in this context. Please re-check the definition. Thank you.
     
  9. Jan 5, 2007 #8
    Yeah, sometimes I don't know if the vocabulary I've learned makes sense in what I say/write. Here is the definition: Being beyond what is required or sufficient. How come it doesn't work with chroot's replies? :biggrin:
     
  10. Jan 5, 2007 #9

    chroot

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    Well, the word "superfluous" usually means that something was pointless or needless, not that it was unexpectedly generous.

    You might say "Man, this textbook is horrible! It's full of superfluous sidebars," in order to disparage it. You probably wouldn't tell your mom "Thanks for all these superfluous Christmas gifts!"

    - Warren
     
  11. Jan 5, 2007 #10
    Oh I see why berkeman said that to me then. Sorry if you thought of it as the other way. My vocabulary isn't that great and it's still developing. I mean I know the definitions to some words but I just haven't had much experience with them. That's why some of my sentences are weird lol. Alright, thanks for the vocab. lesson xD I'll try not to use the word superfluous as something that is too super. Thanks to both of you.
     
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