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Using Kepler's constant for Jupiter

  1. Aug 3, 2009 #1
    1. The problem statement, all variables and given/known data
    Given that Jupiter has a mass of 1.9*10^27kg, and the sun has a mass of 1.99*10^30kg:

    a)Calculate the value of Kepler's constant for Jupiter


    2. Relevant equations

    k = r^3/T^2
    K = G(M+m)/4pi^2


    3. The attempt at a solution

    Well since the question only gave me the mass of Jupiter and the Sun I'm assuming that I should use Gm/4pi^2 instead even though this formula should only be used when the masses of the two objects (Jupiter and Sun) are close to each other......right?

    Let G = Universal gravitational constant
    Let M = mass of the Sun
    Let m = mass of Jupiter

    K = G(M+m)/4pi^2
    k = (6.67*10^-11)(1.99*10^30 + 1.9*10^27)/4pi^2
    =3.365*10^18, I'm not sure about the units for this formula

    I'm pretty sure I can't use k = r^3/T^2 because the question didn't provide me with enough information to do so, but out of curiosity I googled the radius of Jupiter along with its period and plugged it in the formula:

    radius (r) is in meters and period (T) is in days (should it be in days or years?)
    k = r^3/T^2
    = 71492000^3/4331.57^2
    = 1.9475*10^16 m/d

    If T was in years (11.89):
    k = r^3/T^2
    = 71492000^3/11.89^2
    = 2.584*10^21 m/y

    Ok, so clearly these numbers aren't close to the one above. Can you guys point out what I'm doing wrong?

    also another confusion that I have, according to other sources(Wikipedia) Kepler's constant should be k = T^2/r^3 rather than(my textbook) k = r^3/T^2.

    and if I use K = T^2/r^3 (for T is in days)
    K = 5.134*10^-17 d/m

    for T in years:
    k = 3.868*10^-22 y/m

    Please help me out on this
     
  2. jcsd
  3. Aug 3, 2009 #2

    kuruman

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    Homework Helper
    Gold Member

    Your dimensions need fixing. The dimensions of K (or k) are [Length]3/[Time]2. If you choose the formula that has G in it, it is less confusing to do all calculations expressing the period in seconds, then convert to m3/yr2 for K if you wish. That's because G has "seconds" buried in it as its units are N.m2/kg2 and N = kg.m/s2.
     
  4. Aug 3, 2009 #3
    Thanks for the reply.
    according to what you said, in this case, would it be better to use the formula with G in it?

    also, Should the calculation be done in years or days?
     
  5. Aug 3, 2009 #4

    kuruman

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    Homework Helper
    Gold Member

    Either way should give you the same number. However, if you use the formula with G, the period must be expressed in seconds.
     
  6. Aug 3, 2009 #5
    ok, I'll try that.
    thanks a bunch for the help. I'll come back to this topic again next week as I'm going on a short trip.
     
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