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Homework Help: 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


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    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


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    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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