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Attn. Oh great helpers

  1. May 12, 2006 #1
    Simple Physics question.

    I thInk this is physics? anyhow.. :confused:

    I found that the distance from the earth to the sun is 1.5*10^11m , and
    that the mass of the sun is 1.9*10^30kg,

    and i found Keplers's Law of Periods here

    but I do not understand how to do the rest... :shy: :yuck: :confused: :cry:

    1.use Kepler’s Law of Periods to find the period of the earth’s orbit, recorded in seconds
    2.show how to express your answer in years; and
    3.if your answer does not agree with the accepted earth orbital period of 365.25 days, explain the discrepancy.
    Last edited: May 12, 2006
  2. jcsd
  3. May 12, 2006 #2


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    Homework Helper

    1. Just apply the formula. To help you out, write down the values of the variables and constants.

    2. How many seconds in a year? (1 Year = 365 days*24 hours/day*60 minutes/hour*60 seconds/minute)

    3. Think of the simplifications involved. Some hints : what's the reason for the leap year? Make a recalculation taking the leap year cycle into account. Can that alone explain the discrepancy? Why does the mass of the Earth not appear in the formula? Is that an acceptable simplification? Do you think these astronomical measurements of distance and mass are made or given to the requisite precision? Justify your concern about data precision by estimating the relative uncertainty in the final answer caused by relative uncertainties in each of the variables.
    Last edited: May 12, 2006
  4. May 12, 2006 #3
    thanks for your help but i dont know how to do any of that, its ok
  5. May 12, 2006 #4


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    Homework Helper

    Why is it OK? Isn't this stuff you're supposed to learn to do?:confused:

    Start with the formula.[tex]T^2 = \frac{4\pi^2}{GM}a^3[/tex]. You're required to find the value of T. If you use standard (SI) units for all the other constants and variables, you'll get T in seconds as required for the first part.

    G is the Universal Constant of Gravitation, given by 6.673*10^(-11) Nm^2/kg^2.

    M is given as 1.9*10^30 kg. a (technically the semi-major axis of an elliptical orbit, but don't worry about that now) is given as 1.5*10^11 m. [tex]\pi[/tex] is approx. 3.1416, or you can use the calculator value.

    So where's the difficulty, at least in the first part?
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