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Orbital period equation

  1. Jun 2, 2010 #1
    Ok just a quick question
    trying to work out the orbital period of an asteroid around the sun which has an orbital period 4times that of the earth which is 1.5*10^11m
    i worked out the radius of the asteroid to be 6*10^11 but cant find the equation for the orbital period.
    Does anyone know it?

    thanks
     
  2. jcsd
  3. Jun 2, 2010 #2
    Use 3rd Kepler's Law:

    [tex]
    \frac{R^{3}}{T^{2}} = \textup{const}
    [/tex]

    for all bodies orbiting the Sun. You do not even need the value of the radius of the Earth's orbit, but just their ratio, which is already given.
     
  4. Jun 2, 2010 #3
    haha thanks, your helping me a lot today!!

    so im guessing as its 4 times that of earth its 4AU?

    what is the constant?
     
  5. Jun 2, 2010 #4
    It's irrelevant. The ratio on the lhs has the same value for the Earth and for the asteroid. Equate the left hand sides for both the bodies and you will have an equation which you will need to solve for the orbital period of the asteroid.
     
  6. Jun 2, 2010 #5
    ok im getting..........T=[tex]\sqrt{4^3}[/tex] = 8.....which is one of the answers on my multiple choice :)
     
  7. Jun 2, 2010 #6
    this would be true if you specified the units in which the orbital period is measured.
     
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