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Planet is 2.00 a.u. when closest to sun. eccentricty is .500. find period

  1. Apr 28, 2008 #1
    1. The problem statement, all variables and given/known data

    a planet is 2.00 a.u. when it is closest to the sun. its eccentricity is .5000. Find its period in years.

    2. Relevant equations



    3. The attempt at a solution

    i got this so far... don't know if it is right

    PD + AD = A
    -AD - PD = F <SUBTRACTING THESE TWO>

    2PD = A - F
    2PD = A - (EA)
    2PD = A(1-E)
    2PD/(1-E) = A

    A = 2(2)/(1.-.5)
    A= 8 a.u.

    then for a

    a= 8.00/2
    a= 4.00 a.u.

    T^2 = a^3

    T = sqrt(a^3)
    T = 8

    im lost T = 8 what? years?
     
  2. jcsd
  3. Apr 28, 2008 #2
    kepler's 3rd law says Period( in years) is directly proportional to semi axis (a) cubed. correct?

    so was that last line:

    T^2 = a^3

    correct?
     
  4. Apr 28, 2008 #3
    anyone i got a 7 15 dead line
     
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