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Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting .

  1. Nov 24, 2012 #1
    Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting.....

    1. The problem statement, all variables and given/known data

    Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting. This moon makes one complete orbit every 27 days. What is the orbital period of a satellite orbiting the planet at a distance of 4.10 x 10^6 km?

    a)96.3 d
    b)181.8 d
    c)63 d
    d)4.01 d
    I was given a formula T^2/r^3 = k constant

    So i set up the equation (27)^2/(1.15 x 10^6)^3= T^2/ (4.10 x 10^6)^3 and got T^2= 3.30 x 10^22 and take square root of that which gives me T=1.87151521 x 10^11. However my answer is not a choice above. Did i do something wrong with my calculation?
     
  2. jcsd
  3. Nov 24, 2012 #2

    lewando

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    Gold Member

    Re: Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting.....

    Yes. T2 is not equal to 3.3x1022.
     
  4. Nov 24, 2012 #3
    Re: Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting.....

    :O Thx I'll go back and see what I did wrong
     
  5. Nov 24, 2012 #4
    Re: Consider a moon that is 1.15 x 10^6 km from the planet it is orbiting.....

    Oh I made a careless computation error I input all the values at once in my calculator and got 181.75 ≈181.8.
    (27^2)(4.10x10^6)^3/ (1.15 x 10^6)^3 = 33035.85699

    radical of 33035.85699 = 181.75 = 181.8 Cheers! =)
     
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