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Homework Help: Rotation of a planet, kinetic energy

  1. Mar 5, 2010 #1
    1. The problem statement, all variables and given/known data

    Calculate the ratio between the kinetic energy of rotation of a planet (mass=4.30E+24 kg, radius=7.60E+6 m) to the kinetic energy of its center of mass orbiting around its sun at a distance of 1.20E+12 m. Like the Earth, it has a day lasting 24 hours and a year lasting 365.25 days.


    2. Relevant equations

    KErot= (1/2)Iw^2 and I believe KE=(1/2)mv^2

    3. The attempt at a solution

    For the KErot:
    KErot= (1/2)((2/5)(4.3E24kg)(7.6E6m)(2pi/(24hr*60min*60sec))^2
    and I got 2.62698301E29 J

    for the other one:
    KE=(1/2)(4.30E24kg)(2pi*1.2E12m/(365.25days*24hrs*60min*60sec))
    and got 1.22730571E35 J

    I devided KE by KErot and got 467192 for the ratio. but it's wrong and I know I'm doing something wrong but can't figure it out. any help will be great!
     
  2. jcsd
  3. Mar 5, 2010 #2

    tiny-tim

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    Welcome to PF!

    Hi Dotty21690! Welcome to PF! :smile:

    (have an omega: ω and a pi: π and try using the X2 tag just above the Reply box :wink:)
    Did you remember to square the radius in both cases? :wink:
     
  4. Mar 5, 2010 #3
    Why would you square the radius in the second part? because isn't the V=Rw?? so the radius wouldn't be squared?
     
  5. Mar 5, 2010 #4

    Char. Limit

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    It's still mv2, which is mR2w2.
     
  6. Mar 5, 2010 #5
    i'm still getting the wrong answer and I don't know why.
     
  7. Mar 5, 2010 #6
    am I using the wrong equation to find the kinetic energy of the center of mass for the orbit??
     
  8. Mar 5, 2010 #7

    tiny-tim

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    You should be using 1/2 mv2 which is the same as 1/2 m (2πr/T)2 :smile:
     
  9. Mar 5, 2010 #8
    so when I plug in my information it will look like this?:

    (1/2)(4.3E24kg)((2π*1.2E12m)/31557600sec)^2)

    I get 1.22730571E35. but when I do the ratio, I'm getting the answer wrong.
     
  10. Mar 6, 2010 #9

    tiny-tim

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    Hi Dotty21690!

    (please use the X2 tag just above the Reply box :wink:)

    Show us your full calculations (and cancel out the common factors before you start). :smile:
     
  11. Mar 6, 2010 #10
    well I showed the work earlier to get the answers for the KErot and KE, then I divided. I'll post it again though:

    For the KErot:
    KErot= (1/2)((2/5)(4.3E24kg)(7.6E6m)(2pi/(24hr*60min*60sec))^2
    and I got 2.62698301E29 J

    for the other one:
    KE=(1/2)(4.30E24kg)(2pi*1.2E12m/(365.25days*24hrs*60min*60sec))
    and got 1.22730571E35 J

    then I divided 1.22730571E35 J/2.62698301E29 J and got 467192.1. I don't even know if I am doing the problem right.
     
  12. Mar 6, 2010 #11
    turns out the answer was 2.140*10-6, which I had got at one time, but it said it was wrong :( but I don't know how to get that answer again. so can someone help me with this one?
     
  13. Mar 6, 2010 #12

    Char. Limit

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    I think you divided your ratio the wrong way. After all, the answer's power is suspiciously close to your answer's power.
     
  14. Mar 6, 2010 #13
    turns out I did divide the wrong way!! but I wonder why I was told the answer was wrong when I did put in 2.140*10-6, hmm. But thanks guys!!
     
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