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Change in planet's day length

  1. Oct 28, 2006 #1
    This is regarding that planet question. I set up the equation as:
    [tex]L_{pi} = L_a + L_{pf} = L_a + I_p \omega_{pf} \Longleftrightarrow \omega_{pf} = \frac{L_{pi} - L_a}{I_p}[/tex]
    where [tex]I_p = \frac{2 M R^2}{5}[/tex]
    so [tex]\omega_{pf} = \frac{L_{pi} - L_a}{\frac{2 M R^2}{5}}[/tex]

    In the original problem, we're given T = 13 hours, so
    [tex]T_i = \frac{13 "hours"}{"rev"} \cdot \frac{60 "min"}{1 "hour"} \cdot \frac{60 "sec"}{1 "min"} = \frac{46800 "sec"}{"rev"}[/tex]

    Using the final [tex]\omega[/tex] I came up with [tex]T_f = \frac{46801.3 "sec"}{"rev"}[/tex] - isn't that a longer day? Yet the question states "But, thanks to the asteroid's angular momentum, the planet rotates faster after the imapact than it did before."
     
  2. jcsd
  3. Oct 28, 2006 #2
    am I just being stupid? I've done this a few times now and I keep coming up with the same answer. If it takes longer to make a revolution, the day would be longer.
     
  4. Oct 28, 2006 #3
    I guess I set up the equation wrong. Using this:
    [tex]L_{pi} + L_a = L_{pf} = I_p \omega_{pf} \Longleftrightarrow \omega_{pf} = \frac{L_{pi} + L_a}{I_p}[/tex]
    where [tex]I_p = \frac{2 M R^2}{5}[/tex]
    so [tex]\omega_{pf} = \frac{L_{pi} + L_a}{\frac{2 M R^2}{5}}[/tex]

    [tex]T_i = \frac{13 "hours"}{"rev"} \cdot \frac{60 "min"}{1 "hour"} \cdot \frac{60 "sec"}{1 "min"} = \frac{46800 "sec"}{"rev"}[/tex]

    Using this new method (and adding the mass of the asteroid to the mass of the planet), I got [tex]T_f = \frac{46798.7 "sec"}{"rev"}[/tex] which means a day is shorter.

    Thanks for letting me figure it out!!! (I finally got the right answer!) Ya'll are great.
     
  5. Oct 28, 2006 #4

    DaveC426913

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    Gold Member

    You're quite welcome.
     
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