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Rotation of Earth effects the weight of star.

  1. Feb 3, 2009 #1
    A star of mass M and radius r rotates with angular speed of such value that objects on the surface at its equator are just about to experience zero apparent weight. The period of rotation of the star is given by?

    2. Relevant equations
    g' = g- Rw2 (where w is the angular velocity)

    3. The attempt at a solution
    From the above..
    mg' =mg -mRw2
    Since apparent weight is zero,

    mg -mRw2 = 0
    mRw2 = mg
    But w = (2Pi/T)^2
    R ( 4Pi^2/T^2 )
    T^2 = 4Pi^2 R/g
    But g = GM/R2 at the surface of the planet

    Period, T= 2Pi square root r3/G

    But the answer given is
    2Pi square root r3/gM

    Why does the Mass of the star still have in the equation
    I tried so many times the Mass will be canceled out..
    Do you think which one is correct??

  2. jcsd
  3. Feb 4, 2009 #2

    D H

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    Staff Emeritus
    Science Advisor

    The mass of the star, big M, appears in only one of your equations. How can it possibly cancel?

    Another way to look at it: Your result, [itex]2\pi\sqrt{r^3/G}[/itex], doesn't have the right units. G has units length3/mass/time2. Thus [itex]r^3/G[/itex] has units of mass*time2. Your answer has units of time*sqrt(mass), which is meaningless.
  4. Feb 4, 2009 #3
    Okay thanks ...

    Sorry.. I was actually not focusing..
    Thanks.. I got it already.
    Next time I'll check on units..


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