Rotation of Earth effects the weight of star.

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SUMMARY

The discussion centers on the calculation of the rotation period of a star with mass M and radius r, where objects at the equator experience zero apparent weight. The derived formula for the period of rotation is T = 2π√(r³/G), but the correct answer is T = 2π√(r³/gM), highlighting the importance of including the star's mass in the equation. The confusion arose from unit discrepancies and the misunderstanding of gravitational acceleration (g) versus gravitational constant (G). The participants clarified that the mass of the star cannot be disregarded in this context.

PREREQUISITES
  • Understanding of angular velocity and its relation to rotational motion
  • Familiarity with gravitational equations, specifically g = GM/R²
  • Knowledge of dimensional analysis to verify unit consistency
  • Basic concepts of apparent weight in rotating systems
NEXT STEPS
  • Study the derivation of gravitational acceleration (g) and its dependence on mass and radius
  • Learn about the implications of angular velocity in rotational dynamics
  • Explore dimensional analysis techniques to ensure unit consistency in physics equations
  • Investigate the effects of rotation on apparent weight in different celestial bodies
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Students and educators in physics, particularly those focusing on mechanics and gravitation, as well as anyone interested in the dynamics of rotating celestial bodies.

Victorian91
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Homework Statement


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?


Homework Equations


g' = g- Rw2 (where w is the angular velocity)


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

Hence,
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??

Thanks
 
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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, 2\pi\sqrt{r^3/G}, doesn't have the right units. G has units length3/mass/time2. Thus r^3/G has units of mass*time2. Your answer has units of time*sqrt(mass), which is meaningless.
 
Okay thanks ...

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

Thanks
 

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