Inertia and Kinetic energy of the sun

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Homework Help Overview

The original poster attempts to calculate the inertia and kinetic energy of the sun, providing specific values for its mass, radius, and spin period. The problem involves understanding the moment of inertia and kinetic energy formulas in the context of celestial bodies.

Discussion Character

  • Conceptual clarification, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the appropriate formulas for calculating the moment of inertia and kinetic energy, with some questioning the assumptions made about the sun's density distribution and the formulas used.

Discussion Status

Some participants have provided feedback on the original poster's calculations, noting discrepancies in the answers and suggesting that the moment of inertia formula may need adjustment based on the sun's structure. There is acknowledgment of errors in the original poster's notes, but no consensus on a corrected approach has been reached.

Contextual Notes

Participants mention that the sun's core density significantly affects the moment of inertia, and there is discussion about the need for correct units in the calculations. The original poster's confusion regarding the formulas is noted, but specific details about the formulas remain under examination.

Torrenza
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Homework Statement



Ive been trying to find the Inertia and Kinetic energy of the sun. I have a feeling I am using the wrong formulas to obtain the answer.

Mass of sun 2*10^30kg
Radius of sun 700*10^6 m
t spin = 26 days

the answers should be
I = 4.0 * 10^98
K energy = 1.5 * 10^16

Homework Equations



I=(2/5)*M*R^2
K= (1/2)M*(ω^2)*(r^2)
ω = 2pi/t

The Attempt at a Solution

If someone knows how to do this i would really appreciate it.:smile:
 
Last edited:
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The equation for the moment of inertia of a solid sphere is correct. The answer you gave is not correct. (by a factor of more than 10^50)

Another problem is that the core of the sun is so much more dense than the upper layers.
According to this http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html"

the moment of inertia is about 0.059*M*R^2 instead of 0.4*M*R^2


The equaton for kinetic energy is [tex]K = \frac {1}{2}I {\omega}^2[/tex]

where I is the moment of inertia. The answer given is wrong. All of the answers
require units.
 
Last edited by a moderator:
willem2 said:
The equation for the moment of inertia of a solid sphere is correct. The answer you gave is not correct. (by a factor of more than 10^50)

Another problem is that the core of the sun is so much more dense than the upper layers.
According to this http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html"

the moment of inertia is about 0.059*M*R^2 instead of 0.4*M*R^2


The equaton for kinetic energy is [tex]K = \frac {1}{2}I {\omega}^2[/tex]

where I is the moment of inertia. The answer given is wrong. All of the answers
require units.

The real problem here isn't the overly-simplified model of the sun, but the wrong formula given for spin kinetic energy.

Spin kinetic energy is the kinetic energy associated with an object's rotation about its own center of mass.

The formula for it is: [tex]K_{spin}=\frac{1}{2}I\omega^2[/tex] where [tex]I[/tex] is the relevant moment of inertia (Depending on what axis the object is spinning about).

The formula you cited is only true for a point object, and for thin rings, where, as a private case of the general formula, [tex]I=mr^2[/tex], which does not hold in general.
 
Last edited by a moderator:
Thanks for answering the question, I now see what i did wrong. The I = 4.0 * 10^98 was a note taking error, lol just a slight difference from the real answer.
 

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