Inertia and Kinetic energy of the sun

  • Thread starter Torrenza
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  • #1
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:

Answers and Replies

  • #2
willem2
2,103
363
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" [Broken]

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.
 
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  • #3
RoyalCat
671
2
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" [Broken]

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.
 
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  • #4
Torrenza
3
0
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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