Calculating the Moment of Inertia of the Sun for Determining Rotational Energy

AI Thread Summary
To calculate the rotational energy of the Sun, the moment of inertia (I) is crucial, with the formula Erot=1/2*I*ω². Using the sphere's moment of inertia formula, I=2/5*m*r², yields an incorrect value for the Sun due to its non-uniform density. The Sun's core, which contains about half its mass, occupies only 1.5% of its volume, making a uniform density model inadequate. An empirical value for I/MR² for the Sun is 0.059, significantly lower than the 0.4 expected for a uniform sphere. Accurate calculations for the Sun's rotational energy require this empirical approach rather than a simplistic model.
AndersLau
Messages
4
Reaction score
0
I want to find the rotational energy of the sun.
Erot=1/2*I*ω2

m=1.9891*10^30 kg
r=6.955*10^5 km


When I am using the formula for a sphere's moment of inertia: I=2/5*m*r2
I'm getting 3.848671797*1047 km*m2

Can i find the moment of inertia in another way? the moment of inertia needs to be in *1046 , to get the rotational energy right.

Thank you.
 
Physics news on Phys.org
A uniform density sphere is a fairly bad model even for the earth, which is more or less solid. For a gas it is an incredibly lousy model. The sun's core contains about half of the total mass of the sun but only 1.5% of its total volume.

Unless you want to get into astrophysics, you are going to need an empirical value. From http://nssdc.gsfc.nasa.gov/planetary/factsheet/sunfact.html, I/MR2 for the sun is 0.059. Compare that for a uniform density sphere, where I/MR2=2/5 or 0.4.
 
Thank you very much for the answer, that sorts it out. Much appreciated.
I'm waiting with the astrophysics :)
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top