What would be the approximate rotation of the sun if....

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SUMMARY

The discussion centers on calculating the rotation rate of the Sun if it were to collapse into a neutron star with a radius of 10 km. The initial average rotation rate of the Sun is 1 rev/25 days, which translates to approximately 2.99 x 10-11 rev/s. Using the conservation of angular momentum formula I1ωo = I2ωf, the final rotation rate (ωf) was calculated to be 267.7 rev/s after correcting for radius and mass distribution. The discussion highlights the importance of accurate unit conversions and algebraic manipulation in astrophysical calculations.

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  • Understanding of angular momentum conservation principles
  • Familiarity with the moment of inertia for non-solid bodies
  • Basic knowledge of neutron star characteristics
  • Proficiency in unit conversions and dimensional analysis
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Astronomy students, astrophysicists, and anyone interested in stellar dynamics and the physical properties of neutron stars will benefit from this discussion.

Vitani11
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Homework Statement


The sun is a pretty typical star with a mass of 1.99x1030kg and a radius of 6.69x108 m. Since it isn't solid, it doesn't rotate uniformly, but has an average rotation rate of 1rev/25d. A star with a mass about about three times that of the Sun eventually explodes as a supernova and leaves behind its collapsed remnants - a neutron star or a black hole. Neutron stars masses are somewhat lower than the masses of the original stars, but have much smaller radii, less than 10 km. Newly-collapsed neutron stars emit beams of radio waves; since they rotate so quickly, the beams may intersect Earth at regular intervals. To produce the observed pulsar rates, the rotation rates must vary from 1rev/s to 1000rev/s.

What would be the approximate rotation rate of the Sun if it became a neutron star with a radius of 10km? Assume it is spherical, with a uniform mass distribution, and that its moment of inertia if 0.059MR2 (Since it is not a solid you can not use I for a solid sphere).

Homework Equations


I1ωo=I2ωf conservation of angular momentum
ωavg = 1rev/25d
ω0 = 1rev/(25x2x6.69x108m) = 2.99x10-11 rev/s
I1=0.059(1.99x1030kg)(6.69x108m)2
I2 = 0.059(1.99x1030)(10m)2
ωf = ?

The Attempt at a Solution


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I used I1ωo=I2ωf and solved for ωf. After I solved for this I got 133,800rev. I then divided this final answer by 25d, or 25x2x10km since the suns radius is now that of a neutron star instead of it's original 6.69x108 m. This gave me a value of 267.7 rev/s, which according to the problem statement makes sense. I'm not exactly confident in my method though so I would like to know if this is correct.
 
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I used I1ωo=I2ωf and solved for ωf. After I solved for this I got 133,800rev. I then divided this final answer by 25d, or 25x2x10km since the suns radius is now that of a neutron star instead of it's original 6.69x108 m. This gave me a value of 267.7 rev/s, which according to the problem statement makes sense. I'm not exactly confident in my method though so I would like to know if this is correct.
 
Vitani11 said:
1rev/(25x2x6.69x108m)
Why are you dividing the rate by the radius?
Vitani11 said:
I2 = 0.059(1.99x1030)(10m)2
Having collapsed to a neutron star, it is no longer a ball of gas.
Also, the new radius is 10km, not 10m.
 
Okay am I supposed to assume that a neutron star is to be regarded as a point mass? sphere? Also what is d? I saw it as a rotation rate as a function of radius which is why I did that
 
Vitani11 said:
Okay am I supposed to assume that a neutron star is to be regarded as a point mass? sphere? Also what is d? I saw it as a rotation rate as a function of radius which is why I did that

A 10 km radius point would be a good trick :smile: Treat it as a solid sphere. The "d" in 25d is meant to be the units, "days". You could have googled "sun rotation" to pick up on that.
 
Okay. Now I'm getting a number that disturbs me. 30,565 revolutions per second?!
 
Vitani11 said:
Okay. Now I'm getting a number that disturbs me. 30,565 revolutions per second?!
Looks a bit high (a couple of orders of magnitude). Check your calculations and see if you've picked up a factor of 100 somewhere (say via unit conversions or mucking some power of ten exponent work).

Failing that, present some details of your calculation steps.
 
Beautiful thank you. I got the correct answer which I'm confident in. It's reassuring (but also not) that the issue lies in the algebra and interpretation rather than the physics, lol
 

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