# Star diameter - momentum

1. Jan 15, 2010

### nnokwoodeye

1. The problem statement, all variables and given/known data

A star of mass 1.81×1031 kg and diameter 8.10E+9 m rotates with a period of 25.0 days. Suddenly the star changes size, and rotates with a new period of 18.0 days. Assuming a uniform density both before and after the size change, what is the new diameter of the star?

2. Relevant equations

Volume = 4/3*3.14*R^3
Density = M / VOL
I =2/5*M*R^2

3. The attempt at a solution

R(initial) = 8.10*10^9/2=4.05*0^9
Volume(initial) = 4/3*3.14*(4.05*10^9)^3=2.782*10^29
Density= (1.81*10^31/(2.782*10^29)=65.061
I(initial)=2/5*(1.81*10^31)*(4.05*10^9)^2=1.187*10^50
I(final)=2/5*65.061*4/3*3.14*Rf^3*Rf^2=109.01Rf^5
I(initial)*W(initial)=I(final)*W(final)
1.187*10^50*(2*3.14/25)=109.01*(2*3.14/18)*R(final)^5
R(final)^5=1.187*10^50*(2*3.14/25)/(109.01*[2*3.14/18])
R=(2.983*10^49/38.051)*10^(-5)=3.792*10^9
K=2R=7.58*10^9

the computer said that this answer is wrong and i don't know why

2. Jan 15, 2010

### tiny-tim

Hi nnokwoodeye!

(try using the X2 tag just above the Reply box )

It's very difficult to read what you've done, but I think you're misreading the question …

the mass is constant, not the density …

the question isn't saying that density is the same before as after, only that the mass is always evenly distributed.

3. Jan 15, 2010

### Gear300

I did not do the calculations but I noticed the units you used for the period were days (as given in the problem); converting to seconds might also make a difference.

4. Jan 15, 2010

### nnokwoodeye

o.k, so how do i calculate the new density? i need it to find the radius and solve the question.

to Gear300: you are right i forgot to convert the days to seconds, but it dosn't matter because i am using days in both side of the equation so the unit conversion would have been canceled out.

5. Jan 15, 2010

### nnokwoodeye

o.k. I succeeded in solving the question
Thanks for the help

Last edited: Jan 15, 2010