A star the size of our Sun runs out of nuclear fuel and with

A star the size of our Sun runs out of nuclear fuel and, without losing mass, collapses to a white dwarf star the size of our Earth. The radius of our Sun is 6.96×108m , the radius of Earth is6.37×106m .

If the star initially rotates at the same rate as our Sun, which is once every 25 days, determine the rotation rate of the white dwarf.

Solution

from the law of conservation of angular momentum, we have

   I(sun)wi =  I(earth)] wf

(2/5m1r1^2)wi =  (2/5m2r2^2)wf

we have, m1=m2= m, thus we get

            (2/5mr1^2)wi = (2/5mr2^2)wf

                       r1^2wi = r2^2wf

hence, the the rotation rate of the white dwarf is,

            wf = wi*(r1^2/r2^2)

                 = (1rev/25days)*[(6.96×108)^2/(6.37×106)^2]

                  = 477.5 rev/day