Question

A star with an initial radius of 1.0 x 10^8 m and period of 30.0 days collapses suddenly to a radius of 1.0 x 10^4 m. (a) Find the period of rotation after collapse. (b) Find the work done by gravity during the collapse if the mass of the star is 2.0 x 10^30 kg. (c) What is the speed of the person standing on the equator of the collapsed star? (Neglect any relativistic or thermal effects, and assume the star is spherical before and after it collapses.

Answer 1

R = 1•10^8 m, r =1•10^4 m, m = 2•10^30 kg. T1 = 30 days

Since there are no mass loss and no external torques, the angular moment of the star is conserved.
I1•ω1 = I2•ω2.

ω=2•π/T
I1•2•π/T1 = I2•2•π/T2.
For uniform sphere of constant mass I1= 2•m•R²/5, I2 = 2•m•r²/5,
2•m•R²•2•π /5 •T1 = 2•m•r²•2•π /5 • T2.
R² /T1 = r² /T2.
T2 =(r/R)² •T1 = (1•10^4/10^8)² •30 = 30•10^-8 days =0.026 s.
(b) W = KE1 –KE2 = I1• ω1²/2 – I2• ω2²/2 = I1•4•π²/2•T1² - I2•4•π²/2•T2² =
= (2•m•R²/5) •(4•π²/2•T1²) - (2•m•r²/5) •(4•π²/2•T2²) =
= (4•m•π²/5) •{(R/T1)² -( r/T2)²} = .............
(c) v =ω2•r = 2•π•r/T2 = .........

Answer 2

6.3