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.

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 = .........

6.3

(a) To find the period of rotation after the collapse, we can use the principle of conservation of angular momentum. Angular momentum is given by the product of moment of inertia and angular velocity.

The moment of inertia of a sphere is given by:

I = (2/5) * m * r^2

Where m is the mass of the star and r is the radius of the star.

Before the collapse:
Initial moment of inertia (I_initial) = (2/5) * m * (1.0 x 10^8)^2

After the collapse:
Final moment of inertia (I_final) = (2/5) * m * (1.0 x 10^4)^2

Since angular momentum is conserved, I_initial * ω_initial = I_final * ω_final

Where ω is the angular velocity.

The period of rotation is the inverse of the angular velocity:

T = 2π/ω

We can rearrange the conservation equation to solve for ω_final:

I_initial * ω_initial = I_final * ω_final
ω_final = (I_initial * ω_initial) / I_final

Substituting the expressions for moment of inertia:

ω_final = [(2/5) * m * (1.0 x 10^8)^2 * ω_initial] / [(2/5) * m * (1.0 x 10^4)^2]

Simplifying:

ω_final = (1.0 x 10^16) * ω_initial

Now we can calculate the period of rotation after collapse:

T_final = 2π / ω_final
T_final = 2π / [(1.0 x 10^16) * ω_initial]
T_final = (2π) / (1.0 x 10^16) * T_initial

Given that the initial period of rotation (T_initial) is 30.0 days:

T_final = (2π) / (1.0 x 10^16) * 30.0 days

(b) To find the work done by gravity during the collapse, we can use the work-energy theorem. The work done by gravity is equal to the change in gravitational potential energy.

The change in gravitational potential energy is given by:

ΔU = U_final - U_initial

Before the collapse:
Initial gravitational potential energy (U_initial) = -G * (m^2) / r_initial

After the collapse:
Final gravitational potential energy (U_final) = -G * (m^2) / r_final

Where G is the gravitational constant.

The work done by gravity is:

W = ΔU = U_final - U_initial
W = (-G * (m^2) / r_final) - (-G * (m^2) / r_initial)

Substituting the given values:
W = (-6.674 x 10^-11 N*m^2/kg^2) * (2.0 x 10^30 kg)^2 * [(1/r_final) - (1/r_initial)]

(c) To find the speed of the person standing on the equator of the collapsed star, we can use the formula for linear velocity in terms of angular velocity and radius:

v = ω * r

Where v is the linear velocity and r is the radius.

The angular velocity after the collapse is ω_final = (1.0 x 10^16) * ω_initial, as calculated in part (a).

Substituting the given radius of the collapsed star (1.0 x 10^4 m), we can calculate the speed:

v = (1.0 x 10^16) * ω_initial * (1.0 x 10^4 m)

To solve this problem, we will use the principles of conservation of angular momentum, gravitational potential energy, and rotational kinetic energy.

(a) Find the period of rotation after collapse:
The angular momentum of a rotating object is given by the equation L = Iω, where L is the angular momentum, I is the moment of inertia, and ω is the angular velocity.

Before the collapse, the star's radius is R1 = 1.0 x 10^8 m, and its period of rotation is T1 = 30.0 days. After the collapse, the star's radius is R2 = 1.0 x 10^4 m. The moment of inertia of a sphere is given by the equation I = (2/5)MR^2, where M is the mass of the star.

Using the conservation of angular momentum, we can equate the initial angular momentum (L1) to the final angular momentum (L2):

L1 = L2

I1ω1 = I2ω2

(2/5)M(R1^2)ω1 = (2/5)M(R2^2)ω2

Dividing both sides by (2/5)MR2^2, we get:

ω2 = (R1^2ω1) / (R2^2)

Next, we can calculate the period of rotation after the collapse (T2) using the equation T = 2π / ω:

T2 = (2π) / ω2

Substitute in the value of ω2 calculated earlier to find T2.

(b) Find the work done by gravity during the collapse:
The work done by gravity can be found by calculating the change in gravitational potential energy (ΔUg) during the collapse. The change in gravitational potential energy is given by the equation ΔUg = -GMm / R, where G is the gravitational constant, M is the mass of the star, m is the mass of an infinitesimally small shell of the star (considered to be one planetary mass), and R is the radius of the shell.

Integrate ΔUg from R1 to R2 to find the total work done by gravity during the collapse.

(c) Find the speed of the person on the equator of the collapsed star:
The speed of a person on the equator of a rotating object can be found using the equation v = ωr, where v is the speed, ω is the angular velocity, and r is the distance from the center of rotation.

Substitute the value of ω2 (calculated earlier) and the new radius (R2) into the equation to find the speed of the person on the equator of the collapsed star.

By following these steps, you should be able to find the answers to parts (a), (b), and (c) of the given problem.