A star with the same mass and diameter as the sun rotates about a central axis with a period of about 29.2 days. Suppose that the sun runs out of nuclear fuel and collapses to form a white dwarf star with a diameter equal to that of the earth. Assume the star acts like a solid sphere and that there is no loss of mass in the process. You will need some data from the inside front cover of you text. (a) What would be the new rotation period (s) of the star? (b) What is the ratio of final to inital kinetic energies

To solve this problem, we need to use the law of conservation of angular momentum, which states that the angular momentum of an object remains constant unless acted upon by an external torque.

(a) The angular momentum of the star before and after the collapse remains the same. We can write this equation as:

L_initial = L_final

The initial angular momentum (L_initial) of the star is given by:

L_initial = I_initial * ω_initial

where I_initial is the moment of inertia of the star before collapse and ω_initial is the initial angular velocity.

The final angular momentum (L_final) of the star is given by:

L_final = I_final * ω_final

where I_final is the moment of inertia of the star after collapse and ω_final is the final angular velocity.

Since the star acts like a solid sphere, the moment of inertia is given by:

I = (2/5) * M * R^2

where M is the mass of the star and R is its radius.

Since the star has the same mass as the sun, we can write:

M = M_sun

And since the star has a diameter equal to that of the Earth, we can write:

R = R_earth/2

Substituting these values into the equation for moment of inertia, we get:

I_initial = (2/5) * M_sun * (R_earth/2)^2

To find the final moment of inertia (I_final), we can use the fact that the mass remains the same:

M_final = M_initial = M_sun

And since the final diameter is equal to the Earth's diameter:

R_final = R_earth/2

Substituting these values into the equation for moment of inertia, we get:

I_final = (2/5) * M_sun * (R_earth/2)^2

Now, we can use the equations for angular momentum to find the relationship between initial and final angular velocities (ω_initial and ω_final):

L_initial = L_final

I_initial * ω_initial = I_final * ω_final

(2/5) * M_sun * (R_earth/2)^2 * ω_initial = (2/5) * M_sun * (R_earth/2)^2 * ω_final

ω_initial = ω_final

Therefore, the rotation period after the collapse is the same as the initial rotation period, which is about 29.2 days.

(b) The ratio of final to initial kinetic energies (KE_final / KE_initial) can be found using the relationship between angular velocity (ω) and kinetic energy (KE):

KE = (1/2) * I * ω^2

Substituting the equations for moments of inertia and angular velocities, we get:

KE_initial = (1/2) * (2/5) * M_sun * (R_earth/2)^2 * ω_initial^2

KE_final = (1/2) * (2/5) * M_sun * (R_earth/2)^2 * ω_final^2

Since ω_initial = ω_final, we can simplify the ratio of kinetic energies:

(KE_final / KE_initial) = [(1/2) * (2/5) * M_sun * (R_earth/2)^2 * ω_final^2] / [(1/2) * (2/5) * M_sun * (R_earth/2)^2 * ω_initial^2]

=(ω_final^2 / ω_initial^2)

= 1

Therefore, the ratio of final to initial kinetic energies is 1, indicating that the kinetic energy remains unchanged during the collapse.

To answer these questions, we need to use two important physical principles: the conservation of angular momentum and the conservation of energy.

(a) To determine the new rotation period of the star, we can use the principle of conservation of angular momentum. Angular momentum (L) is defined as the product of the moment of inertia (I) and the angular velocity (ω). Mathematically, it can be expressed as L = I * ω.

Initially, when the star had the same mass and diameter as the Sun, its moment of inertia (I_initial) can be approximated as a sphere with mass (M_initial) and radius (R_initial). The angular velocity (ω_initial) is given by the period (T_initial) as ω_initial = 2π / T_initial.

Final moment of inertia (I_final) can be approximated as a sphere with mass (M_final) and radius (R_final), which is equal to the Earth's diameter.

To find the new rotation period, we equate the initial angular momentum to the final angular momentum, as angular momentum is conserved:

I_initial * ω_initial = I_final * ω_final

Plugging in the expressions for the moment of inertia and angular velocity, we get:

(M_initial * R_initial^2) * (2π / T_initial) = (M_final * R_final^2) * (2π / T_final)

M_initial and M_final are equal since there is no loss of mass. R_initial is the same as the Sun's radius, and R_final is equal to the Earth's radius, which we can find from a reliable source or use the data from the inside front cover of your text.

We can rearrange the equation to solve for T_final, the new rotation period of the star:

T_final = (T_initial * R_initial^2 * R_final^2) / (R_final^2 * R_initial^2)

(b) To find the ratio of final to initial kinetic energies, we can use the principle of conservation of energy. The kinetic energy of a rotating object can be given as (1/2) * I * ω^2.

The initial kinetic energy (KE_initial) can be calculated as (1/2) * I_initial * ω_initial^2, and the final kinetic energy (KE_final) can be calculated as (1/2) * I_final * ω_final^2.

To find the ratio of final to initial kinetic energies, we divide KE_final by KE_initial:

Ratio = KE_final / KE_initial = [(1/2) * I_final * ω_final^2] / [(1/2) * I_initial * ω_initial^2]

Substituting the expressions for moment of inertia and angular velocity, we get:

Ratio = (I_final * ω_final^2) / (I_initial * ω_initial^2)

Now, we just need to substitute the values of I_initial, I_final, ω_initial, and ω_final that we calculated above into the equation to find the final rotation period and the ratio of final to initial kinetic energies.

Look up the contraction ratio Rsun/Rearth. It is about 100

For an exact answe, you will need the radii or diameters of the sun and the earth, to calculate the contraction ratio Rinitial/Rfinal. It is approximatley 100, but you need a more precise vlaue

Remember that angular momentum, I*w, is conserved. That means that M*R^2/P stays the same, where P is the rotation period.

I*w = Constant*M*R^2/P
You can forget about the Mass m and the constant. They stay the same.

Since R decreases by about 100, P must decrease by a factor of about 10^4. That will make the new period about 29.2 days/10^4 = 4 minutes

For the ratio of kinetic energies, remember that for a rotating sphere, KE = (1/2)I*w^2

The KE ratio will be about
(Iw)final*wfinal/[(Iw)initial*winitial]
wfinal/winitial = 10^4

The increase in KE comes from work done by gravity during contraction.