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 is 6.37×106m

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.

momentum is conserved

initial momentum=finalmomentum
2/5 *M*6.96E8wi=2/3*M*6.37E6*wf

wi=1rev/25days

wf=wi*(6.96E8/6.37E6)
= about 4 revs/day or 27.46rad/day

To determine the rotation rate of the white dwarf star, we need to consider the conservation of angular momentum. The angular momentum of an object is given by the equation:

L = I * ω

Where:
L = Angular momentum
I = Moment of inertia
ω = Angular velocity

The moment of inertia of a rotating object depends on its mass distribution. For simplicity, let's assume that the mass distribution of the star remains the same during the collapse.

Given that the star collapses to a white dwarf with the same mass but a drastically reduced radius, the moment of inertia will decrease. According to the parallel-axis theorem, the moment of inertia of a solid sphere is given by:

I_initial = (2/5) * M * R_initial^2

Where:
M = Mass of the object
R_initial = Initial radius of the object

I_final = (2/5) * M * R_final^2

Where:
R_final = Final radius of the object (radius of Earth)

Since the mass of the star remains the same, we can equate the initial and final moment of inertia:

(2/5) * M * R_initial^2 = (2/5) * M * R_final^2

Simplifying, we find:

R_initial^2 = R_final^2

Taking the square root of both sides:

R_initial = R_final

The radius of the star before and after the collapse are equal. Therefore, the moment of inertia remains the same.

Now, let's consider the rotation rate of the white dwarf star (ω_final). Since the moment of inertia remains constant and the angular momentum is conserved, we can write:

L_initial = L_final

I_initial * ω_initial = I_final * ω_final

Since I_initial = I_final, we have:

ω_final = (I_initial * ω_initial) / I_final

Since I_initial = I_final, the equation becomes:

ω_final = ω_initial

Therefore, the rotation rate of the white dwarf star will be the same as that of our Sun before the collapse, which is once every 25 days.

To determine the rotation rate of the white dwarf, you need to understand the concept of conservation of angular momentum. Angular momentum is the product of an object's moment of inertia and its rotational velocity.

In this scenario, we know that the initial star (with the same mass as our Sun) collapses to a white dwarf without losing mass. Therefore, we can assume that its moment of inertia remains constant since it depends on mass distribution and shape, which in this case doesn't change.

We are given the rotation period of the initial star, which is once every 25 days. To find the rotation rate of the white dwarf, we need to determine its rotation period.

Let's assume the white dwarf's rotation period is T_wd.

According to the conservation of angular momentum:

I_initial * ω_initial = I_wd * ω_wd

where I_initial and ω_initial represent the moment of inertia and rotational velocity of the initial star, and I_wd and ω_wd represent the moment of inertia and rotational velocity of the white dwarf, respectively.

Since we have assumed that the moment of inertia remains constant, we can equate the rotational velocities:

ω_initial = ω_wd

Therefore, the rotation period of the white dwarf, T_wd, will be the same as the initial star, which is 25 days.

So, the rotation rate of the white dwarf will also be once every 25 days.