The typical lifetime of a star like our Sun is 10 billion years. While a star like our Sun is able to generate energy in its core (through fusion) the thermal pressure balances the gravitational pressure from the Sun’s mass. However, once the star rules our of hydrogen, gravitational collapse can occur, and depending on the star’s mass it may become a white dwarf. A white dwarf is a dense stellar remnant with a typical radius of about 104 km.
a) If a star like our Sun (M = 2x1030 kg, R = 750,000 km, rotational period ~25 days) collapsed and became a white dwarf, what would its new rotational period be? Give your answer in minutes or seconds, not days. (Hint: there are no external torques on the system.)
b) Does the kinetic energy of rotation increase, decrease, or stay the same? (Show your work for full credit.)
a) To find the new rotational period of the white dwarf, we can use the conservation of angular momentum.
The angular momentum of a star 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.
Assuming that the star's mass is conserved during the collapse, the moment of inertia will also remain constant.
The initial angular momentum of the star is given by:
L_initial = I_initial * ω_initial
The final angular momentum of the white dwarf is given by:
L_final = I_final * ω_final
Since there are no external torques on the system, the angular momentum is conserved:
L_initial = L_final
Therefore, we have:
I_initial * ω_initial = I_final * ω_final
The moment of inertia of a sphere is given by the equation:
I = (2/5) * m * r^2
Where m is the mass and r is the radius.
For the initial star:
I_initial = (2/5) * M_initial * R_initial^2
And for the final white dwarf:
I_final = (2/5) * M_final * R_final^2
Now, let's substitute the values:
I_initial * ω_initial = I_final * ω_final
(2/5) * M_initial * R_initial^2 * ω_initial = (2/5) * M_final * R_final^2 * ω_final
Since the mass is conserved:
M_initial = M_final
And since the radius changes:
R_initial = 750,000 km = 7.5 x 10^8 m
R_final = 104 km = 1.04 x 10^5 m
We can simplify the equation to:
R_initial^2 * ω_initial = R_final^2 * ω_final
Now, let's solve for ω_final:
ω_final = (R_initial^2 * ω_initial) / R_final^2
ω_final = (7.5 x 10^8 m)^2 * ω_initial / (1.04 x 10^5 m)^2
ω_final = (56.25 x 10^16 * ω_initial) / 1.0816 x 10^10
ω_final = 5.20 x 10^6 * ω_initial
Since the initial rotational period is approximately 25 days, or 2.16 x 10^6 seconds:
Initial rotational period = 2.16 x 10^6 s
Therefore, the new rotational period of the white dwarf in seconds is:
New rotational period = 5.20 x 10^6 * 2.16 x 10^6 s
New rotational period ≈ 1.12 x 10^13 seconds
To convert this to minutes, we divide by 60:
New rotational period in minutes ≈ 1.87 x 10^11 minutes
Therefore, the new rotational period of the white dwarf would be approximately 1.87 x 10^11 minutes.
b) The kinetic energy of rotation is given by the equation:
KE = (1/2) * I * ω^2
Since the moment of inertia (I) is conserved during the collapse, and the new angular velocity (ω_final) is calculated based on the initial angular velocity (ω_initial), we can compare the kinetic energy of rotation.
Let's compare the kinetic energy of rotation before and after the collapse:
KE_initial = (1/2) * I_initial * ω_initial^2
KE_final = (1/2) * I_final * ω_final^2
Since I_initial = I_final:
KE_initial = (1/2) * I_initial * ω_initial^2
KE_final = (1/2) * I_initial * ω_final^2
Dividing the equations:
KE_final / KE_initial = (1/2) * I_initial * ω_final^2 / [(1/2) * I_initial * ω_initial^2 ]
= (ω_final / ω_initial)^2
Now let's substitute the values of ω_final and ω_initial we obtained earlier:
KE_final / KE_initial = [(5.20 x 10^6 * ω_initial) / ω_initial]^2
= (5.20 x 10^6)^2
Therefore, KE_final =KE_initial * (5.20 x 10^6)^2
The kinetic energy of rotation will increase by a factor of (5.20 x 10^6)^2 compared to the initial star.
Hence, the kinetic energy of rotation increases.
To find the new rotational period of the white dwarf, we can use the principle of conservation of angular momentum. According to this principle, the angular momentum of a system remains constant unless acted upon by an external torque. In this case, there are no external torques on the system after the star collapses into a white dwarf, so its angular momentum will remain the same.
The angular momentum (L) of a rotating object can be calculated as the product of its moment of inertia (I) and angular velocity (ω):
L = I * ω
In this case, we want to find the new rotational period, which is the time it takes for one complete rotation. The rotational period (T) is the reciprocal of the angular velocity (ω):
ω = 2π / T
Substituting this into the angular momentum equation, we get:
L = I * (2π / T)
Since angular momentum is conserved, we can equate the initial and final angular momenta:
L_initial = L_final
I_initial * ω_initial = I_final * ω_final
Assuming the moment of inertia remains the same for the white dwarf (which is a reasonable assumption in this context), we can simplify the equation to:
ω_initial = ω_final
2π / T_initial = 2π / T_final
Dividing both sides by 2π, we get:
1 / T_initial = 1 / T_final
Rearranging the equation, we find:
T_final = T_initial
Therefore, the new rotational period of the white dwarf will be the same as the initial rotational period of the star, which is approximately 25 days.
Now, let's move on to part (b) of the question, which asks about the change in kinetic energy of rotation. The kinetic energy of rotation (KE_rot) can be calculated as the product of the moment of inertia (I) and the square of the angular velocity (ω):
KE_rot = 1/2 * I * ω^2
Since the moment of inertia remains the same for the white dwarf, we only need to consider the change in angular velocity.
As we found earlier, the angular velocity remains the same before and after the collapse:
ω_initial = ω_final
Therefore, the square of the angular velocity will also be the same:
ω_initial^2 = ω_final^2
Substituting this into the kinetic energy equation, we find:
KE_rot_initial = KE_rot_final
Therefore, the kinetic energy of rotation stays the same before and after the star collapses into a white dwarf.