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.
Initial moment of inertia = (2/5)mass*R^2=(2/5)mass*10^16 initial angular velocity wi = 2pi /(30*60*60*24)
initial angular momentum =[(2/5)mass*10^16]*(2pi /(30*60*60*24))
final moment of inertia = (2/5)mass*R^2=(2/5)mass*10^8
final angular momentum= [(2/5)mass*10^8]wF
final angular velocity wf=?
wf = 242.406 rad / s
WF = 2PI / (T)
T= 0.026
(a) To find the period of rotation after the collapse, we can use the conservation of angular momentum. The angular momentum of the star before and after the collapse remains constant.
Angular momentum, L = Iω
Where, I is the moment of inertia and ω is the angular velocity.
The moment of inertia for a spherical object is given by:
I = (2/5) * m * R^2
Where, m is the mass of the star and R is the radius.
Before the collapse:
I_initial = (2/5) * m * R_initial^2
After the collapse:
I_final = (2/5) * m * R_final^2
Since the angular momentum is conserved:
L_initial = L_final
Therefore,
I_initial * ω_initial = I_final * ω_final
m * R_initial^2 * ω_initial = m * R_final^2 * ω_final
Rearranging the equation:
(R_initial^2 / R_final^2) = (ω_final / ω_initial)
Now, let's substitute the given values and solve for the period of rotation after the collapse:
R_initial = 1.0 x 10^8 m
R_final = 1.0 x 10^4 m
ω_initial = (2π / T_initial), where T_initial is the initial period of rotation (30.0 days)
(R_initial^2 / R_final^2) = (ω_final / ω_initial)
(1.0 x 10^8 / 1.0 x 10^4)^2 = (ω_final / (2π / T_initial))
(1.0 x 10^4)^2 = (ω_final / (2π / 30.0))
1.0 x 10^8 = ω_final / (2π/30.0)
ω_final = (1.0 x 10^8) * (2π/30.0)
Now we can find the period of rotation after the collapse, T_final:
T_final = (2π) / ω_final
Substituting the value of ω_final:
T_final = (2π) / ((1.0 x 10^8) * (2π/30.0))
Simplifying the expression:
T_final = (30.0 / (1.0 x 10^8))
T_final = 3.0 x 10^(-7) seconds (or s)
Therefore, the period of rotation after the collapse is 3.0 x 10^(-7) seconds.
(b) To find the work done by gravity during the collapse, we can use the formula:
Work = -ΔPE
Where, ΔPE is the change in gravitational potential energy.
The change in gravitational potential energy is given by:
ΔPE = -GMm / (R_final - R_initial)
Where, G is the gravitational constant (6.67 x 10^(-11) Nm^2/kg^2), M is the mass of the star (2.0 x 10^30 kg), m is the mass of an infinitesimal shell, R_final is the final radius, and R_initial is the initial radius.
To find the work done, we need to integrate the equation over the entire star from R_initial to R_final.
Work = ∫[R_initial, R_final] -GMm / (R_final - R) * 4πR^2 dr
Substituting the values:
Work = -GMm * 4π (∫[R_initial, R_final] R / (R_final - R) * R^2 dr)
Simplifying the expression:
Work = -GMm * 4π (∫[R_initial, R_final] R^3 / (R_final - R) dr)
Integrating the expression:
Work = -GMm * 4π ((R_final^4 * ln(R_final - R_initial) - R_initial^4 * ln(R_final - R_initial) - 4 ∫[R_initial, R_final] R^3 ln(R_final - R) dr))
Since the star is spherical, the integral term becomes zero.
Work = -GMm * 4π (R_final^4 * ln(R_final - R_initial) - R_initial^4 * ln(R_final - R_initial))
Substituting the values:
Work = -(6.67 x 10^(-11) Nm^2/kg^2) * (2.0 x 10^30 kg) * 4π ((1.0 x 10^4 m)^4 * ln((1.0 x 10^4 m) - (1.0 x 10^8 m)) - (1.0 x 10^8 m)^4 * ln((1.0 x 10^4 m) - (1.0 x 10^8 m)))
Calculating the numerical value, we can find the work done by gravity during the collapse.
(c) To find the speed of a person standing on the equator of the collapsed star, we can use the formula for the speed of rotation:
Speed = 2πR / T
Where, R is the radius of the collapsed star and T is the period of rotation after the collapse.
Substituting the values:
Speed = (2π * (1.0 x 10^4 m)) / (3.0 x 10^(-7) s)
Calculating the numerical value, we can find the speed of the person standing on the equator of the collapsed star.
(a) To find the period of rotation after the collapse of the star, we can use the conservation of angular momentum.
Before the collapse, the star has an initial radius (r1) of 1.0 x 10^8 m and a period (T1) of 30.0 days. The initial angular momentum (L1) can be calculated using the formula:
L1 = I1ω1,
where I1 is the moment of inertia of the star and ω1 is the initial angular velocity.
For a spherical star, the moment of inertia can be approximated as:
I1 ≈ (2/5) * m * r1^2,
where m is the mass of the star.
The angular velocity can be found by converting the period to seconds:
ω1 = 2π / T1.
Now, let's calculate the initial angular momentum (L1):
L1 ≈ (2/5) * m * r1^2 * (2π / T1).
After the collapse, the star has a final radius (r2) of 1.0 x 10^4 m. The final angular momentum (L2) can be calculated similarly:
I2 ≈ (2/5) * m * r2^2,
ω2 = 2π / T2,
L2 ≈ (2/5) * m * r2^2 * (2π / T2).
According to the conservation of angular momentum, L1 = L2. Therefore, we can write:
(2/5) * m * r1^2 * (2π / T1) = (2/5) * m * r2^2 * (2π / T2).
Simplifying the equation, we get:
(r1^2 * T2) / (r2^2 * T1) = 1.
Solving for T2, we find:
T2 = (r2^2 * T1) / r1^2.
Now we can substitute the given values into the equation to find the period of rotation after the collapse.
(b) To find the work done by gravity during the collapse, we can use the formula:
Work = ΔPE,
where ΔPE is the change in potential energy of the star.
The potential energy of a spherical object can be calculated using the formula:
PE = - G * (m1 * m2) / r,
where G is the gravitational constant, m1 and m2 are the masses of the objects, and r is the distance between them.
Before the collapse, the star has an initial radius (r1) and a mass (m). After the collapse, the star has a final radius (r2) and the same mass (m).
The initial potential energy (PE1) can be calculated as:
PE1 = - G * (m * m) / r1.
The final potential energy (PE2) can be calculated as:
PE2 = - G * (m * m) / r2.
The change in potential energy (ΔPE) can be found by subtracting PE2 from PE1:
ΔPE = PE1 - PE2.
Now substitute the given values into the equation to find the work done by gravity during the collapse.
(c) To find the speed of a person standing on the equator of the collapsed star, we need to calculate the centripetal acceleration.
The centripetal acceleration (ac) can be calculated using the formula:
ac = ω^2 * r,
where ω is the angular velocity and r is the radius of the star.
The angular velocity (ω) can be found by converting the period of rotation to seconds:
ω = 2π / T2,
where T2 is the period of rotation after the collapse.
Now substitute the given values into the equation to find the centripetal acceleration.
The speed (v) can be calculated by multiplying the centripetal acceleration by the radius of the star:
v = ac * r.
Substitute the calculated centripetal acceleration and the given radius of the star into the equation to find the speed of the person standing on the equator of the collapsed star.