Question

A star of mass 1.31x10^{31}1.31×1031 kg and diameter 9.79x10^99.79×109 m rotates with a period of 24.024.0 days. Suddenly, the star changes size and rotates with a new period of 14.014.0 days. The mass of the star is conserved. Assuming a uniform (but different) density both before and after the size change, by what fraction does its diameter change? In other words, what is (new diameter)/(old diameter)?

Answer 1

The fraction by which the star's diameter changes is approximately 1.52.

Step-by-step explanation:

To answer this question, we can use the concept of angular momentum. Angular momentum is proportional to the square of the object's size (diameter) divided by its period of rotation (D²/P).

Since the mass of the star is conserved, we can use the equation D²/P = constant to find the ratio of the new diameter to the old diameter.

Plugging in the values, we have (new diameter)/(old diameter) = (√(old period/new period))2.

By substituting the given values, we get (new diameter)/(old diameter) = (√(24.0/14.0))2 ≈ 1.52.

Answer 2

The star's diameter increases by a factor of approximately 1.40 after the change in its rotation period.

To solve this problem, we can leverage the principle of conservation of angular momentum and the relationship between a star's density, diameter, and angular velocity. Here's how we can approach it:

Define the initial and final states:

Initial period (T_i) = 24.0 days

Final period (T_f) = 14.0 days

Initial mass (M) = 1.31 x 10^31 kg

Initial diameter (D_i) = 9.79 x 10^9 m

Calculate the initial and final angular velocities:

Angular velocity (ω) = 2π / T

Initial angular velocity (ω_i) = 2π / T_i

Final angular velocity (ω_f) = 2π / T_f

Apply the conservation of angular momentum:

Angular momentum (L) = Iω, where I is the moment of inertia

Assuming a uniform density, I ∝ R^2 (for a sphere, I = (2/5)MR^2)

Therefore, L ∝ R^2ω

Since mass is conserved, L_i = L_f

Substitute and rearrange: R_i^2ω_i = R_f^2ω_f

Solve for R_f: R_f = R_i * √(ω_i / ω_f)

Find the final diameter (D_f):

D_f = 2 * R_f

Calculate the ratio of the new diameter to the old diameter:

Diameter ratio = D_f / D_i

Plugging in the given values, we get:

ω_i = (2π) / 24.0 days ≈ 0.262 rad/day

ω_f = (2π) / 14.0 days ≈ 0.443 rad/day

R_f = 9.79 x 10^9 m * √(0.262 rad/day / 0.443 rad/day) ≈ 6.88 x 10^9 m

D_f = 2 * 6.88 x 10^9 m ≈ 13.76 x 10^9 m

Diameter ratio = D_f / D_i ≈ 13.76 x 10^9 m / 9.79 x 10^9 m ≈ 1.40