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