Math Problem Statement

The star 18 Scorpii's mass is 2.0 ✕ 1030 kg, its radius is 7.0 ✕ 105 km, and it has a rotational period of approximately 23 days. If 18 Scorpii should collapse into a white dwarf of radius 6.5 ✕ 103 km, what would its period (in s) be if no mass were ejected and a sphere of uniform density can model 18 Scorpii both before and after?

Solution

To calculate the new rotational period of the star, we can apply the conservation of angular momentum principle. Angular momentum is conserved because there is no external torque acting on the system. The formula for angular momentum is:

L=IωL = I \cdot \omega

Where:

  • II is the moment of inertia of a sphere, I=25MR2I = \frac{2}{5} M R^2,
  • ω\omega is the angular velocity, ω=2πT\omega = \frac{2\pi}{T}, with TT as the rotational period.

Since angular momentum is conserved: Iinitialωinitial=IfinalωfinalI_{\text{initial}} \cdot \omega_{\text{initial}} = I_{\text{final}} \cdot \omega_{\text{final}}

Expanding this: (25MRinitial2)2πTinitial=(25MRfinal2)2πTfinal\left(\frac{2}{5} M R_{\text{initial}}^2\right) \cdot \frac{2\pi}{T_{\text{initial}}} = \left(\frac{2}{5} M R_{\text{final}}^2\right) \cdot \frac{2\pi}{T_{\text{final}}}

The mass MM and the constants 25\frac{2}{5} and 2π2\pi cancel out, leaving: Rinitial2Tinitial=Rfinal2Tfinal\frac{R_{\text{initial}}^2}{T_{\text{initial}}} = \frac{R_{\text{final}}^2}{T_{\text{final}}}

Rearranging for TfinalT_{\text{final}}: Tfinal=TinitialRfinal2Rinitial2T_{\text{final}} = T_{\text{initial}} \cdot \frac{R_{\text{final}}^2}{R_{\text{initial}}^2}

Substituting the known values:

  • Tinitial=23days=23×24×3600s=1,987,200sT_{\text{initial}} = 23 \, \text{days} = 23 \times 24 \times 3600 \, \text{s} = 1,987,200 \, \text{s},
  • Rinitial=7.0×105km=7.0×108mR_{\text{initial}} = 7.0 \times 10^5 \, \text{km} = 7.0 \times 10^8 \, \text{m},
  • Rfinal=6.5×103km=6.5×106mR_{\text{final}} = 6.5 \times 10^3 \, \text{km} = 6.5 \times 10^6 \, \text{m}.

Tfinal=1,987,200(6.5×1067.0×108)2T_{\text{final}} = 1,987,200 \cdot \left(\frac{6.5 \times 10^6}{7.0 \times 10^8}\right)^2

Tfinal=1,987,200(6.5700)2T_{\text{final}} = 1,987,200 \cdot \left(\frac{6.5}{700}\right)^2

Tfinal=1,987,200(657000)2T_{\text{final}} = 1,987,200 \cdot \left(\frac{65}{7000}\right)^2

Tfinal=1,987,200(657000)2=1,987,2000.000086T_{\text{final}} = 1,987,200 \cdot \left(\frac{65}{7000}\right)^2 = 1,987,200 \cdot 0.000086

Tfinal170.3sT_{\text{final}} \approx 170.3 \, \text{s}

Final Answer:

The new rotational period would be approximately 170.3 seconds.

Would you like a deeper explanation or have additional questions?


Related Questions:

  1. How is angular momentum conserved in astrophysical phenomena?
  2. What assumptions are made when modeling stars as spheres of uniform density?
  3. How does the conservation of angular momentum apply to other collapsing objects like neutron stars?
  4. What would happen if the star ejected mass during collapse?
  5. How do scientists measure the rotational period of stars?

Tip: Always check units when solving astrophysical problems to ensure accuracy!

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Math Problem Analysis

Mathematical Concepts

Physics
Angular Momentum Conservation
Rotational Dynamics

Formulas

L = I ⋅ ω
I = (2/5) M R^2
ω = 2π/T
R_initial^2 / T_initial = R_final^2 / T_final

Theorems

Conservation of Angular Momentum

Suitable Grade Level

Undergraduate Physics or Advanced High School Physics