Qn on Rotational Kinetic Energy

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SUMMARY

The discussion centers on calculating the moment of inertia (I) of a neutron star based on its rotational kinetic energy and energy release rate. The Crab Nebula emits energy at a rate of 5x1031 W, with the neutron star's rotation period being 0.0331 seconds and an increase in period of 4.22 x 10-13 seconds per second. The relevant equation used is dK/dt = -(4π2I/T3)(dT/dt). The participant's calculated moment of inertia was 1.1 x 1038, which differs significantly from the expected answer of 1.1 x 1030.

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  • Understanding of rotational kinetic energy principles
  • Familiarity with differential equations in physics
  • Knowledge of neutron star characteristics
  • Proficiency in using the power equation related to energy release
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  • Explore the implications of energy release in astrophysical phenomena
  • Learn about the Crab Nebula and its neutron star dynamics
  • Investigate common errors in solving differential equations in physics
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Homework Statement


Crab Nebula releases energy at a rate of 5x1031 W, obtaining its energy from the rotational kinetic energy K of a neutron star at its centre.
Period of neutron star's rotation = 0.0331s. T increases by 4.22 x 10-13s per second.

What is the moment of inertia of the neutron star?

Homework Equations


dK/dt = -(4[tex]\pi[/tex]2I/T3) (dT/dt)
This was the first part of the question which I managed to prove already.
*Note: the pi is not supposed to be in superscript.

The Attempt at a Solution


I equated the power of the star to the above differential equation. Solving for I should be quite easy, but somehow my answer (1.1x 1038) is different from the answer provided (1.1 x 1030).
 
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Your result looks correct.

ehild
 

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