An Introduction to Modern Astrophysics · Problem 21.3

The Milky Way Galaxy — Carroll & Ostlie

Physics — Astrophysics · jupiter-moment-of-inertia-polytropic

Problem

(a) Assuming spherical symmetry and the density distribution for the polytropic model of Jupiter given in Problem 21.2b, show that the moment of inertia is given by I=8ρc3π(1−6π2)RJ5.I = \frac{8\rho_c}{3\pi}\left(1 - \frac{6}{\pi^2}\right)R_J^5. Hint: Since this analytical model does not assume constant density, you will need to integrate over concentric rings to find the moment of inertia about Jupiter's rotation axis. Recall that I≡∫vola2 dmI \equiv \int_{\rm vol} a^2\,dm, where a=rsin⁡θa = r\sin\theta is the distance from the rotation axis to the ring of mass dmdm, and dm=ρ(r) dV=ρ(r) 2πar dθ dr.dm = \rho(r)\,dV = \rho(r)\,2\pi ar\,d\theta\,dr. (b) Using the estimated value for the central density of Jupiter obtained in Problem 21.2e, calculate the planet's moment-of-inertia ratio. Hint: The moment-of-inertia ratio was first introduced in Problem 20.12. (c) Compare your answer to part (b) with the measured value given in Table 21.2. What does this result say about the true density distribution within the planet relative to the analytical model?

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