Worksheet 8, Problem 2: Revisiting Derivations

Introduction: This week we are learning about star formation, which leads to some of the most beautiful images in astronomy.

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From the worksheet:
One of the most useful equations in astronomy is an extremely simple relationship known as the Virial Theorem. It can be used to derive Kepler’s Third Law, measure the mass of a cluster of stars, or the temperature and brightness of a newly-formed planet. The Virial Theorem applies to a system of particles held together by a force that varies according to the inverse central-force law $F \alpha \frac{1}{r^{\alpha}}$

The problem: Consider a spherical distribution of particles, each with a mass mi and a total (collective) mass $\sum_{i}^{N} m_i = M$, and a total (collective) radius R. Convince yourself that the total potential energy, U, is approximately $$U \approx - \frac{GM^2}{R}$$ You can derive or look up the actual numerical constant out front. But in general in astronomy, you don’t need this prefactor, which is of order unity.

Solve: We can think about this sphere of particles with a uniform density in terms of thin shells, each with a width of dR.

We also know that $$U = - \frac{GMm}{R}$$ We can treat the entire sphere as our mass, M, and the shell as our mass, m. Then, we can rewrite each mass in terms of density: $$\rho = \frac{M}{V}$$ Where $V = \frac{4}{3} \pi R^3$ So $$\rho = \frac{M}{\frac{4}{3} \pi R^3}$$ And we can rewrite the equation for potential energy of one shell as: $$dU = - \frac{GMdm}{R} = - \frac{G(\frac{4}{3} \pi R^3 \rho) (4 \pi R^2 \rho) dR}{R}= - G \frac{16}{3} \pi^2 R^4 \rho^2 dR$$ But now we have dU in terms of dR, so we can integrate to get U: $$\int_{0}^{R} - \frac{16}{3} G \pi^2 R^4 \rho^2 dR =  - \frac{16}{3} G \pi^2 \rho^2  \int_{0}^{R} R^4 dR = - \frac{16}{3} \frac{R^5}{5} G \pi^2 \rho^2 - 0$$ But this still is not the answer we want. We need U in terms of M, not density so we can reuse   $$\rho = \frac{M}{ \frac{4}{3} \pi R^3 }$$ to get: $$U = - \frac{16}{3} \frac{R^5}{5}\left( \frac{M}{\frac{4}{3} \pi R^3 } \right)^2 G \pi^2$$ Amazingly enough, this simplifies to the much nicer expression: $$U = - \frac{4GM^2}{5R}$$ The actual answer is $$U = - \frac{3GM^2}{5R}$$ So something must have gone wrong with our constants, but either way both simplify to our original expression: $$U \approx - \frac{GM^2}{R}$$

Acknowledgements: I worked with Barra and April on this worksheet.