## Tuesday, October 25, 2011

### Hydrostatic Equilibrium and the Virial Theorem

Author: Iryna Butsky
Co-authors: Juliette Becker and Monica He

This is a write up of question 1 of the Hydrostatic Equilibrium and the Virial Theorem worksheet.

• (a) We're looking for the total potential energy contained in the sun. We know that potential energy is expressed by U = - GMm/r. More specifically, in this case                                $U = \frac{GM_{shell}M_{int}}{r}$ where M_shell is the mass and M_int is the mass of the sphere inside that shell. We also have $dM_{shell}= \rho(r)4\pi r^2dr$ and  $dM_{int}= \rho(r)\frac{4}{3}\pi r^3$.The change in potential energy, dU: $dU = \frac{GdM_{shell}dM_{int}}{r}$ therefore, after combining the terms we have that $U =\int_{0}^{R_{\odot}} \frac{16}{3}G\rho^2\pi^2 r^4dr = \frac{16}{15}G\rho^2\pi^2R_{\odot}^5$.  Keeping in mind that  $\rho = \frac{M}{V} = \frac{M_{\odot}}{\frac{4}{3}\pi R_{\odot}^3}$we can simplify the expression to $U = \frac{3GM_{\odot}^2}{5R_{\odot}}$

• (b) Potential energy is related to force in the following expression:                            $U = \int F\cdot dr = \int m\frac{d^2r}{dt^2}$                                                                                                                                  By doing dimensional analysis, we know that U has units of cm^3g^-1s^2 and dr/dt has units of cm/s. Therefore the velocity of a free-falling element would have the following relation to potential energy:              $\frac{dr}{dt}=\sqrt{\frac{U}{M}}$
• (c) We can replace dr/dt in the equation above with $\small R_{\odot}/\tau_{ff}$. We're assuming that the free-fall velocity will be roughly constant. Therefore, if we define tau to be the time it takes a particle to fall from the surface of the star to the center and R_solar as the radius of the Sun, then we can make this substitution to an order of magnitude.

• (d) Plugging in our values into this expression $\small \frac{R_{\odot}}{\tau_{ff}}=\sqrt{\frac{U}{M}}$ we find that $\tau_{ff} = 2065 \;\mathrm{seconds}\simeq 34 \;\mathrm{minutes}$. This means that if the Sun were truly undergoing free fall collapse, it would collapse in 34 minutes, which is surely a time-scale that we would notice.

• (e)

#### 1 comment:

1. (b): That was clever to use dimensional analysis! Another way of looking at it: if a particle starts very very far away (R=infinity) and falls to R_sun, what will its kinetic energy and hence its velocity be?

(c): A particle in free-fall will definitely be accelerating. If our test particle is accelerating, what effect does that have on your estimate of the timescale for free-fall? How does it affect your argument that we would notice if the sun were in free-fall?

(e) Awwwww!