Virial Theorem

The virial theorem, maybe better described as bound body energetics, requires that an object has an inward central force $r^{-a}$ such as a gravitational force or an electromagnetic force. The theorem also requires a constant moment of inertia on average so that the object maintains its rough shape over whatever timescale is relevant. Most structures can be virialized such as Earth, the sun, and the galaxy. For virialized structures we can say that Kinetic energy (K)$= \frac{-1}{2}$ Potential Energy (U)

Worksheet 8.1 #1

For a planet of mass m orbiting a star of mass M, at a distance a, start with the Virial Theorem
and derive Kepler’s Third Law of motion. Assume that m is much less than M 

$U_g = \frac{-GmM}{a}$

$K=\frac {U_g}{2}=\frac{-GmM}{2a}$=$\frac {1}{2} mv_p^2$

$v_p= \frac{2 \pi a}{T}= (\frac{GM}{a})^{\frac{1}{2}}$

$\frac{4 \pi ^2 a^2}{T^2}= \frac{GM}{a}$

$T^2=\frac{4 \pi ^2a^3}{GM}$

#2

Consider a spherical distribution of particles, each with a mass $m_i$ and a total (collective) mass
$\qquad \sum_i^N m_i = M $ and a total (collective) radius R. Convince yourself that the total potential energy, U is $U= \frac{-3GM^2}{5R}$

We can relate $m_i$ and M using the number of particles N

$m_i = \frac{M}{N}$

$U_{ij}= \frac{-Gm_i^2}{r_j}$

We can relate $\rho$ and M since $\rho = M/V= \frac{M}{\frac{4}{3} \pi R^3}$

Therefore M= $\rho \times 4 \pi R^3$

$\frac{dU}{dr}= \frac{-G \rho^2 16\pi^2r^4}{3}$

$dU= \frac{-G \rho^2 16 \pi ^2}{3}$

We take the integral to solve for U and we get

U=$\frac{-16G \rho^2 \pi ^2 R^5}{15}$ which simplifies to

$U= \frac{-3GM^2}{5R}$

3. If the average speed of a star in a cluster of thousands of stars is v, give an expression for the total
mass of the cluster in terms of v, the cluster radius R, and the relevant physical constants.

The virial theorem states that $K= \frac{-1}{2}U$

$K= 1/2Mv^2$ and in question 2 above we found that $U= \frac{-3GM^2}{5R}$

Therefore, $1/2Mv^2 = \frac{-3GM^2}{5R}$

When we solve for M we get $M= \frac{5Rv^2}{3G}$

4. The cluster M80 has an angular diameter about 10 arcminutes and resides about 10 4 parsecs from
the Sun. The average speed of the stars in the cluster is v =10 km $s^{-1}$ . Approximately how much
mass, in solar masses ($M_o$), does the cluster contain?

$v=10kms^{-1}$

$\theta = 10 arcmin$

$d=10^4 parsecs$

In part 3 we found that $M= \frac{5Rv^2}{3G}$

R=$\frac{d \theta}{2}$

$M=\frac{5d\theta v^2}{6G}$

M=$10^{36}kg$

$\theta = 10 arcmin= \frac{1}{360}$rad

$M_o= 2 \times 10^{30}$

M= $5 \times 10^5 M_o$

Worksheet 8.2 #2

Forming Stars Giant molecular clouds occasionally collapse under their own gravity (their own “weight”) to form stars. This collapse is temporarily held at bay by the internal gas pressure of the cloud, which can be approximated as an ideal gas such that $P = n_k T$, where n is the number density ($cm^3$ ) of gas particles within a cloud of mass M comprising particles of mass m (mostly hydrogen molecules, H2), and k is the Boltzmann constant, $k =1.4 \times 10^{16} erg K^{-1}$

(a) What is the total thermal energy, $E_{th}$, of all of the gas particles in a molecular cloud of total mass M? (HINT: a particle moving in the $j_{th}$ direction has$ E_{th,j} = 1/2mv_j ^2= 1/ 2 kT$. This fact is a consequence of a useful result called the Equipartition Theorem.) 

We know the thermal energy of a molecule moving in the j direction, but molecules also move in two other directions

In the j direction $ E_{th,j} = 1/2mv_j ^2 = 1/ 2 kT$

To get the 3D version, we can multiply the thermal energy by 3 to get $\frac{3kT}{2}$ which is the thermal energy of 1 particle.

To find the thermal energy for N particles we use the information that N= $\frac{M}{m}$

$E_{th,N}= \frac{3MkT}{2}$

(b) What is the total gravitational binding energy, $U_g$ of the cloud of mass M? 

As we found above in Worksheet 8.1, $U_g= \frac{-3GM^2}{5R}$

(c) Relate the total thermal energy to the binding energy using the Virial Theorem. 

Using the virial theorem we know that $E_{th}=-\frac{1}{2}U_g$

(d) If the cloud is stable, then the Virial Theorem will hold. What happens when the gravitational binding energy is greater than the thermal (kinetic) energy of the cloud? Assume a cloud of constant density ρ. What is the critical mass, $M_J$ , beyond which the cloud collapses? This is known as the “Jeans Mass.” 

The virial theorem states that $E_{th}=-\frac{1}{2}U_g$ and we know that

$E_{th,N}= \frac{3M_JkT}{2}$ and $U_g= \frac{-3GM_J^2}{5R_J}$ so we get the equation

$E_{th,N}= \frac{3M_JkT}{2}$= $U_g= \frac{-3GM_J^2}{10R_J}$

When we solve for $M_J$ we find that $M_J= \frac{5R_JKT}{Gm}$

(e) What is the critical radius, $R_J$ , that the cloud can have before it collapses? This is known as the “Jeans Length.”

We can use $M_J$ which we just found in part d to solve for $R_J$

We can use the density formula to get rid of $M_J$ since $M_J= \rho \times 4/3\pi R_J^3$

$M_J= \frac{5R_JKT}{Gm}$ so $\rho \times 4/3\pi R_J^3=\frac{5R_JKT}{Gm}$

When we solve for $R_J$ we get $R_J= (\frac{15KT}{4\pi Gm \rho})^{\frac{1}{2}}$

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