(a) What is the total thermal energy, $K$, of all the gas particles in a molecular cloud of total mass $M$? (HINT: a particle moving in the $i^{th}$ direction has $E_{thermal} = \frac{1}{2}mv_i^2 = \frac{1}{2}kT$. This fact is a consequence of a useful result called the Equipartition Theorem.)
Okay, so we want to know the total thermal energy, $K$ of the all the gas particles in the molecular cloud. Well let's look at just one gas particle first. Since the gas particle exists in a three-dimentional space, the thermal energy of the single gas particle is three times that of the thermal energy in any given direction, and can be expressed as:
\begin{align}
E_{thermal_{particle}} = \frac{3}{2}mv_i^2 = \frac{3}{2}kT
\end{align}
Okay, so we now have the thermal energy of a single gas particle. To find the thermal energy of all the gas particles in a molecular clouds, we need to know just how many gas particles there are. Since we know the mass of the entire molecular cloud to be $M$, and the mass of a single gas particle to be $\overline{m}$, we can get the total number of gas particles by the expression $\frac{M}{\overline{m}}$. Therefore, we can multiply the thermal energy of a single particle by the number of particles $\frac{M}{\overline{m}}$ to get the thermal energy of the entire molecular cloud, as such:
\begin{align}
E_{thermal_{cloud}} = \frac{3}{2} \frac{ktM}{\overline{m}}
\end{align}
The total gravitational binding energy of a cloud is simply the gravitational potential energy, $U$ of all the particles in a system, which we derived in Worksheet 8, Problem 2, to be:
\begin{align}
U = -\frac{GM^2}{R}
\end{align}
(c) Relate the total thermal energy to the binding energy using the Virial Theorem, recalling that you used something similar to kinetic energy to get the thermal energy earlier.
We can relate the total thermal energy $E_{thermal}$ of a system to the total binding energy $U$ using the Virial Theorem. The Virial Theorem states:
\begin{align}
K = -\frac{1}{2}U
\end{align}
Since the thermal energy is a result of the kinetic energy, and the binding energy is a result of the gravitational potential energy, we can substitute the kinetic energy and the potential energy of the Virial Theorem with the values we got for the thermal energy and gravitational binding energy and simplify to get the relation between the two:
\begin{align}
\frac{3}{2} \frac{ktM}{\overline{m}} &= -\frac{1}{2}(-\frac{GM^2}{R})\\
\frac{3kT}{\overline{m}} &= \frac{GM}{R}\\
3kT &= \frac{GM \overline{m}}{R}
\end{align}
(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 $\rho$.
In order to understand this question, let's think about the two components of the Virial Theorem. When a cloud is stable, the Virial Theorem holds true, such that $K = -\frac{1}{2}U$. In a stable system where the Virial Theorem holds true, the kinetic energy of the entire system balances with the total gravitational potential energy of the system. Kinetic energy $K$ has the tendency to make things spread apart, since the more kinetic energy there is, the more particles move around and disperse radially outward in a diffusion-like pattern. Gravitational potential energy $U$ has the tendency to bring particles come together. Under the Universal Law of Gravitation, any two objects with mass will attract one another. Therefore, in a stable molecular cloud, the individual gas particles are in a tug-of-war between the outward expansion caused by thermal (kinetic) energy, and the inward pull of gravitational potential energy that wants to bring all the particles together.
When the gravitational potential energy (binding energy) is greater than the thermal (kinetic) energy, the tug-of-war is over in favor of gravity. As a result, the inward pull by gravitational potential energy overcomes the outward expansion by kinetic energy, and the gas particles have a net movement inwards. This is when a molecular cloud "collapses" under the weight of its own gravity.
(e) What is the critical mass, $M_J$, beyond which the cloud collapses? This is known as the "Jeans Mass."
In the case of a cloud collapsing under the weight of its own gravity, we know that the gravitational potential energy of the system is greater than the kinetic energy of the system. We can use the equation from part (c) to model this scenario, except instead of having a stable system where the kinetic energy is equal to the gravitational potential energy, in this case, the gravitational potential energy is greater than the kinetic energy:
\begin{align}
3kT &< \frac{GM \overline{m}}{R}
\end{align}
Now, based on this equation, the gravitational potential energy is only determined by the radius $R$ of the molecular cloud, since the mass of the molecular cloud $M$, and the mass of the individual gas particles $\overline{m}$ are constant along with the gravitational constant.
So now that we know that the gravitational potential energy is determined only by the radius of the molecular cloud. But if the molecular cloud changes its radius because the gas particles are coming closer together since the gravitational potential energy is greater than the kinetic (thermal) energy, then the density of particles in the cloud $\rho$, changes! Therefore, we can relate density and radius using the definition of density as mass over unit volume: $\rho = \frac{mass}{volume}$.
\begin{align}
\rho = \frac{M}{\frac{4}{3} \pi R^3}
\end{align}
And solving for radius, we get:
\begin{align}
R = \left(\frac{3M}{4\rho \pi}\right)^{\frac{1}{3}}
\end{align}
Since we are looking for the critical mass, $M_J$, we can rearrange the equation from part (c), and substitute radius $R$, to solve for the mass:
\begin{align}
3kT &= \frac{GM_J \overline{m}}{R}\\
M_J &= \frac{3kTR}{G\overline{m}}\\
M_J &= \frac{3kT}{G\overline{m}} \cdot \left(\frac{3M}{4\rho \pi}\right)^{\frac{1}{3}}\\
M_J^3 &= \left(\frac{3kT}{G\overline{m}}\right)^3 \left(\frac{3M}{4\rho \pi}\right)\\
M_J^2 &= \left(\frac{3kT}{G\overline{m}}\right)^3 \left(\frac{3}{4\rho \pi}\right)\\
M_J &= \left(\frac{3kT}{G\overline{m}}\right)^{\frac{3}{2}} \left(\frac{3}{4\rho \pi}\right)^{\frac{1}{2}}
\end{align}
(f) What is the critical radius, $R_J$, that the cloud can have before it collapses? This is known as the "Jeans Length."
To find the critical radius, we simply need to solve for the radius $R_J$ from the equation we derived in part (c), and substitute using the density equation.
So, first, let's rearrange the equation from part (c) to solve for $R_j$:
\begin{align}
3kT &= \frac{GM \overline{m}}{R_J}\\
R_J &= \frac{GM \overline{m}}{3kT}
\end{align}
Then, let's rearrange the density equation to solve for $M$ so we can substitute it into the previous equation:
\begin{align}
\rho &= \frac{M}{\frac{4}{3} \pi R_J^3}\\
M &= \frac{4}{3} \rho \pi R^3
\end{align}
Now we can substitute $M$ for the first equation, and solve for the critical radius, $R_J$:
\begin{align}
R_J &= \frac{GM \overline{m}}{3kT}\\
R_J &= \frac{G \left(\frac{4}{3} \rho \pi R_J^3\right) \overline{m}}{3kT}\\
3kTR_J &= G \left(\frac{4}{3} \rho \pi R_J^3\right) \overline{m}\\
3kT &= G \left(\frac{4}{3} \rho \pi R_J^2\right) \overline{m}\\
R^2 &= \frac{3kT}{\frac{4}{3} G \rho \pi \overline{m}}\\
\sqrt{R^2} &= \sqrt{\frac{9kT}{4 G \rho \pi \overline{m}}}\\
R &= \frac{3}{2} \sqrt{\frac{kT}{G \rho \overline{m}}}
\end{align}
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