# Why does a hot cloud need more mass to collapse?

I was wondering why does a hot cloud need more mass to collapse than a cold cloud to form a protostar? Is it because there's a higher thermal pressure inside the hotter cloud than it is in a colder one?

• In layman's terms higher temperature = faster molecules .. if you want those fast molecules to clump together .. you need higher gravitational pull = more mass – eagle275 Mar 10 '20 at 14:42

## 1 Answer

It's more do do with having a higher pressure gradient than a higher pressure, though for a cloud of a set size, the two are equivalent.

For a cloud to be in equilibrium requires $$\frac{dP}{dr} = - G\frac{M\rho}{R^2},$$ where $$dP/dr$$ is the pressure gradient and $$M$$ and $$R$$ are the mass and radius of the cloud and $$\rho$$ is its density.

It we just make a rough approximation of a linear pressure gradient and that the pressure on the outside is zero, then $$P_c \simeq G\frac{M\rho}{R},$$ where $$P_c$$ is the central pressure, which is in turn, for a perfect gas $$P_c = \rho k_B T_c/\mu$$ where $$T_c$$ is the temperature and $$\mu$$ is the average mass of a particle in the cloud.

In order to collapse, the RHS has to get bigger than the LHS. For a given mass, radius and density, then this means the temperature of the cloud must be less than some critical value. i.e. $$G \frac{M\rho}{R} > \frac{\rho k_B Tc}{\mu}$$ $$T_c < \left(\frac{G\mu}{k_B}\right)\left( \frac{M}{R}\right)$$

Thus a hot cloud won't collapse and a cold one will. But one way to get the hotter cloud to collapse would be to increase $$M$$ (while keeping $$R$$ constant).

• Somehow math always makes it better. :-) – StephenG Mar 10 '20 at 22:35