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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?

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    $\begingroup$ In layman's terms higher temperature = faster molecules .. if you want those fast molecules to clump together .. you need higher gravitational pull = more mass $\endgroup$
    – eagle275
    Mar 10, 2020 at 14:42

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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).

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    $\begingroup$ Somehow math always makes it better. :-) $\endgroup$ Mar 10, 2020 at 22:35

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