New answers tagged star-formation
13
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 ...
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