Class Activity: Formation of Stars Problem 2

If we have a small particle in an orbit of eccentricity 1 (essentially a straight line with the mass around which the particle is orbiting at one end of the line--see picture), then half of the period of this orbit is the time it takes the particle to "fall" in toward the center, where the mass is. This is a decent approximation of how long it takes a cloud of interstellar gas to collapse into a star. The mass about which the particle is orbiting is actually the point mass that comes from considering the mass of the cloud as a point mass, which is valid because as the particle falls in, all the mass inside of it falls as well, so the total mass does not change. We can represent this mass as the volume times the initial average density, then we just need Newton's version of Kepler's third law to obtain the free fall time.

where P is the period of the "orbit," a is the semi-major axis of the orbit, and ρ is the average density of the mass M. The free fall time is half the period, and the radius R is two times the semi-major axis a.

The assumption used to obtain this final step is that the cloud that is collapsing is spherical, which is a valid assumption because oftentimes only part of a cloud will collapse to a star. The part that collapses is a spherical part of the whole, even if the whole cloud isn't spherical.


We know that if the free fall time is longer than the time it takes for a pressure wave (like sound) to traverse the cloud, then the cloud won't be unstable enough to collapse. This is because the beginning of collapse will create a pressure wave, and if that moves through the cloud quickly enough, it will push out the other side of the cloud when it reaches it. This won't cause a collapse, since the cloud isn't pushing inward from all directions. However, if the pressure wave can't make it to the other side of the cloud before that side of the cloud has already collapsed inward somewhat, the cloud will be unstable enough to collapse fully.


The time it takes for a pressure wave to traverse the cloud is the width of the cloud, 2R, over the average speed of sound, c_s. Equating this to the free fall time and solving for the radius R will give the minimum possible size of a cloud in order for it to be unstable enough to collapse. If R were any smaller, the sound wave would pass through the cloud in less time, and it will not be unstable. For an isothermal gas of constant density, the speed of sound is constant.

This is an order-of-magnitude estimate of the Jean's Length, which, again, is the minimum size for a cloud to be unstable enough to collapse.

If we consider a spherical cloud collapsing isothermally with an initial radius of the Jean's Length, we can discover what the new Jean's Length would be after the cloud has collapsed to half of its original radius. At half of the original radius, the average density will be 8 times greater than what it was before, since density depends on the inverse of radius cubed. The speed of sound squared is dependent on the pressure over the density, and since pressure is dependent on inverse surface area of the cloud, the pressure will increase by a factor of 4 when the radius decreases by a factor of 2.

So we see that the radius of the cloud changing by a factor of 1/2 makes the Jean's Length change by a factor of 1/4. This means that once the cloud contracts to half its size, the new-sized cloud can fragment into pieces and still be unstable enough to collapse. This means that a single cloud that is just barely large enough to collapse can form more than one star when it collapses, since parts of it will collapse independently once it begins to contract.


Secondary authors: Joanna Robaszewski, Daniel Lo, Melodie Kao