Changes between Version 25 and Version 26 of u/erica/UniformCollapse

Timestamp:

06/29/15 15:53:15 (10 years ago)

Author:

Erica Kaminski

Comment:

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u/erica/UniformCollapse

@@ -8 +8 @@
 This acceleration is due to all of the mass ''inside'' of the sphere, i.e. the mass contained within a radius r, [[latex($M_r$)]]. Since this equation governs the local acceleration of the gas at any radius, we can consider how the acceleration of any shell within the sphere behaves over time, once we know r(t) of course. Initially (t=0), however, we do know the radius, so we can check the behavior of this equation at t=0 for various shells within the sphere. 
 
-Consider first the outer most radius. A sphere with radius [[latex($r=r_0$)]] has within it mass that goes like [[latex($M_{r0} \propto r_0^3$)]]. Therefore, we see that initially, the acceleration of the gas is proportional to [[latex($r_0$)]]. The density of the sphere is uniform, and this means repeating this procedure for any smaller radii shows that the acceleration of interior shells decreases as r decreases. In other words, the furthest shells have the highest acceleration. They speed up from an initial velocity of 0 the fastest. This has to be the case, as 
+Consider first the outer most radius. A sphere with radius [[latex($r=r_0$)]] has within it mass that goes like [[latex($M_{r0} \propto r_0^3$)]]. Therefore, we see that initially, the acceleration of the gas is proportional to [[latex($r_0$)]], that is, [[latex($\frac{dr^2}{dt^2}\propto r_0$)]]. The density of the sphere is uniform, and this means repeating this procedure for any smaller radii shows that the acceleration of interior shells decreases as r decreases. In other words, the furthest shells have the highest acceleration. They speed up from an initial velocity of 0 the fastest. This has to be the case, as 
 
 - all shells reach the origin at the same time (so outer shells have furthest to go in the same amount of time, i.e. they must travel faster)