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The Hernquist model for spherical galaxies states that the mass density at a distance s to the galactic center is given by

 {\displaystyle \rho(s) = \frac{1}{s(1+s)^3} }.

(Here, s represents the variable radius to the center of the galaxy.)  This model was chosen so that the density has the behavior \rho \sim 1/s as s \rightarrow 0 and \rho \sim 1/s^4 as s \rightarrow \infty.  Let M(r) represents the total mass contained within a sphere of radius r, centered at the galactic center.    (I.e., M(r) is a cumulative mass function–it tells the total mass that lies within a distance r to the galaxy’s center.)  Determine a formula for M(r).  Hint: Use spherical shells.

Guest Dec 14, 2014

Best Answer 

 #1
avatar+26399 
+10

In a spherical shell of thickness ds at a radius s, the approximate volume, dV is

$$dV=4\pi s^2ds$$

 

Hence the mass, dM in this shell is given by ρ*dV or:

$$dM=4\pi \frac{s^2}{s(1+s)^3}ds$$

 

Integrating this from 0 to r we get:

$$M(r)=4\pi \int_0^r \frac{s^2}{s(1+s)^3}ds=2\pi \frac{r^2}{(r+1)^2}$$

 

Alan  Dec 14, 2014
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4+0 Answers

 #1
avatar+26399 
+10
Best Answer

In a spherical shell of thickness ds at a radius s, the approximate volume, dV is

$$dV=4\pi s^2ds$$

 

Hence the mass, dM in this shell is given by ρ*dV or:

$$dM=4\pi \frac{s^2}{s(1+s)^3}ds$$

 

Integrating this from 0 to r we get:

$$M(r)=4\pi \int_0^r \frac{s^2}{s(1+s)^3}ds=2\pi \frac{r^2}{(r+1)^2}$$

 

Alan  Dec 14, 2014
 #2
avatar+91435 
0

Thanks Alan,

I think I have got my head around that.  :)

How would you do that integration by hand?

Melody  Dec 15, 2014
 #3
avatar+26399 
+5

 Integration:

.

Alan  Dec 15, 2014
 #4
avatar+91435 
0

Thank you Alan, this was a great question and answer :))

Melody  Dec 15, 2014

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