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how do you find a total mass using spherical shells?, for example the total mass of a spherical Galaxy with a mass density at a distance "s"

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how do you find a total mass using spherical shells?, for example the total mass of a spherical Galaxy with a mass density at a distance "s" to the galactic center, which is given by: 1/s(1+s)^3?

Oct 20, 2014

#1
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If mass density is ρ then you need to find ∫ρ.dV  where dV is infinitesimal volume and the integral goes from zero to infinity.

dV = 4pi.s2.ds, so

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

or

$$mass = 2\pi$$

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Oct 21, 2014
#2
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Thanks Alan,  I am trying to understand this.

this is what I do understand.

$$Mass density = \frac{mass}{unit\; \;of\; \;volume}=\frac{1}{s(1+s)^3}\\\\ going by your answer I assume the (1+s)^3 is meant to be in the denominator.\\\\ Volume=\frac{4}{3}\pi\;s^3=\int_0^\infty \;4\pi s^2 \;ds$$

So obviously,  Mass = density * Volume

BUT

Now, I don't understand how you have put these together.   I mean one is presented as an integral and the other is not.  How does that work?

I am sttuggling with some of the basic concepts of calculus I think.

Oct 21, 2014
#3
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Oct 22, 2014
#4
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If the (1+s)3 term is in the numerator, the density would rapidly increase as s increased and the total mass would be infinite (unless a maximum limit for the size of s were supplied), so I assumed it was in the denominator.

The volume, dV, of a "thin" spherical shell, of thickness, ds, is given by the surface area of a sphere of radius s, namely 4.pi.s2, multiplied by the small thickness ds.  So dV = 4.pi.s2.ds.  The mass per unit volume within this thin shell is 1/(s.(1+s)3), so the mass within this shell is [4.pi.s2/(s.(1+s)3)].ds.  The total mass of the galaxy is obtained by "summing" all these shell masses from zero to infinity.  In other words, we let the shell thickness tend to zero and integrate from zero to infinity.

Does that help?

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Oct 22, 2014
#5
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Thank you Alan, that does help.  I can follow your logic without to much problem but I know that when I try to do it myself tomorrow I will not be able to.  You have explained things like this to me before and although i can follow your logic I do not seem to 'understand' it well it well enough to reproduce it on my own.  It is really frustrating. Do you have any suggestions  :(

Oct 23, 2014
#6
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Another way to think of this question is as follows:

Imagine a sphere of radius, s.  The volume of this sphere is (4/3).pi.s3.

Imagine it is surrounded by a larger sphere of radius s+δs.  The volume of this sphere is (4/3).pi.(s+δs)3.

The volume of the shell between them is therefore (4/3).pi.(s+δs)- (4/3).pi.s3.

This shell volume is therefore: (4/3)*pi*(3.s2.δs + 3.s.δs2 + δs3)

Now we are going to shrink the thickness of this shell until we can ignore terms involving higher than linear powers of δs.  So an infinitesimal thickness shell will have a volume of 4.pi.s2.ds  (where I've replaced δs by ds for the infinitesimal limit). Now the mass in this infinitesimally thin shell is just density*volume or

(1/s(1+s)3)*4.pi.s2.ds

This is an infinitesimal mass, of course, so to get a non-infinitesimal mass we have to sum (integrate) this between one radius and another. The second radius is a non-infinitesimal distance away from the first. In this problem the question is clearly asking for the first radius to be zero and the second to be infinity. So the total mass is given by the integral shown in my first reply above.

I'm not sure that I've got any more general advice on how to approach these sort of problems, except for: do a lot of examples.  I must have done hundreds (thousands?) of these over the years!

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Oct 23, 2014
#7
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Thanks Alan,

I am sure I like your first explanation better LOL.

Do you have some of these handy that you can throw at me.

OR what should i google to find them.  Any ideas?

I am sure you are right about the practice bit.

Oct 23, 2014
#8
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I'll have a look for some practice problems.

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Oct 23, 2014
#9
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Thanks Alan.

Oct 23, 2014
#10
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Alan Oct 23, 2014
#11
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I don't think that they are Alan.

They all look too simple.

I've got a couple that you have already given me.  I can play around with those some more.

But a stack more (with worked answers ) would be really good.

Oct 23, 2014