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# Considering a spherical galaxy, with a volumetric mass density s from the center of the galaxy, given by p(s)= c/(1+s^3)

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Considering a spherical galaxy, with a volumetric mass density s from the center of the galaxy, given by p(s)= c/(1+s^3)

where c is a constant. let c=27.4

how do I determine the  total mass M enclosed within a distance r to the galactic center using spherical shells?

Guest Oct 21, 2014

#3
+80804
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We probably don't have enough bananas to satisfy Chimp Ayumu....however....we DO stock quite a quantity of peanuts......as long as she promises not to get any  (spherical) shells on the floor......

CPhill  Oct 22, 2014
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#1
+91412
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This question was similar - it was probably yours.  Does it help?

http://web2.0calc.com/questions/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

If not you will probably waits till Alan is on.  He is the only one on here that can usually answer questions at this level.  I would not expect him to be on for several hours.

Melody  Oct 22, 2014
#2
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Hay, wait just a minute . . .

That might have been true until I (we) joined the forum.

This might take me a few (dozen) hours to figure out. Maybe only a couple of hours if we chimps brainstorm.

If we really get in a pinch, Chimp Ayumu is skilled in high-level maths, and she helps us sometimes.

Chimp Ayumu

I like her, but some of the board members don’t because she makes us look kind of dumb.

We really need a fulltime, high-level mathematician, but she won’t join our board. She says we couldn’t grow enough bananas to get her to join.

We know how to count bananas, though.

#3
+80804
+10

We probably don't have enough bananas to satisfy Chimp Ayumu....however....we DO stock quite a quantity of peanuts......as long as she promises not to get any  (spherical) shells on the floor......

CPhill  Oct 22, 2014
#4
+26397
+5

Without the extra s in the denominator the mass is given by the uglier expression:

More explanation of the process is given in the answer at the link given above by Melody.

Alan  Oct 22, 2014

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