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Outkast3r09
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Find the depth at which the Gfield of a star is 25% with a density given by p(x), given radius is 4000km and mass at the surface is 3000kg.
 one year ago
 one year ago
Outkast3r09 Group Title
Find the depth at which the Gfield of a star is 25% with a density given by p(x), given radius is 4000km and mass at the surface is 3000kg.
 one year ago
 one year ago

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Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
@UnkleRhaukus
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
p=m/v now i know that the change in density is going to be due to the change i mass and volume
 one year ago

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so would it be \[dp=\frac{dm}{dv}\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
my teacher had his second equation in a similar problem as \[pdv=dm\] which i can only get if i let \[p*dv=\frac{dm}{dv}*dv\]
 one year ago

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p is a linear function also
 one year ago

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dw:1358678795614:dw
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
What subject is this?
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
it's calculusbased physics
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
the only unrelevant equation that makes it physics is \[g=\frac{Gm}{r^2}\] where G is the gravitational constant \[6.67*10^{11}\] m is the mass below and r is the radius to the center
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
so since we're given a radius and a mass you can find out what the Gfield is at the surface and solve for what 25% of it is. However to find the unknowns depth and mass , we need calculus
 one year ago

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calculus is needed due to the fact that the mass and volume are changing as you get farther into the core which changes the density
 one year ago

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dw:1358679271874:dw
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.0
\[dp \neq \frac{ dm }{ dv }\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
\[p=\frac{dm}{dv}\]
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.0
\[dp = \frac{ vdm  mdv }{ v^2 }\]
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
correct so where does \[pdv=dm\] come from
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
by any chance, is the density function given? p(x)?
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
ehh i think it was something like (.0000007+10^4)r
 one year ago

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it is (.000005r+10^4)
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
So p(x) is (.000005r+10^4)?
 one year ago

bhaskarbabu Group TitleBest ResponseYou've already chosen the best response.0
pdv = dm is wrong
 one year ago

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p(x) given?
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
p=.0000004r+10^4
 one year ago

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i need somehow to relate the change of mass nad volume to the change in radius
 one year ago

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or change in density
 one year ago

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dw:1358680137198:dw
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
what my teacher wrote for a very similar problem was \[pdv=dm\] \[p(4\pi r^2)=dm\]
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
dw:1358680208001:dw
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
dw:1358680330123:dw
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
now is it adding 10^4 or multiplying 10^4?
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
adding
 one year ago

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this is what he hsa i just need help understanding where the dm is coming from
 one year ago

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dw:1358680376740:dw
 one year ago

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only difference is he has Rd at top
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
dw:1358680521783:dw
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
What do you mean?
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
on the integral?
 one year ago

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dw:1358680673477:dw
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
i found this \[M=\int pdV\] which would make sense but how does pdv=dm relate. Afterthis it makes sense \[\int dm=\int pdV\] \[m=pdV\]
 one year ago

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forogot int sign
 one year ago

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dw:1358680956774:dw
 one year ago

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dw:1358681105717:dw
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
\[ p=\frac{m}{v}\] is it because p is essentially the change in density
 one year ago

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so \[p=\Delta p=\frac{\Delta m}{\Delta v}\]
 one year ago

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\[p=\frac{dm}{dv}\]
 one year ago

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\[pdv=dm\]
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
dw:1358681240467:dw
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
Yes, p is the density function, and since it has r in it, it changes depending on the radius. In the previous steps, I showed the mass by integrating p over the volume of the star... so that's like rearranging the equation to read m = p*v. But since p changes, I had to do an integral.
 one year ago

Outkast3r09 Group TitleBest ResponseYou've already chosen the best response.0
or would it actually be \[pdr=\frac{dm}{dv}\] \[pdvdr=dm\]
 one year ago

zordoloom Group TitleBest ResponseYou've already chosen the best response.0
Alright, I hope some of my work was of some use.
 one year ago
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