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anonymous
 4 years ago
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.
anonymous
 4 years ago
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.

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anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0p=m/v now i know that the change in density is going to be due to the change i mass and volume

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so would it be \[dp=\frac{dm}{dv}\]

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0p is a linear function also

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358678795614:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0What subject is this?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it's calculusbased physics

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0the 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so 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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0calculus is needed due to the fact that the mass and volume are changing as you get farther into the core which changes the density

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358679271874:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[dp \neq \frac{ dm }{ dv }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[dp = \frac{ vdm  mdv }{ v^2 }\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0correct so where does \[pdv=dm\] come from

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0by any chance, is the density function given? p(x)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0ehh i think it was something like (.0000007+10^4)r

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0it is (.000005r+10^4)

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0So p(x) is (.000005r+10^4)?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i need somehow to relate the change of mass nad volume to the change in radius

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358680137198:dw

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

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358680208001:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358680330123:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0now is it adding 10^4 or multiplying 10^4?

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0this is what he hsa i just need help understanding where the dm is coming from

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358680376740:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0only difference is he has Rd at top

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358680521783:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358680673477:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0i found this \[M=\int pdV\] which would make sense but how does pdv=dm relate. Afterthis it makes sense \[\int dm=\int pdV\] \[m=pdV\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358680956774:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358681105717:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0\[ p=\frac{m}{v}\] is it because p is essentially the change in density

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0so \[p=\Delta p=\frac{\Delta m}{\Delta v}\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0dw:1358681240467:dw

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Yes, 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.

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0or would it actually be \[pdr=\frac{dm}{dv}\] \[pdvdr=dm\]

anonymous
 4 years ago
Best ResponseYou've already chosen the best response.0Alright, I hope some of my work was of some use.
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