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Find the depth at which the G-field of a star is 25% with a density given by p(x), given radius is 4000km and mass at the surface is 3000kg.

Mathematics
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p=m/v now i know that the change in density is going to be due to the change i mass and volume
so would it be \[dp=\frac{dm}{dv}\]

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Other answers:

my teacher had his second equation in a similar problem as \[pdv=dm\] which i can only get if i let \[p*dv=\frac{dm}{dv}*dv\]
p is a linear function also
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What subject is this?
it's calculus-based physics
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
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
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
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\[dp \neq \frac{ dm }{ dv }\]
\[p=\frac{dm}{dv}\]
\[dp = \frac{ vdm - mdv }{ v^2 }\]
correct so where does \[pdv=dm\] come from
by any chance, is the density function given? p(x)?
ehh i think it was something like (-.0000007+10^4)r
it is (-.000005r+10^4)
So p(x) is (-.000005r+10^4)?
pdv = dm is wrong
p(x) given?
p=-.0000004r+10^4
i need somehow to relate the change of mass nad volume to the change in radius
or change in density
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what my teacher wrote for a very similar problem was \[pdv=dm\] \[p(4\pi r^2)=dm\]
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now is it adding 10^4 or multiplying 10^4?
adding
this is what he hsa i just need help understanding where the dm is coming from
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only difference is he has R-d at top
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What do you mean?
on the integral?
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i found this \[M=\int pdV\] which would make sense but how does pdv=dm relate. Afterthis it makes sense \[\int dm=\int pdV\] \[m=pdV\]
forogot int sign
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\[ p=\frac{m}{v}\] is it because p is essentially the change in density
so \[p=\Delta p=\frac{\Delta m}{\Delta v}\]
\[p=\frac{dm}{dv}\]
\[pdv=dm\]
|dw:1358681240467:dw|
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.
or would it actually be \[pdr=\frac{dm}{dv}\] \[pdvdr=dm\]
Alright, I hope some of my work was of some use.

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