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.

- anonymous

- jamiebookeater

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- anonymous

@UnkleRhaukus

- anonymous

p=m/v
now i know that the change in density is going to be due to the change i mass and volume

- anonymous

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

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## More answers

- anonymous

my teacher had his second equation in a similar problem as
\[pdv=dm\]
which i can only get if i let
\[p*dv=\frac{dm}{dv}*dv\]

- anonymous

p is a linear function also

- anonymous

|dw:1358678795614:dw|

- anonymous

What subject is this?

- anonymous

it's calculus-based physics

- anonymous

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

- anonymous

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

- anonymous

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

- anonymous

|dw:1358679271874:dw|

- anonymous

\[dp \neq \frac{ dm }{ dv }\]

- anonymous

\[p=\frac{dm}{dv}\]

- anonymous

\[dp = \frac{ vdm - mdv }{ v^2 }\]

- anonymous

correct so where does \[pdv=dm\] come from

- anonymous

by any chance, is the density function given? p(x)?

- anonymous

ehh i think it was something like (-.0000007+10^4)r

- anonymous

it is (-.000005r+10^4)

- anonymous

So p(x) is (-.000005r+10^4)?

- anonymous

pdv = dm is wrong

- anonymous

p(x) given?

- anonymous

p=-.0000004r+10^4

- anonymous

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

- anonymous

or change in density

- anonymous

|dw:1358680137198:dw|

- anonymous

what my teacher wrote for a very similar problem was
\[pdv=dm\]
\[p(4\pi r^2)=dm\]

- anonymous

|dw:1358680208001:dw|

- anonymous

|dw:1358680330123:dw|

- anonymous

now is it adding 10^4 or multiplying 10^4?

- anonymous

adding

- anonymous

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

- anonymous

|dw:1358680376740:dw|

- anonymous

only difference is he has R-d at top

- anonymous

|dw:1358680521783:dw|

- anonymous

What do you mean?

- anonymous

on the integral?

- anonymous

|dw:1358680673477:dw|

- anonymous

i found this
\[M=\int pdV\] which would make sense but how does pdv=dm relate. Afterthis it makes sense
\[\int dm=\int pdV\]
\[m=pdV\]

- anonymous

forogot int sign

- anonymous

|dw:1358680956774:dw|

- anonymous

|dw:1358681105717:dw|

- anonymous

\[ p=\frac{m}{v}\]
is it because
p is essentially the change in density

- anonymous

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

- anonymous

\[p=\frac{dm}{dv}\]

- anonymous

\[pdv=dm\]

- anonymous

|dw:1358681240467:dw|

- anonymous

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.

- anonymous

or would it actually be
\[pdr=\frac{dm}{dv}\]
\[pdvdr=dm\]

- anonymous

Alright, I hope some of my work was of some use.

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