Outkast3r09
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.
Delete
Share
This Question is Closed
Outkast3r09
Best Response
You've already chosen the best response.
0
@UnkleRhaukus
Outkast3r09
Best Response
You'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
Outkast3r09
Best Response
You've already chosen the best response.
0
so would it be
\[dp=\frac{dm}{dv}\]
Outkast3r09
Best Response
You'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
\[p*dv=\frac{dm}{dv}*dv\]
Outkast3r09
Best Response
You've already chosen the best response.
0
p is a linear function also
Outkast3r09
Best Response
You've already chosen the best response.
0
|dw:1358678795614:dw|
zordoloom
Best Response
You've already chosen the best response.
0
What subject is this?
Outkast3r09
Best Response
You've already chosen the best response.
0
it's calculus-based physics
Outkast3r09
Best Response
You'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
Outkast3r09
Best Response
You'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
Outkast3r09
Best Response
You've already chosen the best response.
0
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
Outkast3r09
Best Response
You've already chosen the best response.
0
|dw:1358679271874:dw|
bhaskarbabu
Best Response
You've already chosen the best response.
0
\[dp \neq \frac{ dm }{ dv }\]
Outkast3r09
Best Response
You've already chosen the best response.
0
\[p=\frac{dm}{dv}\]
bhaskarbabu
Best Response
You've already chosen the best response.
0
\[dp = \frac{ vdm - mdv }{ v^2 }\]
Outkast3r09
Best Response
You've already chosen the best response.
0
correct so where does \[pdv=dm\] come from
zordoloom
Best Response
You've already chosen the best response.
0
by any chance, is the density function given? p(x)?
Outkast3r09
Best Response
You've already chosen the best response.
0
ehh i think it was something like (-.0000007+10^4)r
Outkast3r09
Best Response
You've already chosen the best response.
0
it is (-.000005r+10^4)
zordoloom
Best Response
You've already chosen the best response.
0
So p(x) is (-.000005r+10^4)?
bhaskarbabu
Best Response
You've already chosen the best response.
0
pdv = dm is wrong
bhaskarbabu
Best Response
You've already chosen the best response.
0
p(x) given?
Outkast3r09
Best Response
You've already chosen the best response.
0
p=-.0000004r+10^4
Outkast3r09
Best Response
You've already chosen the best response.
0
i need somehow to relate the change of mass nad volume to the change in radius
Outkast3r09
Best Response
You've already chosen the best response.
0
or change in density
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358680137198:dw|
Outkast3r09
Best Response
You've already chosen the best response.
0
what my teacher wrote for a very similar problem was
\[pdv=dm\]
\[p(4\pi r^2)=dm\]
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358680208001:dw|
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358680330123:dw|
zordoloom
Best Response
You've already chosen the best response.
0
now is it adding 10^4 or multiplying 10^4?
Outkast3r09
Best Response
You've already chosen the best response.
0
adding
Outkast3r09
Best Response
You've already chosen the best response.
0
this is what he hsa i just need help understanding where the dm is coming from
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358680376740:dw|
Outkast3r09
Best Response
You've already chosen the best response.
0
only difference is he has R-d at top
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358680521783:dw|
zordoloom
Best Response
You've already chosen the best response.
0
What do you mean?
zordoloom
Best Response
You've already chosen the best response.
0
on the integral?
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358680673477:dw|
Outkast3r09
Best Response
You'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
\[\int dm=\int pdV\]
\[m=pdV\]
Outkast3r09
Best Response
You've already chosen the best response.
0
forogot int sign
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358680956774:dw|
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358681105717:dw|
Outkast3r09
Best Response
You've already chosen the best response.
0
\[ p=\frac{m}{v}\]
is it because
p is essentially the change in density
Outkast3r09
Best Response
You've already chosen the best response.
0
so \[p=\Delta p=\frac{\Delta m}{\Delta v}\]
Outkast3r09
Best Response
You've already chosen the best response.
0
\[p=\frac{dm}{dv}\]
Outkast3r09
Best Response
You've already chosen the best response.
0
\[pdv=dm\]
zordoloom
Best Response
You've already chosen the best response.
0
|dw:1358681240467:dw|
zordoloom
Best Response
You'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.
Outkast3r09
Best Response
You've already chosen the best response.
0
or would it actually be
\[pdr=\frac{dm}{dv}\]
\[pdvdr=dm\]
zordoloom
Best Response
You've already chosen the best response.
0
Alright, I hope some of my work was of some use.