## Outkast3r09 Group Title 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. one year ago one year ago

1. Outkast3r09 Group Title

@UnkleRhaukus

2. Outkast3r09 Group Title

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

3. Outkast3r09 Group Title

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

4. Outkast3r09 Group Title

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

5. Outkast3r09 Group Title

p is a linear function also

6. Outkast3r09 Group Title

|dw:1358678795614:dw|

7. zordoloom Group Title

What subject is this?

8. Outkast3r09 Group Title

it's calculus-based physics

9. Outkast3r09 Group Title

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

10. Outkast3r09 Group Title

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

11. Outkast3r09 Group Title

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

12. Outkast3r09 Group Title

|dw:1358679271874:dw|

$dp \neq \frac{ dm }{ dv }$

14. Outkast3r09 Group Title

$p=\frac{dm}{dv}$

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

16. Outkast3r09 Group Title

correct so where does $pdv=dm$ come from

17. zordoloom Group Title

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

18. Outkast3r09 Group Title

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

19. Outkast3r09 Group Title

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

20. zordoloom Group Title

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

pdv = dm is wrong

p(x) given?

23. Outkast3r09 Group Title

p=-.0000004r+10^4

24. Outkast3r09 Group Title

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

25. Outkast3r09 Group Title

or change in density

26. zordoloom Group Title

|dw:1358680137198:dw|

27. Outkast3r09 Group Title

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

28. zordoloom Group Title

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29. zordoloom Group Title

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30. zordoloom Group Title

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

31. Outkast3r09 Group Title

32. Outkast3r09 Group Title

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

33. zordoloom Group Title

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34. Outkast3r09 Group Title

only difference is he has R-d at top

35. zordoloom Group Title

|dw:1358680521783:dw|

36. zordoloom Group Title

What do you mean?

37. zordoloom Group Title

on the integral?

38. zordoloom Group Title

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39. Outkast3r09 Group Title

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

40. Outkast3r09 Group Title

forogot int sign

41. zordoloom Group Title

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42. zordoloom Group Title

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43. Outkast3r09 Group Title

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

44. Outkast3r09 Group Title

so $p=\Delta p=\frac{\Delta m}{\Delta v}$

45. Outkast3r09 Group Title

$p=\frac{dm}{dv}$

46. Outkast3r09 Group Title

$pdv=dm$

47. zordoloom Group Title

|dw:1358681240467:dw|

48. zordoloom Group Title

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.

49. Outkast3r09 Group Title

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

50. zordoloom Group Title

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