## anonymous 3 years ago 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.

1. anonymous

@UnkleRhaukus

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

p is a linear function also

6. anonymous

|dw:1358678795614:dw|

7. anonymous

What subject is this?

8. anonymous

it's calculus-based physics

9. 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

10. 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

11. 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

12. anonymous

|dw:1358679271874:dw|

13. anonymous

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

14. anonymous

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

15. anonymous

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

16. anonymous

correct so where does $pdv=dm$ come from

17. anonymous

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

18. anonymous

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

19. anonymous

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

20. anonymous

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

21. anonymous

pdv = dm is wrong

22. anonymous

p(x) given?

23. anonymous

p=-.0000004r+10^4

24. anonymous

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

25. anonymous

or change in density

26. anonymous

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27. anonymous

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

28. anonymous

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29. anonymous

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30. anonymous

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

31. anonymous

32. anonymous

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

33. anonymous

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34. anonymous

only difference is he has R-d at top

35. anonymous

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36. anonymous

What do you mean?

37. anonymous

on the integral?

38. anonymous

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39. anonymous

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

40. anonymous

forogot int sign

41. anonymous

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42. anonymous

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43. anonymous

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

44. anonymous

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

45. anonymous

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

46. anonymous

$pdv=dm$

47. anonymous

|dw:1358681240467:dw|

48. 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.

49. anonymous

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

50. anonymous

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