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Outkast3r09

  • 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.

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  1. Outkast3r09
    • 3 years ago
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    @UnkleRhaukus

  2. Outkast3r09
    • 3 years ago
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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

  3. Outkast3r09
    • 3 years ago
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    so would it be \[dp=\frac{dm}{dv}\]

  4. Outkast3r09
    • 3 years ago
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    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
    • 3 years ago
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    p is a linear function also

  6. Outkast3r09
    • 3 years ago
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    |dw:1358678795614:dw|

  7. zordoloom
    • 3 years ago
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    What subject is this?

  8. Outkast3r09
    • 3 years ago
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    it's calculus-based physics

  9. Outkast3r09
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    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
    • 3 years ago
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    |dw:1358679271874:dw|

  13. bhaskarbabu
    • 3 years ago
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    \[dp \neq \frac{ dm }{ dv }\]

  14. Outkast3r09
    • 3 years ago
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    \[p=\frac{dm}{dv}\]

  15. bhaskarbabu
    • 3 years ago
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    \[dp = \frac{ vdm - mdv }{ v^2 }\]

  16. Outkast3r09
    • 3 years ago
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    correct so where does \[pdv=dm\] come from

  17. zordoloom
    • 3 years ago
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    by any chance, is the density function given? p(x)?

  18. Outkast3r09
    • 3 years ago
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    ehh i think it was something like (-.0000007+10^4)r

  19. Outkast3r09
    • 3 years ago
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    it is (-.000005r+10^4)

  20. zordoloom
    • 3 years ago
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    So p(x) is (-.000005r+10^4)?

  21. bhaskarbabu
    • 3 years ago
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    pdv = dm is wrong

  22. bhaskarbabu
    • 3 years ago
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    p(x) given?

  23. Outkast3r09
    • 3 years ago
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    p=-.0000004r+10^4

  24. Outkast3r09
    • 3 years ago
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    i need somehow to relate the change of mass nad volume to the change in radius

  25. Outkast3r09
    • 3 years ago
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    or change in density

  26. zordoloom
    • 3 years ago
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    |dw:1358680137198:dw|

  27. Outkast3r09
    • 3 years ago
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    what my teacher wrote for a very similar problem was \[pdv=dm\] \[p(4\pi r^2)=dm\]

  28. zordoloom
    • 3 years ago
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    |dw:1358680208001:dw|

  29. zordoloom
    • 3 years ago
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    |dw:1358680330123:dw|

  30. zordoloom
    • 3 years ago
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    now is it adding 10^4 or multiplying 10^4?

  31. Outkast3r09
    • 3 years ago
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    adding

  32. Outkast3r09
    • 3 years ago
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    this is what he hsa i just need help understanding where the dm is coming from

  33. zordoloom
    • 3 years ago
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    |dw:1358680376740:dw|

  34. Outkast3r09
    • 3 years ago
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    only difference is he has R-d at top

  35. zordoloom
    • 3 years ago
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    |dw:1358680521783:dw|

  36. zordoloom
    • 3 years ago
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    What do you mean?

  37. zordoloom
    • 3 years ago
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    on the integral?

  38. zordoloom
    • 3 years ago
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    |dw:1358680673477:dw|

  39. Outkast3r09
    • 3 years ago
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    i found this \[M=\int pdV\] which would make sense but how does pdv=dm relate. Afterthis it makes sense \[\int dm=\int pdV\] \[m=pdV\]

  40. Outkast3r09
    • 3 years ago
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    forogot int sign

  41. zordoloom
    • 3 years ago
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    |dw:1358680956774:dw|

  42. zordoloom
    • 3 years ago
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    |dw:1358681105717:dw|

  43. Outkast3r09
    • 3 years ago
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    \[ p=\frac{m}{v}\] is it because p is essentially the change in density

  44. Outkast3r09
    • 3 years ago
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    so \[p=\Delta p=\frac{\Delta m}{\Delta v}\]

  45. Outkast3r09
    • 3 years ago
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    \[p=\frac{dm}{dv}\]

  46. Outkast3r09
    • 3 years ago
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    \[pdv=dm\]

  47. zordoloom
    • 3 years ago
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    |dw:1358681240467:dw|

  48. zordoloom
    • 3 years ago
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    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
    • 3 years ago
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    or would it actually be \[pdr=\frac{dm}{dv}\] \[pdvdr=dm\]

  50. zordoloom
    • 3 years ago
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    Alright, I hope some of my work was of some use.

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