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Homework Help: Gravitational force

  1. Feb 5, 2010 #1
    Gravitational force on mass m outside a sphere with mass M is given by [tex]F=G\frac{mM}{r^2}[/tex], where r is the distance to the center of mass. Gravity inside the sphere surface because of the mass distribution, since only the portion of the sphere mass is inside r contributes to the attraction. If the Earth's density is constant (which it definitely is not), so given the mass inside r of [tex]M'=\frac{Mr^3}{R^3} [/tex], where R is Earth radius. Gravitational force in the earth's surface (r less than R) thus becomes: [tex]F=G\frac{mMr}{R^3}[/tex]

    my question is how can we prove/shown formula [tex]M'=\frac{Mr^3}{R^3} [/tex]??
     
  2. jcsd
  3. Feb 5, 2010 #2

    Doc Al

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    Hint: What fraction of the total spherical volume does the mass < r occupy?
     
  4. Feb 5, 2010 #3
    what does mean with occupy?
     
  5. Feb 5, 2010 #4

    Doc Al

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    Occupy means 'take up', but I'll restate it differently. Since you assume uniform density, the mass is proportional to the volume. Compare the volume of a sphere of radius = r to one of radius = R.
     
  6. Feb 6, 2010 #5
    I can not really understand how to get
    [tex]M’= \frac{Mr^3 }{R^3}[/tex]
    “the mass is proportional to the volume” = [tex]M(V)=\rho \frac{4\pi r^2}{3}[/tex]
    if we write [tex]M(V)= \frac{\rho4\pi }{3}\cdot r^2[/tex]
    [tex] \frac{\rho4\pi }{3}=k[/tex]
    k=constant
    [tex]M(r)=k\cdot r^2[/tex]

    can you show me mathematical how I can get [tex]M’= \frac{Mr^3 }{R^3}[/tex]
     
  7. Feb 6, 2010 #6

    Doc Al

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    That should be:

    [tex]M(V)= \frac{\rho4\pi }{3}\cdot r^3[/tex]

    Compare the total mass MR (where radius = R) to the partial mass Mr (where radius = r).
     
  8. Feb 6, 2010 #7
    but there is no evidence that the mass is [tex]M'=\frac{Mr^3}{R^3}[/tex]
    can you show me mathematical, so I can understand
     
  9. Feb 6, 2010 #8

    Doc Al

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    What does M equal? (Use the formula for density times volume.)
    What does M' equal? (Use the formula for density times volume.)

    Then just divide M' by M and see what you get.
     
  10. Feb 6, 2010 #9


    [tex]M=\rho \frac{4\pi \cdot R^3}{3}\\ [/tex]
    [tex]M'=\rho \frac{4\pi r^3}{3}[/tex]
    [tex]\frac{M'}{M}=\frac{\rho \frac{4\pi r^3}{3}}{\rho \frac{4\pi \cdot R^3}{3}}[/tex]

    this does not give us the mass, this give is the share
     
  11. Feb 6, 2010 #10

    Doc Al

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    Finish the division--canceling things that can be canceled--and you'll get the formula you want.
     
  12. Feb 6, 2010 #11
    [tex]\frac{M'}{M}=\frac{r^3}{R^3}[/tex]
    but that's not what I want, this give me the share, not the new mass
     
  13. Feb 6, 2010 #12

    Doc Al

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    Multiply both sides by M.
     
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