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The gravitational field g due to a point

  1. Jul 16, 2013 #1
    The gravitational field g due to a point mass M may be obtained by analogy with the electric field by writing an expression for the gravitational force on a test mass, and dividing by the magnitude of the test mass, m. Show that Gauss' law for the gravitational field reads:

    phi = oint g*dA=-4*pi*GM

    where G is the gravitational constant.


    Use this result to calculate the gravitational acceleration g at a distance of R/2 from the center of a planet of radius R = 6.15 x 1006 m and M = 4.25 x 1024 kg.

    -------------------------------------------

    Hi, I got the final equation to be g= G M / r^2. My final result was -29.9 m/s^2. But thats not correct. I dont see what i did wrong. Anyone has any ideas?
     
  2. jcsd
  3. Jul 16, 2013 #2
    That equation is correct outside the planet. But you need the acceleration at R/2 from the center.
     
  4. Jul 16, 2013 #3
    does that mean i would have to take a fraction of the mass, because of the radius?
     
  5. Jul 16, 2013 #4
    It probably does, but then you need to know the distribution of mass inside the planet.
     
  6. Jul 16, 2013 #5
    I tried to divide the mass by 2, and the answer is still wrong. Do you know any way to solve this problem T-T
     
  7. Jul 16, 2013 #6
    Why would you divide the mass by 2, and not by 123, for example? How did you use Gauss's law to obtain the equation?
     
  8. Jul 16, 2013 #7
    I got g * Integral of dA = -4*pi G*M

    g * 4 * pi * r^2= -4* pi* G*M

    g= - GM/ r^2

    Maybe the integral of dA is something else?
     
    Last edited: Jul 16, 2013
  9. Jul 16, 2013 #8
    First of all, what is ## \int dA ## if you are asked for ##g## at ##R/2##?

    Second, what is ##M## in this case?
     
  10. Jul 16, 2013 #9
    ∫ dA is your gaussian surface. Would that be 4/3 pi r^3 ? Since our radius is shortened. M is the Mass of the whole planet.
     
  11. Jul 16, 2013 #10
    And what is the surface here?

    Even including that outside the surface?
     
  12. Jul 16, 2013 #11
    So the gaussian surface is 4 pi r^2 still, but the mass is a ratio between V and m?
     
    Last edited: Jul 16, 2013
  13. Jul 16, 2013 #12
    You did not answer the question. Explain the shape of the surface, and then what its area is.

    Assuming V is volume and m is mass, how can mass be a ratio between volume and mass? That is dimensionally impossible.
     
  14. Jul 16, 2013 #13
    I think its m = M ( r^3/ R^3)
     
  15. Jul 16, 2013 #14
    R is the radius of the planet, and M is its mass. What are r and m?
     
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