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Find the force of attraction for a particle outside sphere

  1. Oct 25, 2016 #1
    1. The problem statement, all variables and given/known data
    A sphere of radius R contains two spherical cavities. Each cavity has a radius of R/2 and touches both the outside surface of the sphere and its center as shown. The mass of a similar sphere without the cavities is M. Find the force of attraction on a small particle of mass m located on the x - axis a distance d > R from the center of the sphere.

    2. Relevant equations
    g=GMm/R2

    3. The attempt at a solution
    Will upload picture and solution.
     
  2. jcsd
  3. Oct 25, 2016 #2
  4. Oct 25, 2016 #3
    I am sorry that I don't quite understand your work. You defined g = GMm/R^2, g is a quantity with unit of force, but in your work, g seems to be acceleration.

    Here I have an idea:
    $$ \frac{GMm}{d^2} - 2\frac{GMm}{8r^2}cosθ = ma =F $$
    The first term is the force by a fully filled sphere and the second term is the force by two small spheres. As the radius is halved, mass is 1/8. r is the distance between the small sphere and the particle.
    After that, you can write
    $$ r^2 = d^2 + (\frac R 2)^2$$
    $$ cosθ = \frac {d}{\sqrt{(\frac R 2)^2+d^2}} $$

    You have enough information for the solution. I hope it helps
     
  5. Oct 25, 2016 #4
    Okay. I see how the mass is 1/8 when the radius becomes 1/2 by writing out the equation for density. But why can you just put an 8 in the denominator? I know it makes sense conceptually, but I mean mathematically (without introducing density into the equation)
     
  6. Oct 25, 2016 #5
    PS. the drawing should have the r vectors pointing towards the center of each small sphere from d
     
  7. Oct 25, 2016 #6
    Um.. I mean
    $$ \frac{Gm\frac M 8 }{r^2}$$
    You can directly know it is 1/8 by the argument learnt in math class. For similar objects ,
    $$ \frac {V_1}{V_2} = (\frac {l_1}{l_2})^3$$
     
  8. Oct 25, 2016 #7
    Okay then. Thank you, that helped
     
  9. Oct 25, 2016 #8
    No problem
     
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