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Gravitational field problem

  1. Dec 30, 2013 #1
    1. The problem statement, all variables and given/known data
    A thin hemispherical shell of mass M and radius R is placed as shown in figure. The magnitude of gravitational field at P due to the hemispherical shell is ##I_0##. The magnitude of gravitational field at Q due to thin hemispherical shell is given by

    A)##I_0/2##

    B)##I_0##

    C)##\frac{2GM}{9R^2}-I_0##

    D)##\frac{2GM}{9R^2}+I_0##


    2. Relevant equations



    3. The attempt at a solution
    I tried the problem using spherical coordinates and ended up with some messy integrals. Since this is an exam problem, I wonder if I really need to solve those integrals as it would take a lot of time. (I solved the integrals using Wolfram Alpha and the result was not nice so I immediately dropped the approach.) I believe there is a shorter way to solve this.

    Any help is appreciated. Thanks!
     

    Attached Files:

  2. jcsd
  3. Dec 30, 2013 #2

    Dick

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    There is a much faster way. You can picture the field due to the hemisphere as the sum of the field from a whole sphere and the field from an imaginary hemisphere of mass -M covering the lower half of the sphere.
     
  4. Dec 30, 2013 #3
    Hi Dick! :)

    I consider a sphere of radius R and mass 2M.

    The field at P is given by:

    $$\frac{2GM}{9R^2}+E_{-M}=I_0=E_M$$

    where ##E_{-M}## represents field at P due to the imaginary hemisphere of mass -M and ##E_M## represents the field at P due to hemisphere of mass M.

    The field at Q is given by:

    $$\frac{2GM}{9R^2}+E'_{-M}$$

    where ##E'_{-M}## is the field at Q due to imaginary hemisphere of mass -M.

    Since ##E'_{-M}=-E_M=-I_0##, the field at Q is given by:

    $$\frac{2GM}{9R^2}-I_0$$

    Is this correct?
     
  5. Dec 30, 2013 #4

    Dick

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    Hi Pranav-Arora! Yes, that's correct. This sort of a method is called 'using superposition'. For sort of obvious reasons.
     
  6. Dec 30, 2013 #5
    Yes, I have heard of this method, thanks a lot Dick! :smile:
     
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