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Homework Help: Gravitational force between a particle and sphere

  1. Apr 28, 2005 #1
    the question is http://home.earthlink.net/~urban-xrisis/phy002.jpg [Broken]

    I thought that...

    is this correct?
    Last edited by a moderator: May 2, 2017
  2. jcsd
  3. Apr 28, 2005 #2
    The only correct one is c.

    For a and b, note that the mass is inside the shell, consider the acceleration from the mass ofthe shell closest to it in comparison to the mass farthest of it. Is there a net effect?
    Last edited: Apr 28, 2005
  4. Apr 28, 2005 #3
    when the mass is in the middle of the shell, it has a net force of zero, but I dont know how to calculate the force when it is not in the center of the shell
  5. Apr 28, 2005 #4
    It's very similar to the reasoning behind the gravitational effects of a mass inside a solid sphere. Tell me what the gravitational effect for a mass inside a sphere is, and think of the reasoning why it is so. Take a guess as to how this applies for a shell.
  6. Apr 28, 2005 #5
    for gravitational effect inside a shere...


    where r<R

    I'm not sure how this relates to a shell
  7. Apr 28, 2005 #6
    Ugh, I had this whole thing typed out and something went wrong and lost it.

    Anyway, for a particle (at radius [itex] r [/itex]) inside a solid sphere of radius [itex] R [/itex], the net force of gravitaiton on the particle results only from the mass contained inside the sphere with radius [itex] r [/itex].
  8. Apr 28, 2005 #7
    what about the gravity inside a shell? it's not the same as a sphere is it?
  9. Apr 28, 2005 #8
    Same principle applies, except theres no mass in the smaller sphere
  10. Apr 28, 2005 #9
  11. Apr 28, 2005 #10
    Reconsider what I said about a particle inside a hollow shell
  12. Apr 28, 2005 #11
    And for B, check your radius, R_1 + b doesnt make sense.
  13. Apr 28, 2005 #12
    are you saying it should be:
  14. Apr 28, 2005 #13
    Yeah, do you understand why?
  15. Apr 29, 2005 #14
    but the formula: [tex]F=G\frac{mMr}{R^3}[/tex]

    it doesnt include the mass of the inner sphere...so it cant equal zero

    m- is the point mass
    M- mass of sphere/shell
    R- radius of sphere/shell
    r- little radius

    also, what about the force pulling it towards the shell? as in since it's in the shell, would the shell exert a force towards the shell's surface? hence the negative force? (that's why I subtracted)
  16. Apr 29, 2005 #15
    The force pulling it towards the shell cancels in the horizontal direction due to symmetry, in the vertical direction, the forces cancel since theres a certain amount of mass on top of the particle at a small radius away, but theres also alot more mass at a farther radius. These two end up cancelling also, so everything cancels.

    That formula is for a uniform sphere. The shell isnt a uniform sphere.

    It is also analogous to

    [tex] F = G\frac{mM_{contained}}{R^2} [/tex]
  17. Apr 29, 2005 #16
    [itex]M_{contained}[/itex] being the sphere that is below the point mass? which in my case would = 0?
  18. Apr 29, 2005 #17
    Yes, with respect to the hollow shell, it contains no mass inside the sphere with radius b, therefore exerts no net force.
  19. Apr 29, 2005 #18
    Mass is proportional to radius. So M2/radius 2=Mcountained/b. So the force that the big mass exerts is

    if M=Particle mass
    M2=Big sphere mass


    That is why you get this equation.
    Last edited: Apr 29, 2005
  20. Apr 30, 2005 #19
    The big sphere density is:

    [tex]\frac{3M_{2}} {4 r^3 {\pi}}[/tex]

    and since the volume of the sphere with radius b is

    [tex]\frac{4 b^3 {\pi}} {3}[/tex]

    So the mass is

    [tex]M_{countained} = \frac{M_{2}b^3} {r^3}[/tex]


    [tex]a = \frac{M_{countained}G} {b^2}[/tex]

    [tex]a = \frac{G M_{2} b^3} {r^3 b^2}[/tex]


    [tex]a= \frac{G M_{2} b} {r^3}[/tex]
    Last edited: Apr 30, 2005
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