Example 5.1 Worked on the Board!

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    09-Feb-2016

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Example 5.1 Worked on the Board!. Find the gravitational potential inside & outside a spherical shell, inner radius b , outer radius a . (Like a similar electrostatic potential problem!) - PowerPoint PPT Presentation

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  • Example 5.1 Worked on the Board!Find the gravitational potential inside & outside a spherical shell, inner radius b, outer radius a. (Like a similar electrostatic potential problem!) This is an important & fundamental problem in gravitational theory! If you understand this, you understand the gravitational potential concept! (& probably the electrostatic potential concept as well!) .Could approach the spherical shell using forces directly (Prob. 5.6), but its MUCH easier with the potential!

  • Find inside & outside a spherical shell of mass M, mass density , inner radius b & outer radius a.

    = -G[(r)dv/r]. Integrate over V. The difficulty, of course, is properly setting up the integral! If properly set up, doing it is easy!

  • Recall Spherical Coordinates

  • Outline of Calculation!

  • Summary of Results M (4)(a3 - b3) outside the shell R > a, = -(GM)/R (1) The same as if M were a point mass at the origin!completely inside the shell R < b, = -2G(a2 - b2) (2) = constant, independent of position. within the shell b R a, = -4G[a2- (b3/R) - R2] (3) Also, is continuous! If R a, (1) & (3) are the same! If R b, (2) & (3) are the same!

  • These results are very important, especiallythose for R > a, = -(GM)/R This says: The potential at any point outside a spherically symmetric distribution of matter (shell or solid; a solid is composed of infinitesimally thick shells!) is independent of the size of the distribution & is the same as that for a point mass at the origin. To calculate the potential for such a distribution we can consider all mass to be concentrated at the center.

  • Also, the results are very important for

    R < b, = -2G(a2 - b2) The potential is constant anywhereinside a spherical shell. The force on a test mass m inside the shell is 0!

  • Given the results for the potential , we can compute the GRAVITATIONAL FIELD g inside, outside & within the spherical shell: g - depends on R only g is radially directed g = g er = - (d/dR)er [M (4)(a3 - b3)] outside the shell R > a, g = - (GM)/R2The same as if M were a point mass at the origin! completely inside the shell R < b, g = 0 Since = constant, independent of position.within the shell b R a, g = (4)G[(b3/R2) - R]

  • Plots of the potential & the field g inside, outside & within aspherical shell. g - g = - (d/dR)

    g = - (GM)/R2g = 0 = -(GM)/R = constant

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