Example 5.1 Worked on the Board! Find the gravitational potential Φ 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 Φ inside & outside a spherical shell of mass M, mass density ρ, inner radius b & outer radius a. Φ = -G∫[ρ(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  (4π)ρ(a3 - b3) outside the shell R > a, Φ = -(GM)/R (1) The same as if M were a point mass at the origin! completely inside the shell R < b, Φ = -2πρG(a2 - b2) (2) Φ = constant, independent of position. within the shell b  R  a, Φ = -4πρG[a2- (b3/R) - R2] (3) Also, Φ is continuous!  If R  a, (1) & (3) are the same! If R  b, (2) & (3) are the same!

 This says: The potential at any point outside These results are very important, especially those for R > a, Φ = -(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, Φ = -2πρG(a2 - b2)  The potential is constant anywhere inside a spherical shell.  The force on a test mass m inside the shell is 0!

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

Plots of the potential Φ & the field g inside, outside & within a spherical shell. g  - Φ g = - (dΦ/dR)  Φ = constant Φ = -(GM)/R  g = - (GM)/R2   g = 0