2. 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!

3. 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!

4. Recall Spherical Coordinates

5. Outline of Calculation!

7. 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!

8.  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.

9. 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!

10. 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]

11. 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