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!)
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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!
Summary of ResultsM(4π)ρ(a3 - b3)
outside the shell
R > a, Φ = -(GM)/R (1)
The same as ifM 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)
If R a, (1) & (3) are the same! If R b, (2) & (3) are the same!
those forR > 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.
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!
g = g er = - (dΦ/dR)er [M(4π)ρ(a3 - b3)]
outside the shellR > a, g = - (GM)/R2The same as ifM 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]
Φ & the field g
inside, outside &
g - Φ
g = - (dΦ/dR)
Φ = constant
Φ = -(GM)/R
g = - (GM)/R2
g = 0