2. Example 5.2 Astronomical measurements show that the orbital speed of
masses in spiral
galaxies rotating
about their centers
is approximately
constant as a
function of distance R from the galaxy center, as in the figure (for the Andromeda Galaxy). Show that this is inconsistent with the galaxy having its mass concentrated at its center & can be explained if the galaxy mass increases with distance R from the center.

3. Solution: Assume that the galaxy mass M has a spherically symmetric distribution ρ. We can solve the problem, despite the fact that the radius R might be hundreds of light years! The orbital speed v of a small mass m in orbit at R about the galaxy center is found (Phys. I!) by setting the gravitational force = (mass)  (centripetal acceleration) or: [(GMm)/)(R2)] = [m(v2)/R]
 v = (GM/R)½ (1)
or: v R-½ (2)
If (2) is true, v vs. R would
be like the dashed curve! For the
solid (expt.) curve to be consistent
with (1) requires M = M(R)  R  CONCLUSION: There must be more matter in a galaxy than is observed (“dark matter” = ~90% of mass in the universe). At the forefront of research.

4. Example 5.3 Consider a thin, uniform circular ring of radius a &
mass M. A small
mass m is placed
in the plane of the
ring. Find a
position of
equilibrium &
determine if it is
stable.

5. Use the result for a Line Distribution:
(M = ∫ρ(r)ds, Φ = - G∫[ρ(r)ds/r]
Here, ρ = const. = ρ  M = 2πaρ
The potential at the position of a
point mass m a distance r from
the ring center & a distance b from
the differential mass dM:
Φ(r) = -ρaG∫(dφ/b)
φ = angle shown in the figure, limits: (0  φ  2π)
From the figure, b2 = a2 + (r)2 – 2arcosφ
 Φ = - ρaG∫dφ[a2 + (r)2 – 2arcosφ]-½
= - ρG∫dφ[1 + (r/a)2 – 2(r/a)cosφ]-½

6. Φ = - ρG∫dφ[1 + (r/a)2 – 2(r/a)cosφ]-½ (1)
Consider positions close to the center
point (near r = 0): Expand (1) for
r << a & integrate term by term:
Φ(r)  -ρG∫dφ[1 + (r/a)cosφ
+ (r/a)2(3cos2φ -1)+….]
(For those who know & care, the
integrand in this form is a series of Legendre Polynomials!)
Φ(r)  -(MG/a)[1 + (r/a)2 +…]
The potential energy of mass m at r is:
U(r) = mΦ(r)  -m(MG/a)[1 + (r/a)2 +…]

7. U(r) = mΦ(r)  -m(MG/a)[1 + (r/a)2 +…]
The equilibrium position is given by
(dU/dr) = 0 = -m(MGr)/(2a3)
 r = 0 is an equilibrium position!
This should be obvious by symmetry!
Stability (or not) is obtained from the
2nd derivative: (d2U/dr2)0 = -m(MG)/(2a3) < 0
 r = 0 is a position of unstable equilibrium!
(“Not obvious”!)