Planetary orbits around a star can be modeled with the following potentialU (r) = ar+br2 (1) (a) Show that the equilibrium position for this potential is equal tore = 2b=a. (b) Use the Taylor expansion on the potential about the equilibrium position to show thatthe / spring" constant of small oscillations around this equilibrium position isa4=8b3

a) r eq = - a / (2b)

b) k = a/r eq = - 2b

Explanation:

since

U (r) = ar + br²

a) the equilibrium position dU/dr = 0

U (r) = a + 2br = 0 → r eq = - a/2b

b) the Taylor expansion around the equilibrium position is

U (r) = U (r eq) + ∑ Un (r eq) (r - r eq) ^n / n!

, where Un (a) is the nth derivative of U respect with r, evaluated in a

Since the 3rd and higher order derivatives are = 0, we can expand until the second derivative

U (r) = U (r eq) + dU/dr (r eq) (r - r eq) + d²U/dr² (r eq) (r - r eq) ² / 2

since dU/dr (r eq) = 0

U (r) = U (r eq) + d²U/dr² (r eq) (r - r eq) ² / 2

comparing with an energy balance of a spring around its equilibrium position

U (r) - U (r eq) = 1/2 k (r-r eq) ² → U (r) = U (r eq) + 1/2 k (r-r eq) ²

therefore we can conclude

k = d²U/dr² (r eq) = - 2b, and since r eq = - a/2b → - 2b=a/r eq

thus

k = a/r eq