Question

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

Answer 1

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

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

Step-by-step 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