# Problem GPS 3.13 If the center of force is located at the origin of an

## Transcription

Problem GPS 3.13 If the center of force is located at the origin of an

Goldstein Problem 3-6 Problem GPS 3.13 If the center of force is located at the origin of an x-y coordinate system, then the orbit is a circle that passes through the origin as, for example, drawn below. This circle has radius a, and its origin is at (a, 0). The equation that describes it is (x!a)2 + y2 = a2 . In terms of polar coordinates measured from the center of force, we have x = r cosθ and y = r sinθ . After substituting for x and y in the equation of the circle, we obtain the representation of the orbit in polar coordinates as r = 2a cosθ . Note that a full orbit is traversed as θ varies from !π/2 to π/2 instead of the usual 0 to 2π. Problem Goldstein 3-6 (a) Start with the diﬀerential equation of the orbit in terms of the variable u = 1/r, l2 2 u m µ ¶ 1 d2 u 2 + u = −f ( u ) , dθ (1) where l is the angular momentum, m is the mass of the particle, and u = (2a cos θ)−1 . (2) tan θ du = , dθ 2a cos θ (3) · ¸ 2 tan2 θ 1 d2 u 1 + = . 2a cos θ cos θ dθ2 (4) The required derivatives are and After we substitute Eqs.(2) and (4) into (1), we obtain l2 8a2 l2 5 l2 1 1 1 (1 + tan2 θ) = − u2 = − u . f( ) = − u2 u m a cos θ m a cos3 θ m (5) Thus, in terms of r, the force is k , r5 (6) 8a2 l2 . m (7) f(r) = − where the force constant k is k= (b) The potential corresponding to Eq.(6) is V (r) = − k . 4r4 (8) Thus, the total energy of this orbit is E= ·2 k m ·2 (r + r 2 θ ) − 4 . 2 4r (9) With r = 2a cos θ , 1 (10) we find · · r = −2aθ sin θ . (11) After substituting Eqs.(10) and (11) into Eq.(9), we find ·2 E = 2ma2 θ − k , 4r4 (12) but from the conservation of angular momentum we also know that · l = mr2 θ . (13) Thus, Eq.(12) simplifies to E= 1 4r4 µ 8a2 l2 −k m ¶ =0, (14) which is seen to equal zero after substituting for k using Eq.(7). (c) The period of the motion may be found from the conservation of angular momentum, Eq.(13), in the form · l dA r2 θ = = , 2 2m dt (15) where A is the area of the orbit, in this case, πa2 . Integrating Eq.(15) over a full period τ gives τl = πa2 , 2m (16) τ = 2πma2 /l . (17) and solving for τ gives (d) Start with the definitions of x and y in polar coordinates, to obtain · · · · · · x = r cos θ − rθ sin θ , (18) and y = r sin θ + rθ cos θ . · (19) · Then, substitute for r and θ, using Eqs.(11) and (13) to find · x=− l tan θ , ma 2 (20) and · y= l (1 − tan2 θ) . 2ma (21) We can find the speed either from ·2 ·2 v = (x + y )1/2 , (22) or, equivalently from ·2 ·2 · v = (r + r2 θ )1/2 = 2aθ. · (23) · · After substituting for x and y or for θ, we obtain v= l sec2 θ . 2ma (24) At the center of force θ equals either π/2 or −π/2, for which sec θ and tan θ are · · infinite. Thus, it is clear that x, y, and v are also infinite at this point. 3