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- Sketch the graph of Veff (r) when V = − kr and describe what a particle in this potential would do, depending on it’s energy E.
Solution: We have Veff (r) = (^2) mrj^22 − kr = k
( (^) j 2 2 mkr 2 −^ r
. This is minimized when r = (^) mkj^2 , and
then Veff = − k 22 jm 2. Since^12 m r˙^2 must be nonnegative, we know that a particles total energy can never be less than − k 22 jm 2. If a particles total energy is positive, then r can go to infinity. If E < 0, then the particle will be stay in between to values of r, determined by when the energy equals Veff.
- Show θ = θ 0 + arccos mr^ j −^ kj √ (^2) E m +^ k j^22
Solution:
∫ (^) (j/mr (^2) )dr √ (^2) m (E^ −^ Veff^ (r))
∫ (^) (j/mr (^2) )dr √ (^2) m (E^ +^ kr −^2 mrj^22 )
∫ (^) dw √ (^2) Em j^2 +^2 kmj^2 w^ −^ w^2
, where
w = (^1) r. Using the fact
∫ (^) dx √ax (^2) + bx + c = − √^1 −a arccos
( (^) − 2 ax − b √b (^2) − 4 ac
when a < 0, we have
−
∫ (^) dw √ (^2) Em j^2 +^2 kmj^2 w^ −^1 j w^2
= arccos
(^) √ (^42) kw (^2) m^ − 2 2 kmj^2 j^4 + 4^2 Emj^2
(^) = arccos
√mr jk 2 −^ kj j^2 +^2 mE
. So θ = θ 0 +
arccos mr^ j −^ kj √ (^2) E m +^ k j^22
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- Letting p = (^) kmj^2 and e =
1 +^2 mkEj 22 , show θ = θ 0 + arccos
( (^) p/r − 1 e
Solution: θ = θ 0 + arccos mr^ j −^ kj √ (^2) E m +^ k j^22
= θ 0 + arccos mr^ j −^ kj jk^ √^2 mkEj 22 + 1=^ θ^0 + arccos √mr^ j jk^ −^1 (^2) mkEj 22 + 1= θ 0 + arccos
( (^) p/r − 1 e
. Solving for r gives r = (^) 1 + e cos(pθ − θ 0 ).
- The equation r = (^) 1 + e cos(pθ − θ 0 ) describes an ellipse, parabola or hyperbola based on the value of e. Solution: By a rotation, we can assume θ 0 = 0. So we have p = r +er cos θ or p = √x^2 + y^2 +ex. So x^2 + y^2 = p^2 − 2 epx + x^2 , or (1 − e^2 )x^2 + 2epx + y^2 = p^2. If e = 0, we have p^2 = x^2 + y^2 , the equation of a circle. If 0 < e < 1, let k = (^1) −ep e 2. Then we have x^2 + 2kx + (^1) −y^2 e 2 = (^1) −p^2 e 2 = (x − k)^2 + (^1) −y^2 e 2 = p^2 1 − e^2 +^ k^2 , and ellipse. If e = 1, we have 2px = −y^2 + p^2 , a parabola. If e > 1, let k = (^1) −ep e 2. Then we have x^2 + 2kx + (^1) −y^2 e 2 = (^1) −p^2 e 2 = (x − k)^2 − (^) e 2 y (^) −^2 1 = p^2 1 − e^2 +^ k^2 , a hyperbola.
- How are the 3 kinds of orbits related to energy? Solution: The hyperbola corresponds to e > 1, which requires the energy to be positive. The parabola correspond to e = 1, which requires energy to be 0. The ellipse corresponds to 0 < e < 1 which corresponds which requires energy to be negative.
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