The Kepler Problem Revisited
1. Define J=mq ×˙q. Then, using m¨q=−kˆq/q2, we have ¨q×J=k˙
ˆq
Solution: Using the vector identities given, we have ¨q×J= ¨q×(mq ×˙q) = −(kˆq/q2)×(q×˙q) =
(−k/q2) ˆq×(q×˙q) = (−k/q2)(( ˆq·˙q)q−(ˆq·q) ˙q) = −k(q·˙q)q−(q·q) ˙q
q3=k˙
ˆq
2. Using the above d
dt( ˙q×J) = k˙
ˆq.
Solution: We have d
dt( ˙q×J) = ¨q×J+ ˙q×˙
J= ¨q×J=k˙
ˆqsince ˙
J= 0.
3. Using the above ˙q×J=kˆq+x, where xis independent of time.
Solution: Since d
dt( ˙q×J−kˆq) = d
dt( ˙q×J)−k˙
ˆq= 0, we have ˙q×J−kˆq=x, where xis
independent of time.
4. Define A=x
k=˙q×J
k−ˆq. Then we have A·q=J·J
km − |q|
Solution: Since ˆq·q=|q|, we have A·q=q·˙q×J
k−|q|=J·q×˙q
k−|q|=J·mq ×˙q
mk −|q|=
J·J
km − |q|, using the identity a·(b×c) = c·(a×b).
5. If θis the angle between Aand q, we have A·q=|A||q|cos θ. Then |q|=J·J
km
1
1 + |A|cos θ.
Solution: We have J·J
km =A·q+|q|=|A||q|cos θ+|q|=|q|(1 + |A|cos θ). So
|q|=J·J
km
1
1 + |A|cos θ.
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