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r(t) is a position vector. v(t) is r'(t) or the velocity vector, and a(t) is the acceleration vector. u(t) is a unit r(t) vector. Also, F(t)=ma(t)=GMmu(t)/r^2
h(t)= r(t) x v(t)=r^2 u(t) x u'(t)
show that (v(t)xh(t))'=GMu'(t)
somehow I keep getting GMu(t) or (a(t)xh(t)) instead...
 one year ago
 one year ago
r(t) is a position vector. v(t) is r'(t) or the velocity vector, and a(t) is the acceleration vector. u(t) is a unit r(t) vector. Also, F(t)=ma(t)=GMmu(t)/r^2 h(t)= r(t) x v(t)=r^2 u(t) x u'(t) show that (v(t)xh(t))'=GMu'(t) somehow I keep getting GMu(t) or (a(t)xh(t)) instead...
 one year ago
 one year ago

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MuH4hABest ResponseYou've already chosen the best response.1
By you getting a(t)xh(t), you mean you keep getting\[a \times h =  G\,M\, \dot u\] ???
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
wait sorry, it's reversed. the question wants (v(t)xh(t))'=GMu˙but i keep getting (a(t)xh(t))=GMu' instead
 one year ago

MuH4hABest ResponseYou've already chosen the best response.1
Well, then you are fine, cause\[(v \times h)' = v' \times h = a \times h = G\, M\, \dot u\]That's because you showed earlier, that h is constant (second law of Kepler) and thus the derivative of it is 0.
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
so i guess the book's wrong...
 one year ago

MuH4hABest ResponseYou've already chosen the best response.1
Why would it be wrong? What does the book say? o.O
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
show that (v(t)xh(t))'=GMu(t) this is the original question... lol
 one year ago

MuH4hABest ResponseYou've already chosen the best response.1
Well .. that is what you showed, isn't it? (sry, I don't see your problem ... please elaborate)
 one year ago

ShadowysBest ResponseYou've already chosen the best response.0
I'm trying to seek confirmation that the book is missing a particular ' sign. :P it seems so. lol thanks.
 one year ago
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