Here's the question you clicked on:
Shadowys
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...
By you getting a(t)xh(t), you mean you keep getting\[a \times h = - G\,M\, \dot u\] ???
wait sorry, it's reversed. the question wants (v(t)xh(t))'=GMu˙but i keep getting (a(t)xh(t))=-GMu' instead
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.
so i guess the book's wrong...
Why would it be wrong? What does the book say? o.O
show that (v(t)xh(t))'=GMu(t) this is the original question... lol
Well .. that is what you showed, isn't it? (sry, I don't see your problem ... please elaborate)
I'm trying to seek confirmation that the book is missing a particular ' sign. :P it seems so. lol thanks.