A community for students.
Here's the question you clicked on:
 0 viewing
anonymous
 3 years 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...
anonymous
 3 years 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...

This Question is Closed

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0By you getting a(t)xh(t), you mean you keep getting\[a \times h =  G\,M\, \dot u\] ???

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

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well, 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.

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0so i guess the book's wrong...

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Why would it be wrong? What does the book say? o.O

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0show that (v(t)xh(t))'=GMu(t) this is the original question... lol

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0Well .. that is what you showed, isn't it? (sry, I don't see your problem ... please elaborate)

anonymous
 3 years ago
Best ResponseYou've already chosen the best response.0I'm trying to seek confirmation that the book is missing a particular ' sign. :P it seems so. lol thanks.
Ask your own question
Sign UpFind more explanations on OpenStudy
Your question is ready. Sign up for free to start getting answers.
spraguer
(Moderator)
5
→ View Detailed Profile
is replying to Can someone tell me what button the professor is hitting...
23
 Teamwork 19 Teammate
 Problem Solving 19 Hero
 Engagement 19 Mad Hatter
 You have blocked this person.
 ✔ You're a fan Checking fan status...
Thanks for being so helpful in mathematics. If you are getting quality help, make sure you spread the word about OpenStudy.