anonymous
  • anonymous
hey guys a question in physics, the angular momentum of the earth revolving around the sun is proportional to R^n where R is the distance between the earth and the sun. the value of n is?
Mathematics
jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
L=mvr
anonymous
  • anonymous
v=2pi*R/(365*24*3600)
anonymous
  • anonymous
thats what i thought...but unfortunately my book says it 1/2. i dont know how!

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anonymous
  • anonymous
\[R^{2}{} (2 \pi)\over(365*24*3600)*\]
anonymous
  • anonymous
the value of n should be 2 ryt?
anonymous
  • anonymous
that's what I think
anonymous
  • anonymous
http://www.livephysics.com/problems-and-answers/classical-mechanics/find-earth-angular-momentum.html here its solved the same way, so i think its plausible to go with 2
anonymous
  • anonymous
When I was taking physics I found this site very useful www.physicsfourm.com
anonymous
  • anonymous
did you miss an "s" by any chance?
anonymous
  • anonymous
yes http://www.physicsforums.com/
anonymous
  • anonymous
ok..
anonymous
  • anonymous
1/2 is actually correct. I'll prove it for you: L = m (r x v), which means that L = mrv (because in this case, our angle is 90) Now, we know that F (gravity) = GMm/r^2 and that F = ma assuming uniform circular motion, a = v^2/r, which means that v^2 = GM/r^3 Plugging this back into our angular momentum equation, we get \[L = mr^2*\sqrt{GM}*r^{-1.5} = m*\sqrt{GM} * r ^.5\]
anonymous
  • anonymous
hey how did you get this, v^2 = GM/r^3 ?? m totally clueless.
anonymous
  • anonymous
mv^2/r=GM/r^2 =>v^2=GM/r plugn ds in ang. momntm eq. - L=mvr=mr(GM/r)^(1/2)=m(GMr)^(1/2) so n should be 1/2
anonymous
  • anonymous
i have a question, why wouldnt replacing v by \[\omega r\] work??
anonymous
  • anonymous
It does work, you just need to do an extra step to convert a= v^2/r to w^2r
anonymous
  • anonymous
can you work it out. i am not able to comprehend. :-(

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