anonymous
  • anonymous
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?
OCW Scholar - Physics I: Classical Mechanics
chestercat
  • chestercat
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anonymous
  • anonymous
You can treat the earth as a point particle, so by definition I = mr^2... If you want to do it by parallel axis theorem, you get \[I _{\parallel} = I _{cm} + mR ^{2} = 2/5 mr _{earth}^{2} + mR ^{2}\] and you can see that the term involving the radius of the earth is insignificant... so the answer is n = 2 in either case
anonymous
  • anonymous
my book says its n = 1/2 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, 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 is there something wrong with dis?
anonymous
  • anonymous
no - nothing wrong with it - I was giving you moment of inertia, not angular momentum. I should read your question more carefully!!!

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anonymous
  • anonymous
ok here is another way i worked it out. we know, L=mvr and v=wr right? if i plug in the value of v in the momentum equation, L=mwr^2 here n=2 so which one is correct, this or the one i did before? and also i want to know what is wrong with the other.

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