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?

Question

anonymous · Answer

L=mvr

anonymous · Answer

v=2pi*R/(365*24*3600)

anonymous · Answer

thats what i thought...but unfortunately my book says it 1/2.
i dont know how!

anonymous · Answer

$$R^{2}{} (2 \pi)\over(365*24*3600)*$$

anonymous · Answer

the value of n should be 2 ryt?

anonymous · Answer

that's what I think

anonymous · Answer

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 · Answer

When I was taking physics I found this site very useful
www.physicsfourm.com

anonymous · Answer

did you miss an "s" by any chance?

anonymous · Answer

yes http://www.physicsforums.com/

anonymous · Answer

ok..

anonymous · Answer

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 · Answer

hey how did you get this,  v^2 = GM/r^3 ?? m totally clueless.

anonymous · Answer

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 · Answer

i have a question,
why wouldnt replacing v by $$\omega r$$ work??

anonymous · Answer

It does work, you just need to do an extra step to convert a= v^2/r to w^2r

anonymous · Answer

can you work it out. i am not able to comprehend. :-(