## anonymous 5 years ago 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?

1. anonymous

L=mvr

2. anonymous

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

3. anonymous

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

4. anonymous

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

5. anonymous

the value of n should be 2 ryt?

6. anonymous

that's what I think

7. 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

8. anonymous

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

9. anonymous

did you miss an "s" by any chance?

10. anonymous
11. anonymous

ok..

12. 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 $L = mr^2*\sqrt{GM}*r^{-1.5} = m*\sqrt{GM} * r ^.5$

13. anonymous

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

14. 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

15. anonymous

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

16. anonymous

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

17. anonymous

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