I'm not that familiar with stellar formation, so you might be right. According to this website:
http://www.josleys.com/show_gallery.php?galid=313
the earth is malleable because of its liquid core and tectonic plates on the surface (quite interestingly, according to that website, if the...
Thanks for your help. I appreciate it. This is not my area either, but I teach lab courses to pay for my tuition, and I wanted to give my students examples of some good applications of centripetal acceleration and I thought what could be a more grand example of centripetal acceleration than the...
Actually, according to Wikipedia:
http://en.wikipedia.org/wiki/Clairaut's_theorem
the gravity is modified by:
g[1+(\frac{5m}{2}-f)\sin^2 \varphi]
where m is the ratio of the centrifugal force to gravity at the equator (which should be really small), and f is proportional to the difference...
Thanks. I appreciate it. I wish I could just slip in a factor of 2 on
m\omega^2(r \cos \varphi) \sin\varphi = -mg \frac{1}{2}\frac{1}{r}\frac{dr}{d\varphi}
and say that the 2 comes from considering mass spread over the entire ellipsoid. That would produce the right answer. But...
If you can get the right answer, let us know! I've been trying for hours and I've given up. I teach a lab course and I'm trying to give my students some information on centripetal acceleration, but I can't figure out how to calculate the bulge - like the original poster I get half the value.
I thought the formula should be:
m\omega^2(r \cos \varphi) \sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}
but maybe I'm wrong.
You need the centripetal force along the tangential direction, so m\omega^2r \cos \varphi gets multiplied by \sin\varphi.
When you integrate:
m\omega^2(r \cos...
Shouldn't your r in the LHS of:
m\omega^2r\sin\varphi = -mg\frac{1}{r}\frac{dr}{d\varphi}
be r*cos(phi), or distance from the rotation axis? If you do that, then you get the original poster's value of half the bulge.
I think it's fairly obvious to prove that if the potential U(r) satisfies:
U(\kappa r)=\kappa^\nu U(r)
then the period 'T' is related to the size of the orbit 'a' by:
T \mbox{ } = \mbox{ } a^{-\frac{\nu}{2}+1}
In particular for the Kepler orbit \nu=-1 which reproduces his 3rd law, and for...