# Search results

1. ### Question about Earth's equatorial bulge-derivation

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...
2. ### Question about Earth's equatorial bulge-derivation

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...
3. ### Question about Earth's equatorial bulge-derivation

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...
4. ### Question about Earth's equatorial bulge-derivation

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...
5. ### Question about Earth's equatorial bulge-derivation

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.
6. ### Question about Earth's equatorial bulge-derivation

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...
7. ### Question about Earth's equatorial bulge-derivation

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.
8. ### The reason behind the inverse proportionality r^2 to the force of attraction

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...
9. ### Mark Srednicki's QFT Textbook

It's back! I thought maybe the publishers got mad at him for offering a preprint so maybe the website went down due to the police or something.