Recent content by RedX

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.

Sure. It's a little bit lengthy though, so it might take some work to read it: P(m)=\frac{\sqrt{2 \pi n}n^ne^{-n}}{\sqrt{2 \pi m}m^me^{-m}*\sqrt{2 \pi (n-m)}(n-m)^{(n-m)}e^{-(n-m)}}p^m(1-p)^{n-m}= \frac{n^{n+1}}{\sqrt{2 \pi n}*m^{m+\frac{1}{2}}*(n-m)^{(n-m)+\frac{1}{2}}}p^m(1-p)^{n-m}...

Normally I would just dismiss the formula, but I found it in two different sources (both particle physics sources though). One book talked about the vacuum bubble expansion of the integral: \int \frac{1}{[k^2-m^2][(k-p)^2-m^2]}=\int \frac{1}{[k^2-m^2]^2} -\int \frac{p^2}{[k^2-m^2]^3}...

This is probably a dumb question, but I have a book that claims that if you have a function of the momentum squared, f(p2), that: \frac{d}{dp^2}f=\frac{1}{2d}\frac{\partial }{\partial p_\mu} \frac{\partial }{\partial p^\mu}f where the d in the denominator is the number of spacetime...

I want to show that the binomial distribution: P(m)=\frac{n!}{(n-m)!m!}p^m(1-p)^{n-m} using Stirling's formula: n!=n^n e^{-n} \sqrt{2\pi n} reduces to the normal distribution: P(m)=\frac{1}{\sqrt{2 \pi n}} \frac{1}{\sqrt{p(1-p)}} exp[-\frac{1}{2}\frac{(m-np)^2}{np(1-p)}]...

For some information on tides, here's a website: http://www.lhup.edu/~dsimanek/scenario/tides.htm For some mathematical detail (just algebra): http://mb-soft.com/public/tides.html You only need gravity to explain the tides. Omega is a constant, equal to the angular velocity of the earth...

I'm pretty confused by the post - I think you might have a lot of misconceptions. First, tidal forces have nothing to do with centripetal acceleration. Tidal forces are due to gravity. Second, if you're standing still on the earth, unless you're at the poles, there is a centripetal force on...

I always thought it'd be dangerous to ground the primary side since it's at such high voltage. So maybe isolation transformers allow you to ground the secondary side, rather than, as you say, allowing the primary to float when you have a grounded secondary?

When lightning strikes near power lines, it breaks down the air which shorts the power lines, but why would this have an effect on the load? The short and the load are in parallel, so the load should not be affected. When you have two things in parallel what happens to one branch should not...