3 ms·
Probably not. You're looking at a height difference of about 9 km [1], at a radius of 6380 km from earth's core [0] versus about an equatorial 6371 km. Rotatio
by HCIdivision17 12y ago
Probably not. You're looking at a height difference of about 9 km [1], at a radius of 6380 km from earth's core [0] versus about an equatorial 6371 km. Rotational velocity at the equator is about 1040 mph [5], while Everest is looking at about 6380 km / 6371 km * 1040 mph = 1041.5 mph, assuming we moved Everest to the equator and kept everything else the same. One or two mph is really small compared to c, so much so WA just sorta says gamma = 1 [4].
And so is a mere 0.15% increase in height. But compare the gravitational acceleration at these two heights:
[2] Everest = 9.76322 m/s^2
[3] Sea Level = 9.831 m/s^2
Sea level is 0.69% stronger. That's somewhat more significant, and given that gravitational energy follows the inverse-square law, it makes some sense that it would amplify differences more than the linear effects of speed.
[0] https://www.google.com/search?q=radius+of+earth https://www.google.com/search?q=radius+of+earth
[1] https://www.google.com/search?q=height+of+everest https://www.google.com/search?q=height+of+everest
[2] http://www.wolframalpha.com/input/?i=gravitational+acceleration+at+the+top+of+mount+everest http://www.wolframalpha.com/input/?i=gravitational+accelerat...
[3] http://www.wolframalpha.com/input/?i=gravitational+acceleration+at+sea+level http://www.wolframalpha.com/input/?i=gravitational+accelerat...
[4] http://www.wolframalpha.com/input/?i=lorentz+factor+at+1.5+mph http://www.wolframalpha.com/input/?i=lorentz+factor+at+1.5+m...
[5] http://www.wolframalpha.com/input/?i=rotational+velocity+of+earth http://www.wolframalpha.com/input/?i=rotational+velocity+of+...
EDIT: forgot the reference to rotational velocity (and spelling).
Note that the point is scale here. Gravity is n^2 versus velocity's n. It'll have a stronger effect in most cases (and any where speed overcomes, well, we call them relativistic speeds, and they'll often be some appreciable fraction of c, like 1% or more).
DOUBLE EDIT: 'AnimalMuppet makes a good point that the entire estimate for Everest's rotational velocity is flat out wrong. So I've changed my premise to match my initial incredibly incorrect assumption.
If I were to attempt to be correct about the problem, we'd see that Everest is cos(27.9881 degrees [10]) = 0.8837 [11] as fast as the equator, which sorta blasts out any height changes by long shot. (Note I'm assuming the earth to be a perfect, frictionless sphere here, and definitely not a oblate spheroid.)
[10] https://www.google.com/search?q=latitude+of+everest https://www.google.com/search?q=latitude+of+everest
[11] https://www.google.com/search?q=cos(27.9+degrees) https://www.google.com/search?q=cos(27.9+degrees)
- AnimalMuppet 12y agoYou're forgetting that Everest isn't on the equator. Distance from the center of the earth doesn't matter; distance from the axis of rotation matters. So you need to multiply Everest's (height + radius) by the cosine of its latitude.
- HCIdivision17 12y agoYou're entirely correct, and I'm going to modify my answer to a convenient lie (in the true style of a physicist). The point stands, but now I'm interested in what kind of cones we could make where relative heights and latitudes yield identical velocities. Prolly as useful a question as most XKCD What If? notes :)