Timeline for Why is the Earth so fat?
Current License: CC BY-SA 4.0
19 events
| when toggle format | what | by | license | comment | |
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| Sep 21 at 7:35 | history | edited | Qmechanic♦ | CC BY-SA 4.0 |
added 3 characters in body; edited tags
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| Jul 27, 2019 at 14:16 | review | Suggested edits | |||
| Jul 27, 2019 at 20:35 | |||||
| Sep 6, 2012 at 0:56 | answer | added | Art Brown | timeline score: 4 | |
| Aug 8, 2012 at 7:36 | comment | added | Ron Maimon | Although it's in my answer, the upshot is that I got nearly the exact right bulge (44.3 km vs. 42.7) from a two-ellipsoid model which has an elliptical core, also an equipotential surface, 9% more eccentric than the surface. So the correction from the naive estimate would be 2.5 for a uniform sphere (as calculated by qmechanic), but is reduced to 2 due to the mass of the core and the ellipticity of the core, which are competing effects of the same order of magnitude. | |
| Aug 6, 2012 at 8:39 | answer | added | Ron Maimon | timeline score: 8 | |
| Aug 3, 2012 at 9:48 | comment | added | Qmechanic♦ | @Marc C: I guess the original motivation for OP and answerers was not just to get the correct result by introducing enough parameters in their models (and optimize them accordingly), but rather (as an academic exercise) to see how close they could get (to the correct result) by using only the most primitive models, i.e. having very few adjustable parameters. | |
| Aug 2, 2012 at 15:22 | answer | added | Alan Rominger | timeline score: 5 | |
| Sep 7, 2011 at 5:05 | comment | added | Mark C | I am curious as to why all of the answers do not consider the problem from view of the crust and mantle of the earth. This seems like a roundabout and difficult way of making a guess. If we liken the earth to a spinning water balloon or mass of clay, we can model the cohesiveness (or bulk modulus and others) of the clay, its gravity, and the forces resulting from its motion. | |
| Apr 22, 2011 at 18:52 | answer | added | Qmechanic♦ | timeline score: 3 | |
| Apr 20, 2011 at 0:05 | answer | added | Qmechanic♦ | timeline score: 24 | |
| Apr 5, 2011 at 18:41 | answer | added | Mark Eichenlaub | timeline score: 13 | |
| Apr 5, 2011 at 18:33 | vote | accept | Mark Eichenlaub | ||
| Apr 5, 2011 at 16:28 | comment | added | Mark Eichenlaub | @Kakemon Good point, but the tides from the sun are smaller than from the moon, and the lunar tides in the ocean are only a few meters high. On solid earth, this would be smaller because you're alternately being squished and raised as the earth spins. | |
| Apr 5, 2011 at 16:27 | comment | added | Mark Eichenlaub | @Martin I don't see what you're getting at. The gradient of the potential times minus one is the force, and the force is $-mg\hat{r}$. | |
| Apr 5, 2011 at 14:40 | comment | added | TROLLHUNTER | If the earth was static and the sun rotating about it, it would still have a bulge due to the suns gravity not exerting equal force on each side of the earth. I dont know how large this contribution is. | |
| Apr 5, 2011 at 12:42 | answer | added | Luboš Motl | timeline score: 56 | |
| Apr 5, 2011 at 11:18 | comment | added | Martin Gales | I think that the gravitational potential per unit mass of the ball is $\frac{hg}{2}$. (Think about the opposite side of the Earth) | |
| Apr 5, 2011 at 9:09 | history | tweeted | twitter.com/#!/StackPhysics/status/55195258136428544 | ||
| Apr 5, 2011 at 8:13 | history | asked | Mark Eichenlaub | CC BY-SA 2.5 |