Timeline for Is it possible that the force due to gravity varies based on density?
Current License: CC BY-SA 3.0
11 events
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Jul 21, 2018 at 13:42 | comment | added | Luboš Motl | I wrote it was an estimate and I insist that $10^{-17}$ is an excellent estimate for the current precision of these experiments, see e.g. arxiv.org/abs/1709.02768 - So despite your apparent critical tone, my number is still more precise than yours. I refuse to discuss particular sources, especially in answers that have nothing to do with the technicalities of these particular experiments. This is a theory question so I won't pretend to be a technician following particular experimental teams. The particular teams aren't really important for the key knowledge here. | |
Jul 1, 2018 at 8:43 | comment | added | Anthony | It was experimentally verified to 10^−15 in 2016, 3 years after this answer claimed 10^-17. This is why citing your sources is important, folks. I'm sure it was an honest mistake, but one of significant magnitude. | |
Sep 16, 2013 at 4:17 | comment | added | abiessu | In other words, the force due to gravitation from the earth on me standing on its surface is exactly identical to the force due to gravitation on me of a point mass equal to earth if I were floating a distance from the point mass equal to earths radius? How can that be when the mass of the earth near the surface to the north, east, south and west of me is all nearly canceling the gravitational force on me from those directions? | |
Sep 9, 2013 at 5:54 | comment | added | Luboš Motl | Dear Abiessu, yes, the shape of the field of course depends on the matter distribution which makes difference especially "inside" the matter or very close to it. But if you have point masses or spherical masses of the same total mass $M$, their gravitational field will actually look completely identical at distance $R\gt R_\text{largest sphere}$ from the center of the object. | |
Sep 7, 2013 at 17:24 | comment | added | abiessu | Just to verify, a spherical mass will have a "diffuse" gravitational field at relatively close distances compared to a point mass which will have no field diffusion, is that correct? | |
Sep 7, 2013 at 12:16 | comment | added | Alexander | There are currently experiments using test masses in satellites planned ( [STEP](en.wikipedia.org/wiki/… ) that supposedly decreases the uncertainty to even lower limits of $10^{-20}$. The outcome of these experiments will limit any deviations even further. | |
Sep 7, 2013 at 8:39 | comment | added | Luboš Motl | Abiessu: well, I am saying it except that the equivalence of the two masses isn't an assumption that "fell from the heaven", it's an assumption of general relativity or its extensions and the ultimate justification for it are the observations of the universal acceleration. Michael: the degree of ugliness may differ in these f(R) theories, much like the severity with which they violate the equivalence principle (in general, they don't have to) but in principle yes, I would consider them as what I wrote, too. | |
Sep 7, 2013 at 8:12 | vote | accept | abiessu | ||
Sep 7, 2013 at 6:25 | comment | added | Michael | To give a concrete example: $f(R)$ theory violates the strong form of the equivalence principle, and is strongly constrained by a bunch of these experiments. You would probably also think it is "an ugly, partially inconsistent, unjustified, and numerically small deformation of a beautiful, consistent, robust, justified theory." :) | |
Sep 7, 2013 at 6:24 | comment | added | abiessu | So, to clarify, you are saying that the inertial mass of C ($m_C$) is equivalent to the gravitational masses $m_{C_A}$ and $m_{C_B}$ by assumption, and this is the basis for the system of relativity we have. Further, you are saying that for up to 17 orders of magnitude of relative density difference, there is no measurable discrepancy between these three mass measurements. Does that roughly equate to what you are saying? | |
Sep 7, 2013 at 5:55 | history | answered | Luboš Motl | CC BY-SA 3.0 |