Timeline for Coordinates of center of gravity in non-uniform gravitational field
Current License: CC BY-SA 4.0
7 events
| when toggle format | what | by | license | comment | |
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| Aug 9, 2023 at 16:16 | comment | added | raphael | This clears things up a lot for me. Thanks, @JEB for the time and effort spent! | |
| Aug 8, 2023 at 23:16 | comment | added | JEB | In newtonian terms, a parallel g-field only comes from an infinite sheet of mass (it's a standard problem treated all over this site), and in relativity is comes from uniform accelerations (Rindler Coordinates). For the infinite sheet, there's no center of gravity or mass because all points are equivalent, so the field is perpendicular to the sheet (it's also doesn't depend on distance...which makes sense because the only way to get weaker without divergence (Poisson's Eq) is by spreading out, but it can't spread out , b/c it's parallel) | |
| Aug 8, 2023 at 20:02 | comment | added | raphael | @JEB thank you for the answer. Poisson equations for now are beyond me, since it's been 30 years since I've done partial differential equations. I also understand that a non-uniform gravitational field cannot be truly parallel. In these respects, the textbook problem is oversimplified in order to get a certain point accross. I don't know. But I'm not sure what you mean by "Not sure how that works". I will do the research, but if you can tell me if I'm interpreting your answer incorrectly, please let me know. Thank you again! | |
| Aug 8, 2023 at 19:47 | history | edited | raphael | CC BY-SA 4.0 |
deleted 22 characters in body
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| Aug 8, 2023 at 14:37 | comment | added | JEB | In the general case, the field isn't parallel, so you're finding the 1st (vector) moment of a scalar distribution weighted by a vector. Not sure how that works. | |
| S Aug 7, 2023 at 19:23 | review | First answers | |||
| Aug 7, 2023 at 21:28 | |||||
| S Aug 7, 2023 at 19:23 | history | answered | raphael | CC BY-SA 4.0 |