Timeline for Gravitational attraction between a particle and a bar
Current License: CC BY-SA 4.0
14 events
| when toggle format | what | by | license | comment | |
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| Nov 15, 2019 at 21:44 | comment | added | Vilius Zem | Thanks that was a huge help. Now I only need to find the expression for the torque. | |
| Nov 15, 2019 at 19:02 | comment | added | Philip Wood | indeed so. And if we're really, really far away from P we can neglect terms in $a/r$ as well, so the force will go to zero – as it should! Thanks for accepting the answer. | |
| Nov 15, 2019 at 18:51 | vote | accept | Vilius Zem | ||
| Nov 15, 2019 at 18:50 | comment | added | Vilius Zem | Why is it ok to neglect $(a/r)^2$ at that point? Because its order of magnitude is smaller than $(a/r)$ and we know that $(a/r)$ tends to zero? | |
| Nov 15, 2019 at 18:39 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Nov 15, 2019 at 18:06 | comment | added | Philip Wood | It is now done. | |
| Nov 15, 2019 at 18:03 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Nov 15, 2019 at 17:43 | comment | added | Philip Wood | Fair comment. I'm about to extend my answer to show what to do with $\vec n_1$. | |
| Nov 15, 2019 at 17:02 | comment | added | Vilius Zem | Well you did it only for the $\overrightarrow{n2}$ term, I wanted to check if it works for the $\overrightarrow{n1}$ as well. (I don't know how you came up to $sin(\psi)\overrightarrow{n1}$ in your last equation, I mean its obvious that it should be $sin(\psi)$ in that direction, but I wanted to prove it). | |
| Nov 15, 2019 at 16:50 | comment | added | Philip Wood | I suspect that what I did was rather simpler. Did you not follow it? | |
| Nov 15, 2019 at 10:05 | comment | added | Vilius Zem | So I tried using binomal series with the $\overrightarrow{n_1}$ term and I came up with this: $$\frac{GmM}{2ar^2sin(\psi)}\left[(rcos(\psi)+a)(1 - (a/r)cos(\psi)-\tfrac12 (a/r)^2)-(rcos(\psi)-a)(1 + (a/r)cos(\psi)-\tfrac12 (a/r)^2)\right]\overrightarrow{n_1}$$ $$\frac{GmM(2(a-acos(\psi)^2)-(a/r)^2)}{2ar^2sin(\psi)}\overrightarrow{n_1}$$ After simplification: $$\frac{GmM(2(1-cos(\psi)^2)-(a/r)^2)}{2r^2sin(\psi)}\overrightarrow{n_1}$$ And when the term $(a/r)^2$ goes to zero, we have: $$\frac{GmMsin(\psi)}{r^2}\overrightarrow{n_1}$$ So I get what I expected and assume it should be correct. | |
| Nov 14, 2019 at 15:27 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Nov 14, 2019 at 12:56 | history | edited | Philip Wood | CC BY-SA 4.0 |
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| Nov 14, 2019 at 12:46 | history | answered | Philip Wood | CC BY-SA 4.0 |