Timeline for Does the trajectory of a body in a central-force field leave the force undetermined if I don't specify the number of bodies interacting?
Current License: CC BY-SA 4.0
5 events
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Dec 4 at 22:10 | comment | added | Triatticus | What I mean is that you should have four situations demonstrating this, why use different numbers of masses for the inverse four vs inverse square and not make all four situations. Two masses in both potentials and three masses in both. | |
Dec 4 at 21:05 | comment | added | Craterus | So yes, in the first case I have the "exotic" ˜ 1/(r^4) force law, in the second the standard inverse square one. But in the first case I have n=2 bodies, and in the second n=3. My question is if it is possible to arrange the 3 bodies affected by the standard force in such a way that the orbit of one of them equals the one it would have in case n=2 and F=K*1/(r^4). Sorry for not having been clear | |
Dec 4 at 20:51 | comment | added | Triatticus | I'm not sure it's clear what you mean, do you mean the first situation has the first force and the second has the standard inverse square force? Shouldn't you also show that the three body's have the same orbit in the inverse four situation? | |
S Dec 4 at 19:56 | review | First questions | |||
Dec 4 at 20:06 | |||||
S Dec 4 at 19:56 | history | asked | Craterus | CC BY-SA 4.0 |