Timeline for Projectile/orbital motion over very long distance
Current License: CC BY-SA 3.0
13 events
| when toggle format | what | by | license | comment | |
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| Nov 28, 2018 at 3:27 | comment | added | Ewoud | @sammygerbil This question specifically asks for the optimum launch angle, while that question does not address that (as far as I could see) | |
| Nov 28, 2018 at 3:24 | answer | added | Ewoud | timeline score: 0 | |
| Oct 28, 2016 at 4:24 | history | protected | Qmechanic♦ | ||
| Oct 27, 2016 at 23:49 | answer | added | sammy gerbil | timeline score: 3 | |
| Oct 27, 2016 at 22:24 | comment | added | sammy gerbil | Possible duplicate of How to calculate a ballistic trajectory for a suborbital flight? | |
| Oct 27, 2016 at 20:25 | answer | added | aventurin | timeline score: 0 | |
| Oct 27, 2016 at 19:48 | answer | added | Gert | timeline score: 2 | |
| Oct 27, 2016 at 19:39 | history | edited | user108787 | CC BY-SA 3.0 |
added mathjax
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| Oct 27, 2016 at 18:40 | comment | added | pathintegral | I don't think you can. For constant gravity all trajectories are of the same shape -- all are parabolas. But with $F=GMm/r^2$ the trajectories are different in shape, even if the shooting angles are the same. I would be very surprised a single "golden" angle will maximize the distance for all the shapes. | |
| Oct 27, 2016 at 18:24 | comment | added | Tammy Chong | what if the speed is not given and we made an assumption it is v?Just refer to the steps used to obtain the longest horizontal distance like what we normally did, can i find the angle then? | |
| Oct 27, 2016 at 18:13 | comment | added | pathintegral | I think it will depends on the speed. For large enough speed, even if you shoot it horizontally, it never falls and the distance is thus $\infty$. | |
| Oct 27, 2016 at 18:10 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
edited tags; edited title
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| Oct 27, 2016 at 18:05 | history | asked | Tammy Chong | CC BY-SA 3.0 |