Timeline for Calculating position of a particle at time $t$ in a gravitational field
Current License: CC BY-SA 3.0
14 events
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Nov 7, 2015 at 17:46 | history | edited | Qmechanic♦ | CC BY-SA 3.0 |
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Nov 7, 2015 at 17:41 | vote | accept | Anton | ||
Nov 7, 2015 at 17:32 | comment | added | Gert | In $F_g = am =\frac{-GMm}{d^2}$, $d$ is the distance between the particle and the centre of the Earth, not 'the distance from the floor'. | |
Nov 7, 2015 at 17:31 | answer | added | CR Drost | timeline score: 2 | |
Nov 7, 2015 at 17:27 | comment | added | Anton | Interesting, I think the leapfrog method is what I'm looking for, but could you tell me where I went wrong when I tried to do this on my own? | |
Nov 7, 2015 at 17:17 | comment | added | Kyle Kanos | Though if you're trying to model this, I'd suggest that you're going about it the wrong way. You are more likely going to want to use the leapfrog method (or something similar). | |
Nov 7, 2015 at 17:15 | comment | added | Anton | @KyleKanos Yes I can. You're right. But, that still leaves me with $2.59$ and $9.44$. | |
Nov 7, 2015 at 17:14 | comment | added | Kyle Kanos | If your lower bound is the floor at $d=0$, you can drop the negative root, no? | |
Nov 7, 2015 at 17:13 | comment | added | Anton | I'm not bouncing from a great or small height. The particle (magically) gets upward velocity when it's on the ground | |
Nov 7, 2015 at 17:12 | history | edited | Anton | CC BY-SA 3.0 |
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Nov 7, 2015 at 16:59 | comment | added | Gert | You can't just drop $G$ (or the minus sign!) because you 'have no use for it'! Also, $a=\frac{d^2 r}{dt^2}$. Your equations make very little sense and you freely mix up $d$ and $r$. You also might want to decide whether you're bouncing from a great height or a small one: the latter case is easier to derive. | |
Nov 7, 2015 at 16:53 | history | edited | Gert |
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Nov 7, 2015 at 16:39 | review | First posts | |||
Nov 7, 2015 at 17:58 | |||||
Nov 7, 2015 at 16:35 | history | asked | Anton | CC BY-SA 3.0 |