Timeline for Elliptical Trajectory, or Parabolic?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
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| Dec 11, 2017 at 8:32 | history | edited | J. Manuel | CC BY-SA 3.0 |
edited body
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| Dec 8, 2017 at 20:37 | comment | added | Ruslan | I didn't edit because the text in the image should also be fixed. It's a bit more work, while you might have the sources for the image, whatever they are, so it's easier for you. | |
| Dec 8, 2017 at 20:22 | comment | added | J. Manuel | @MaxW. When solving physics problems, one always have to make assumptions, and simplify things, otherwise the problem may end up having infinite degrees of freedom becoming impossible to solve. In this particular situation, it may or may not be the case. If you neglect earth’s curvature, the flat earth is the sin of the angle formed by the starting point, the ending point, and the center of earth. For small angles the sin of an angle becomes equivalent to the arc length. | |
| Dec 8, 2017 at 20:00 | history | edited | J. Manuel | CC BY-SA 3.0 |
added 8 characters in body
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| Dec 8, 2017 at 19:55 | comment | added | J. Manuel | @Ruslan. You’re right. Unfortunately this will happen when a person is not a native English speaker, I’ll edit it. But, you can always edit it ;-) | |
| Dec 8, 2017 at 19:05 | comment | added | Ruslan | ellipsis→ellipses. Ellipsis is another thing. | |
| Dec 8, 2017 at 17:17 | comment | added | MaxW | Ah.. It seems that the parabolic flight is a significant figures argument. In other words the ellipse is so elongated and a near earth trajectory is such a small arc of the ellipse that a parabola will do. Correct? | |
| Dec 8, 2017 at 14:34 | vote | accept | stoic-santiago | ||
| Dec 8, 2017 at 14:54 | |||||
| Dec 8, 2017 at 14:31 | history | answered | J. Manuel | CC BY-SA 3.0 |