Timeline for How to calculate the velocity needed for a rocket to get to a L1 point (escape a body without orbiting)?
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
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| Apr 22, 2014 at 9:51 | comment | added | Ehryk | If launched from earth's equator, that would give 459 m/s of angular velocity. The moon rotates with 1024 m/s of velocity, so yes the rocket would need to 'add' 566 m/s in the rotational direction of the moon to simply 'Fall In'. However, I'm looking for the rest of the component, and I suspect that one would not need all 566 m/s if some fancy maneuvers were added. | |
| S Mar 31, 2014 at 7:15 | history | bounty ended | CommunityBot | ||
| S Mar 31, 2014 at 7:15 | history | notice removed | CommunityBot | ||
| Mar 27, 2014 at 19:59 | history | tweeted | twitter.com/#!/StackPhysics/status/449274561750896641 | ||
| Mar 27, 2014 at 18:25 | vote | accept | Ehryk | ||
| Mar 27, 2014 at 16:42 | comment | added | Ross Millikan | Your question misses the point that the L1 point is rotating around the earth at the same angular velocity as the moon. Your image is to get to L1 with zero velocity in inertial space, which is a well-defined operation and DavePhD has shown how to calculate the energy requirement for that. However, it is incorrect to think you will just "fall to the moon" from there. The moon is running away from you with its orbital velocity. You have a lot of sideways velocity relative to the moon, so are likely to miss. | |
| Mar 27, 2014 at 15:27 | answer | added | DavePhD | timeline score: 3 | |
| S Mar 23, 2014 at 6:06 | history | bounty started | Ehryk | ||
| S Mar 23, 2014 at 6:06 | history | notice added | Ehryk | Improve details | |
| Mar 19, 2014 at 23:45 | answer | added | Klaas van Aarsen | timeline score: 0 | |
| Mar 19, 2014 at 23:31 | history | asked | Ehryk | CC BY-SA 3.0 |