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
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| Mar 31, 2014 at 7:15 | history | bounty ended | CommunityBot | ||
| Mar 28, 2014 at 12:08 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 20:23 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 19:29 | comment | added | DavePhD | @Ehryk you would need to slow down to be captured in orbit by the Moon. Also, as Ross Millikan points out, the straight line from the Earth, through L1, to the Moon is rotating. You shouldn't think this corresponds to traveling in an actual straight path in an inertial frame. | |
| Mar 27, 2014 at 19:09 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 19:02 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 18:40 | comment | added | DavePhD | @Ehryk as to the balloon comment, balloons travel with the atmosphere, which rotates at the same angular velocity as the suface, a somewhat faster linear velocity than the surface | |
| Mar 27, 2014 at 18:30 | comment | added | DavePhD | yes(to the "Would this be correct for solar gravity inclusion" comment), but only at a moment that the three bodies are aligned with the moon in the middle (a solar eclipse) | |
| Mar 27, 2014 at 18:29 | comment | added | Ehryk | And, (finalish question) - if I got a small rocket precisely to L1 and used the surface rotation timing, how much velocity/energy would it take for lunar capture? Or would it just speed past the moon unless slowed down significantly? Would it need to get much closer to the moon than L1 to actually be captured with the KE from earth's rotation? | |
| Mar 27, 2014 at 18:25 | vote | accept | Ehryk | ||
| Mar 27, 2014 at 18:25 | comment | added | Ehryk | Also, one of the things I'm confused about is atmospheric orbital speed: if the balloon setup sat at 100km altitude for long enough to synchronize with the atmosphere at that height (what little of it there is); would it speed up to the orbital rotation at that height (I.E. track the same geostationary location once settled), or slow down, or ? | |
| Mar 27, 2014 at 18:20 | comment | added | Ehryk | Nailed it. Would this be correct for solar gravity inclusion: $V_e = −Gm(\frac {M_e} {r_e} + \frac {M_l} {LD−r_e} + \frac {M_s} {SD-r_e})$ and $V_{L1} = −Gm(\frac {M_e} {d_{L1}} + \frac {M_l} {LD−d_{L1}} + \frac {M_s} {SD-d_{L1}})$ ? | |
| Mar 27, 2014 at 18:08 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 18:03 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 17:50 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 16:01 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 15:53 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 15:46 | history | edited | DavePhD | CC BY-SA 3.0 |
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| Mar 27, 2014 at 15:27 | history | answered | DavePhD | CC BY-SA 3.0 |