4 ms·

Without considering the fact that the orbit of the Moon is not circular, this only helps reduce the delta-v needed to go from the surface of the earth to the su

Without considering the fact that the orbit of the Moon is not circular, this only helps reduce the delta-v needed to go from the surface of the earth to the surface of the moon by ~3.2km/s, but we still need ~12-14km/s to go to GTO/GEO.

Barrin92 7y ago
i think the bigger upside here is that getting humans into orbit and then to the moon via elevator is likely safer than repeated landings back and forth.
- t_fatus 7y ago
  Indeed this would avoid the tricky 'land to a place where nobody can help you' part.
  - extropy 7y ago
    Also would introduce the tricky no way to abort if something goes wrong. Unless you are willing to carry a fast abort rocket with you.
t_fatus 7y ago
However since the moon period is not the same as a GEO period, maybe the fact that you could juste 'catch' the end of the cable before falling back to earth without having the speed to maintain at GEO could help you save some delta-v.
- extropy 7y ago
  That's essentially what GTO is - an orbit going close and fast at one side and going high and almost stopped at high side. Going directly up means over 1h of fighting earth gravity. So need Isp > 3600. Much cheaper to go for orbit.
  - logfromblammo 7y ago
    An elliptical orbit that intersects the surface of the massive body at the periapsis side is fine, if you can change the orbit when you reach apoapsis. But if you miss the cable, you're boned.
extropy 7y ago
Useful for refueling GTO sattelites from the Moon however. Just need to build a trans Lunar railway from the poles to supply the water. Or attach the damn thing to one pole because no atmosphere.
emilfihlman 7y ago
You don't need to hold the orbit. Just jump to the elevator and grab on.
- t_fatus 7y ago
  Would be nice to get the savings in delta-v, I've no idea how to compute this.
  - emilfihlman 7y ago
    A ballistic approach is enough so the kinematics are simple, too.
    - extropy 7y ago
      The gravity is only 3% up there, so more tricky. And reducing with square of the distance. Also every second you are accelerating straight up costs you gravity worth of Delta-v because gravity losses. So need to be quite short and intense burn to be worth it.
      - emilfihlman 7y ago
        Eh? You don't need to get sideways velocity at all. It's enough to only reach the distance. This is a major and huge saving in fuel.
  - marcosdumay 7y ago
    My estimate is ~9.5km/s from the Equator to the hook. Add some more for atmospheric losses.
  - s1artibartfast 7y ago
    you can use the partial escape velocity equation. ~10 km/s to get to GEO without the radial velocity. delta-V to the moon is about 18 km/s (one way). Keep in mind that kinetic energy is proportional to velocity squared.

Barrin92 7y ago

i think the bigger upside here is that getting humans into orbit and then to the moon via elevator is likely safer than repeated landings back and forth.

t_fatus 7y ago
Indeed this would avoid the tricky 'land to a place where nobody can help you' part.
- extropy 7y ago
  Also would introduce the tricky no way to abort if something goes wrong. Unless you are willing to carry a fast abort rocket with you.

t_fatus 7y ago

However since the moon period is not the same as a GEO period, maybe the fact that you could juste 'catch' the end of the cable before falling back to earth without having the speed to maintain at GEO could help you save some delta-v.

extropy 7y ago
That's essentially what GTO is - an orbit going close and fast at one side and going high and almost stopped at high side. Going directly up means over 1h of fighting earth gravity. So need Isp > 3600. Much cheaper to go for orbit.
- logfromblammo 7y ago
  An elliptical orbit that intersects the surface of the massive body at the periapsis side is fine, if you can change the orbit when you reach apoapsis. But if you miss the cable, you're boned.

extropy 7y ago

Useful for refueling GTO sattelites from the Moon however. Just need to build a trans Lunar railway from the poles to supply the water. Or attach the damn thing to one pole because no atmosphere.

emilfihlman 7y ago

You don't need to hold the orbit. Just jump to the elevator and grab on.

t_fatus 7y ago
Would be nice to get the savings in delta-v, I've no idea how to compute this.
- emilfihlman 7y ago
  A ballistic approach is enough so the kinematics are simple, too.
  - extropy 7y ago
    The gravity is only 3% up there, so more tricky. And reducing with square of the distance. Also every second you are accelerating straight up costs you gravity worth of Delta-v because gravity losses. So need to be quite short and intense burn to be worth it.
    - emilfihlman 7y ago
      Eh? You don't need to get sideways velocity at all. It's enough to only reach the distance. This is a major and huge saving in fuel.
- marcosdumay 7y ago
  My estimate is ~9.5km/s from the Equator to the hook. Add some more for atmospheric losses.
- s1artibartfast 7y ago
  you can use the partial escape velocity equation. ~10 km/s to get to GEO without the radial velocity. delta-V to the moon is about 18 km/s (one way). Keep in mind that kinetic energy is proportional to velocity squared.