4 ms·

Launching from the moon would be easier and cheaper than launching from the Earth, unless you include the cost of reaching the moon. But according to the table

Launching from the moon would be easier and cheaper than launching from the Earth, unless you include the cost of reaching the moon. But according to the table in [1], going from the Earth's surface to geostationary orbit (to rendezvous with a lunar elevator) requires about 14km/s delta-v, and from there to the moon's surface only about 4km/s delta-v. Of the 18km/s total delta-v required, a lunar space elevator only eliminates the last 4km/s. 1: https://en.wikipedia.org/wiki/Delta-v_budget#Earth%E2%80%93Moon_space%E2%80%94high_thrust https://en.wikipedia.org/wiki/Delta-v_budget#Earth%E2%80%93M...

Ajedi32 7y ago
I haven't done the math, but are you sure that last 4km/s isn't way more significant than you're estimating due to the rocket equation? A small mass savings near the end of the journey might translate to significant mass savings near the start.
raducu 7y ago
Also I think (a series of) rotating elevators to get from LEO to GSO are also possible?
dwaltrip 7y ago
The end of the lunar space elevator wouldn't be traveling at geostationary orbital speed, it would be going 28x slower. A rocket could sync with the end of the elevator at this lower speed, which would significantly reduce the delta-v budget. [1] Not sure how practical all of this, but that is at least the theory. [1] Credit to mnw21cam: https://news.ycombinator.com/item?id=20996246 https://news.ycombinator.com/item?id=20996246
s1artibartfast 7y ago
The really interesting part is that the elevator would not be in a true geostationary orbit traveling at ~3000m/s. instead it would be traveling at 100m/s at the same height. The delta-V to reach GEO without the radial velocity to maintain it is closer to 10 km/s if you use the partial escape velocity equation. This is pretty huge as mass required to achieve delta-V is nonlinear.

Ajedi32 7y ago

I haven't done the math, but are you sure that last 4km/s isn't way more significant than you're estimating due to the rocket equation? A small mass savings near the end of the journey might translate to significant mass savings near the start.

raducu 7y ago

Also I think (a series of) rotating elevators to get from LEO to GSO are also possible?

dwaltrip 7y ago

The end of the lunar space elevator wouldn't be traveling at geostationary orbital speed, it would be going 28x slower. A rocket could sync with the end of the elevator at this lower speed, which would significantly reduce the delta-v budget. [1] Not sure how practical all of this, but that is at least the theory. [1] Credit to mnw21cam: https://news.ycombinator.com/item?id=20996246 https://news.ycombinator.com/item?id=20996246

s1artibartfast 7y ago

The really interesting part is that the elevator would not be in a true geostationary orbit traveling at ~3000m/s. instead it would be traveling at 100m/s at the same height. The delta-V to reach GEO without the radial velocity to maintain it is closer to 10 km/s if you use the partial escape velocity equation. This is pretty huge as mass required to achieve delta-V is nonlinear.