# Elevator to the Moon

Related but too different in details for answers to apply Why not build a space elevator starting from the moon?

As answered in the related question an elevator from surface to surface is impractical. But an elevator does not actually need to come to surface or even within atmosphere to be useful. An elevator built between an anchor point on the Lunar surface and a free hanging end point well within Earth's gravity but also well outside the atmosphere would still require much less energy and propellant to reach than going to moon. And savings in the opposite direction would be even greater.

I am fairly sure that the distance (and variable distance at that) makes this economically insane, but could it work from the engineering standpoint? What would the stresses involved be and how would they compare to available or proposed materials?

EDIT: As Thucydides indirectly implies as initially written this reduces to something that has the same answer as another question. Specifically any practical elevator starting from the moon has to extend into the gravity field of the Earth, but won't extend to the atmosphere. And that has already been answered. So as written first the answer was that it is believed practical.

This makes this question kind of pseudo-duplicate, the concept is different and there would be some resulting practical differences. But I forgot to actually ask about those differences. Fundamentally it would be a question of whether extending the elevator lower down to Earth's gravity to reduce the needed energy (and braking going down) would be practical. And seriously that would be IMHO a question of the relative economics of hypothetical launch system and elevator building. Which is probably unanswerable.

• I received 5 down votes for my question. If you are going to leave this question up, I hope you don't mind losing points... – Lorry Laurence mcLarry Nov 25 '15 at 4:31
• @LorryLaurencemcLarry I don't. I kind of deserve it really. While I had a valid question in mind, I failed at writing it. – Ville Niemi Nov 26 '15 at 3:29
• The moon orbits in the ecliptic plane, not the equatorial plane. Furthermore, it has an inclination in that plane, and that inclination when viewed in the Equatorial plane changes over time. Also, the moon is not geosynchronous (and therefore not geostationary) so your elevator would rarely even be aligned with the moon. The physical issues aside, it just wouldn't line up. – iAdjunct Nov 26 '15 at 4:18
• @iAdjunct You misunderstood the question. Or just commented on the wrong question as those would have been valid points in the question linked as related. The elevator in this question would be free moving at Earth end. You can check the answer by Thucydides for details. – Ville Niemi Nov 26 '15 at 20:39

Jerome Pearson did the first theoretical calculations for a Lunar elevator as far back as the 1970's, so this is a reasonably well understood engineering project from a theoretical perspective.

An elevator starting from the surface of the Moon would have far less stress, although unless it is tensioned by a massive counterweight on the end, it will also be far longer than the corresponding "skyhook" from the Earth's surface. A lunar elevator could be built from currently available materials like Spectra 100 or M5 polymer in terms of static strength. Obviously, a material that can withstand the wild temperature swings, exposure to radiation and vacuum environment would be needed to make a functional elevator.

An introductory article can be read here: https://en.wikipedia.org/wiki/Lunar_space_elevator

and a proposed commercial venture is discussed here: http://lunarelevator.com

• True enough, that it reduces to the lunar elevator as you add to the altitude of the end point enough. So clearly it is possible at some height. Maybe I should edit the question to ask how close to Earth surface can the end be brought with it still being practical... (But I have to go now...) – Ville Niemi Nov 25 '15 at 4:25

For areas near the equator, the speed of the cable end is roughly

$$V = \frac{2\times \pi \times 3960 \text{ miles}}{24} = 1036 \text{mph}$$

So any attempt to use the elevator would require matching speeds and transferring to the end of the cable - at Mach 1.4.

Top Gun, anyone?

• Mach 1.4 would be a huge saving on current spaceflight especially since the end would be high enough for drag not to be an issue. – Ville Niemi Nov 25 '15 at 4:20
• Funny. The person who originally posted that explanation received 7 up votes and 0 comments. – Lorry Laurence mcLarry Nov 25 '15 at 4:25
• Yes, but he was answering a different question. – Ville Niemi Nov 25 '15 at 4:31
• Sure. But his conjecture was no more valid then than it is now. Cable is travelling at the same relative speed. Only now it is higher. – Lorry Laurence mcLarry Nov 25 '15 at 4:34
• Mach 1.4 wouldn't be such a huge saving. See Why aren't all satellite-carrying rockets launched from airplanes? on Space Exploration. – user Dec 8 '15 at 12:10