Space Elevator Question: If You're At The "Top," Are You In Zero-G?

So they print a negative mold, you hot pour a positive mold, then use that to mass produce. Point is, if the process were completely reliable and automated, this additional step could just be queued up in the automation system and completed within a few minutes. It’s the reliability. Current automation isn’t this good, and things will go wrong, requiring manual intervention.

Also don’t forget, by the way, the cost of design, which also must be amortized over an entire production line. CAD software has made that more efficient, too, but you still need highly-skilled humans spending a lot of time on that, even when everything goes right.

So now I can buy a Tesla Model S?!

(Checking….) Ummmm…. Nope. Not yet.

No, but the S is cheaper than the Roader which is cheaper than experimental handmade lithium battery EVs from the 1990s.

I haven’t read EVERY reply, but this fact wasn’t mentioned in the ones I did read.

At 250 miles up, you still feel .88 G. If the car is stopped, you won’t feel *too *much different than 1 G at sea level. If you let go of the car, you will fall (pretty much) straight back to earth.

At the height of geosynchronous orbit, 22,236 miles above sea level, you *still *feel .023 G, but your orbital velocity is exactly enough so that you will remain at the same relative position to the space elevator. Technically, you aren’t weightless, you are in free fall.

J.

p.s., this is an amateur’s view. Some of the rocket scientists on the board will be able to give a much more accurate / correct interpretation of this.