Assume:
- Space elevator extends from Earth's surface to geostationary orbit (GEO) at 35,786 km altitude.
- Maximum acceleration/deceleration: 1 g = 9.81 m/s².
- Maximum speed in the atmosphere: Mach 1 ≈ 343 m/s.
- Atmospheric portion is about 100 km (a generous estimate).
- Above the atmosphere: vacuum, so 1 g acceleration is comfortable and unrestricted.
For the constant-acceleration portions:
Atmospheric segment (0–100 km)
At 1 g, reaching Mach 1 takes:
- Time: 35 s
- Distance covered accelerating: 6.0 km
Then:
- Cruise at Mach 1 for about 88 km
- Decelerate for another 6 km
Total atmospheric travel:
- Acceleration: 35 s
- Cruise: 88,000 m / 343 m/s ≈ 257 s
- Deceleration: 35 s
≈327 seconds = 5.5 minutes
Vacuum segment (100 km → GEO)
Remaining distance:
Accelerate at 1 g halfway, then decelerate the second half.
Half-distance:
Time to midpoint:
[
t=\sqrt{\frac{2d}{a}}
]
- (d = 17,843,000) m
- (a = 9.81) m/s²
This gives:
Peak speed:
Decelerating takes the same amount of time.
Vacuum travel:
Total trip
- Atmosphere: 5.5 min
- Vacuum: 63.6 min
Total ride time: ≈69 minutes (about 1 hour 9 minutes).
Peak speed
The climber would briefly reach about 18.7 km/s, which is:
- Mach 55 (if expressed relative to sea-level speed of sound)
- Faster than low Earth orbital velocity (7.8 km/s)
- Well below Solar System escape velocity from Earth's distance (42 km/s).
This is dramatically faster than conventional space elevator concepts, which typically climb at only 100–300 km/h and therefore require 5–8 days to reach geostationary orbit. The limiting factor in those designs isn't physics in vacuum but the enormous power delivery, cable dynamics, and stresses induced by rapidly accelerating a massive climber. Under your idealized assumptions, however, the trip is only a little over an hour.
