4 ms·

There would still be some delta-v. Anyone have a handle on how much? I'm too lazy to look up the data and do the math.

• persimmon_b 9y ago
  Translunar injection needs about 10.8 km/s, Earth's escape velocity is 11.2 km/s.
  • aptwebapps 9y ago
    I mean to hook up with the elevator.
• ChuckMcM 9y ago
  First, I've neither done the requisite orbital mechanics calculations nor have I simulated this. But with that caveat, this is how I imagine it would work. Assuming your tug is in the translunar orbit (going in a figure 8 around the Moon and Earth with the intersection being the Earth-Moon L1 point. And assuming the tug is significantly more massive than your cargo. As you cross the L1 point you can use a mass driver on the tug to 'shoot' the cargo out behind you which would decelerate it to a low delta-v relative to the elevator base (and slightly boost the tug on its way) and when you came around the other side you hook the cargo through a loop like the old mail planes did [1] and pulling it away (which would impart a small deceleration on the tug as it headed back to earth). Depending on masses and how closely you could match things, the amount of delta-v you would need for the tug to keep its orbit true should be very manageable. [1] https://postalmuseum.si.edu/collections/object-spotlight/skyhooking-container.html https://postalmuseum.si.edu/collections/object-spotlight/sky...
• Dylan16807 9y ago
  http://i.imgur.com/SqdzxzF.png http://i.imgur.com/SqdzxzF.png You might save 2km/s if you can use an elevator on the moon. Which is not a lot compared to a space elevator on Earth, but it would definitely be useful. That could be half your delta-v budget if you're going between the moon's surface and an orbit around Earth/Mars/Venus.