The Space Pier -- Operations


 
A payload launched from the pier at 8km/s, launch altitude 100 km, corrected to a 340-km orbit.


For altitudes in the 100km range, we can assume gravity is constant for estimation purposes. The energy cost of sending a 10-tonne payload to the top of the tower by elevator is about 10 GJ (2778 kWh or $138.89 worth of electricity), which amounts to 1.4 cents per kg. At an express elevator speed of 50 m/s (108 mph) it takes over half an hour to get to the top. It is not clear that there is a compelling reason to go faster, since, unlike rockets, elevators can be made efficient at as slow a climb rate as desired.

Inert freight would presumeably have been packed into projectiles on the ground, but passengers would want to wait until getting to the top. (Nice view!  The horizon is over 1000 km away.) At 10 G's launch acceleration, passengers will be best off ensconced in form-fitting, fluid-filled sarcophagi. Even so the launch will not be particularly pleasant.

The vehicle is then accelerated at 10G along the top of the tower for 80 seconds to 8 km/sec. The track is level and with no lateral acceleration the vehicle is in an orbit with a 100-km perigee and 500-km apogee. (The numbers include an assist from Earth rotation.) The energy required is 300 GJ ($4166 worth of electricity, again for a 10 tonne payload), i.e. 42 cents per kg. Of course, there will be inefficiencies in the conversion of electricity to kinetic energy, so real electricity costs will be higher. The power (as opposed to energy) requirements average 3750 MW for the 80 seconds. A typical suburb on the same land might draw a peak load of 750MW. The tower's power draw increases linearly from 0 to 7500 MW during the 80 seconds of a launch. Local short-term energy storage, in a form conducive to rapid drawdown such as flywheels, will be necessary for load averaging. Energy storage (and release) requirements are constant per track length and amount to 1 MJ per meter. (1 MJ is the energy in an ounce of butter.) It might even be possible to store energy in the magnetic fields of the accelerator coils. Drawing at a typical power station's production of 1000 MW it would take 5 minutes to recharge the tower.

The other major part of the cost of a launch is amortization of the cost of the tower. If we can launch once an hour (and at an interest rate of 8%) we must charge $0.91 per kg per billion dollars of tower cost. If the traffic is there the rate might be increased (and the cost reduced) by a factor of 10 but not 100. (NB: as payload size is increased, power costs go up per launch but not per kg; amortization costs go down per kg; and more kinds of things can be launched in one piece. The main limit is the accelerator. Chances are you could build one to handle 10 tonnes, but that's a guess.)

Also, your payload needs to be a spacecraft capable of some 330 m/s delta-V to circularize its orbit. Orbits decay rapidly at 100 km, which is why the tower is not in much danger from space debris. The best strategy is to inject into an orbit whose energy is the same as the desired circular one, and do a correction at the point the orbits cross which only changes direction. This is not a Hohmann transfer, which optimizes total delta-V. With the launch tower, delta-V at correction is considerably more expensive than at launch, so we minimize it preferentially. Even so, given typical chemical Isp's, propellant for correction is some 10 to 15% of gross vehicle weight. For freight that can be a cheap one-shot strap-on solid rocket. For passengers the vehicle needs to be re-entry capable in case of accidents, and gets expensive.