3 ms·
Well it's technically correct. To get a 100kg object from the surface to a 100km Karman line needs about 30kWh of power. That's the same order of magnitude as t
by iso1631 4y ago
Well it's technically correct. To get a 100kg object from the surface to a 100km Karman line needs about 30kWh of power. That's the same order of magnitude as travelling 100km sideways in an electric vehicle.
At 20c/kWh it would cost $100 to lift a 2 ton car and 4 occupants into space, or $100 each to get to the c. 400km altitude of the ISS.
Of course to stay in space requires far more energy as you also have to go sideways, fast. Very Fast.
But it's still pretty low. Sticking 1kg into Geostationary Orbit and keeping it there would take about 15kWh. A 2 ton vehicle with 4 passengers at 20c/kWh would be $1500 each.
Of course we lack a practical way of doing this other than with rockets, and thus enter the tyranny of the rocket equation. But the fuel cost is still far lower than the cost of the entire shuttle or apollo system, as shown by the efficiency of SpaceX -- fuel costs of $20/kg to LEO, or $1500 for a typical adult.
- Aperocky 4y agokWH of power and its cost does not translate here, your kWh of power does not have any specific impulse.
- randallsquared 4y agoSpace elevators and Lofstrom loops also do not have Isp (or have very, very poor Isp, if you're treating the Earth as reaction mass...).
- iso1631 4y agokWh is a measure of energy, not power. And specific impulse is only of relevance to rockets, not to the amount of energy required to lift. The problem in space access isn't an energy problem, it's a method problem.
- matheusmoreira 4y agoYou're comparing rocket engines designed to operate in a vacuum to electric vehicles designed to operate on the ground. Rocket engines need a lot more fuel to generate that amount of power.
- iso1631 4y agoNo, you're comparing the method of applying the energy required, which is painfully inefficient (rockets), to the energy actually required. But I did touch on that in the last paragraph -- the energy required on a rocket to get into LEO is $20/kg, or $2k/person. It's a tiny amount of the total cost of launch.
- pfdietz 4y agoRocket engines are actually extraordinarily efficient. They're the most efficient heat engines we have. More than 90% of the thermal energy produced in the thrust chamber is converted to kinetic energy of the exhaust jet (especially in high expansion ratio engines). The fraction of chemical energy of the propellants of a launch vehicle that ends up in the kinetic energy of the final stage is surprisingly high (well, it was to me, initially.)
- iso1631 4y agoThey are not efficient at doing their job (moving things upwards), hence we don't use them to get from the bottom of a skyscraper to the top. They have to carry their entire mass of fuel with them. A large crane is far more efficient at lifting items. But even if you accept you have to use a rocket, the cost of the fuel is trivial compared with the cost of launch.
- pfdietz 4y agoOf course they are not efficient for that use case. That use case is not the same as the use case of getting to orbit. When the delta-V involved is similar to the exhaust velocity of the rocket the overall efficiency goes way up. If you could vary the exhaust velocity of a rocket continuously, so it was equal to the total delta-V so far, the efficiency of converting jet kinetic energy to vehicle energy would be 100%, as the jet would be left stationary in the reference frame of the launcher. (This ignores gravitational potential energy and also that the mass ratio would diverge to infinity at zero velocity, but never mind that.) In practice, using a lower Isp first stage and a higher Isp upper stage partially implements this, and the overall efficiency isn't too bad.