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The holy grail of the space industry is to reduce launch cost – the cost of launching a payload into orbit. Right now, this cost is about $1000 to $2000 per kg. The space shuttle was supposed to reduce the cost by an order of magnitude – but it didn't reduce the cost at all. Shuttle costs, averaged over the lifetime of the program, were about sixty thousand dollars per kg. If launch costs were cheap, it would be easy to put biochemical factories into orbit, for making drugs in zero-G. Or materials manufacturing facilities. Or passengers, who just wanted to look out the window and see the earth below. Or space hotels. Or facilities to build spaceships – spaceships that would in turn travel to the Moon, or to Mars, or to the Asteroid belt to do mining. Space advocates believe that, in the same way affordable desktop computers launched whole new industries, or cheap access to the World Wide Web launched new industries, cheap access to space could do the same. Then maybe we'd see – at least within a couple of decades – the Lunar bases and Mars colonies and routine spaceflights those of us who remember the first manned Moon landings thought we'd have by now. So, how realistic is this? Is it possible to cut launch costs dramatically? While one might think that the energy required to put something in orbit is the limiting factor, it's not. I'm going to show that the energy cost is trivial. To do this, I'm going to have to use some (high school senior or college freshman) physics. The math is pretty simple. Let's start with energy. To put something into orbit, we need two types of energy – potential and kinetic. Potential energy (PE) is that required to raise something into the orbit altitude. Kinetic energy (KE) is energy of motion. The faster something moves, the more kinetic energy it has. For something to stay in orbit, it has to have a velocity. Potential energy (PE) is just weight multiplied by distance upward; this is the same as mass multiplied by acceleration of gravity, multiplied by distance: If I weigh 220 pounds, and climb a ten-foot flight of stairs, my potential energy is increased by 220 pounds x 10 feet = 2200 foot-pounds. In metric units, that's 100 kilograms x 10 meters per second squared (written m/sec2) x 3 meters = 3000 kg m2/sec2, or 3000 Joules. Kinetic energy is just half the mass, multiplied by the square of the velocity: Suppose I can run a 100-yard dash in 12 seconds. This is an average speed of 25 feet per second, approximately 7.62 meters per second. Then This is about 2900 kg-m2/sec2, or 2900 Joules. In English units ½ v2 is 312 ft2/sec2. Since ½ mg is 110 lb and g is 32 ft/sec2, m is 220 lb/(32 ft/sec2), about 7 lb - sec2/ft. Then ½ m v2 is 2184 ft lbs. Suppose the orbit altitude we're interested in is 400 kilometers (about 250 miles) in altitude. Then potential energy is: where I've converted 400 km = 400,000 meters = 4x105 meters. The result is 4 x 106 kg-m2/sec2, or 4 million Joules. (This expression for the potential energy will be slightly incorrect, because gravity decreases with increasing altitude. But for the purposes of this calculation, it doesn't make any significant difference.) What's a good number for orbit velocity? Most people probably recall hearing 17,000 or 18,000 miles per hour for an orbit of 100 or 200 miles or so. This is about five miles per second, or eight kilometers per second. (For a 400 km orbit, the velocity will be less. So we will underestimate the energy.) The result is 32 million Joules. If I add together the kinetic and the potential energies, I get 36 million Joules. How much energy is this? It turns out we use Joules as a unit of measurement all the time. A Joule per second is called a Watt. That's right, the same Watt used to measure electric power. A Kilowatt-Hour is 1000 Watts for one hour, which is 1000 Joules per second for 3600 seconds, or 3.6 million Joules. So ten Kilowatt-Hours is 36 million Joules, which is enough. A Kilowatt-Hour of electricity usually costs less than a dime, so ten Kilowatt-Hours costs less than a dollar. So the energy cost of putting a kilogram (slightly more than two pounds) into orbit is less than a dollar. Now if I'd done this calculation exactly, I'd taken account of the fact that gravity decreases with altitude. This would decrease the energy slightly. I'd also have calculated the exact orbit velocity, which is less than eight km/sec. Either way, the energy would be less than 36 million Joules. Suppose I get even more picky. Suppose I say "I can't put a payload directly into orbit. I have to put a rocket, with a payload on top of it, into orbit. So I have to put the rocket mass, including fuel, into the equation as well." Even if I do this, and assume I need ten to twenty kilograms of rocket mass (including fuel) for each kilogram that goes into orbit, I can still put one kilogram into orbit for an energy cost of ten to twenty dollars. So the cost of putting stuff into orbit is unaffected by the energy cost of putting it there. Compared to the current cost of $1000 per kg, the energy cost is nothing. |
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