Here are six or so rocket science insights, some simple, some advanced. It’s a fun area of engineering that touches many areas of science and politics. Besides, some people seem to think I’m a rocket scientist.

A basic question I get asked by kids is how a rocket goes up. My answer is it does not go up. That’s mostly an illusion. The majority of the rocket — the fuel — goes down, and only the light shell goes up. People imagine they are seeing the rocket go up. Taken as a whole, fuel and shell, they both go down at 1 G: 9.8 m/s^{2}, 32 ft/sec^{2}.

Because 1 G ofupward acceleration is always lost to gravity, you need high thrust from the rocket engine, especially at the beginning when the rocket is heaviest. If your engine provides less thrust than the weight of your rocket, your rocket sits on the launch pad, and if your thrust is merely twice the weight of the rocket your waste half of your fuel doing nothing useful. Effectively, the upward acceleration of the shell, a = F/m -1G where F is the force of the engine, and m is the mass of the rocket and whatever fuel is in it, and the 1 G is the upward acceleration lost to gravity. My guess is that you want to design a rocket engine so that the upward acceleration, a, is in the range 8-10 G. This range avoids wasting lots of fuel without requiring you to build the rocket too sturdy. At a = 9G, the rocket engine force, F, has to be about 10 times the rocket weight; it also means the rocket structure must be sturdy enough to support a force of ten times the rocket weight. This can be tricky because the rocket will be the size of a small skyscraper, and the structure must be light so that the majority is fuel. It’s also tricky that this 9-11 times the rocket weight must sit on an engine that runs really hot, about 3000°C. Most engineering projects have fewer constraints than this, and are thus “not rocket science.”

Basic force balance on a rocket going up.

A space rocket has to reach very high speeds; most things that go up, come down almost immediately. You can calculate the minimum orbital speed by balancing the acceleration of gravity, 9.8 m/s^{2}, against the orbital acceleration of going around the earth, a sphere of 40,000 km in circumference (that’s how the meter was defined). Orbital acceleration, a = v^{2}/r, and r = 40,000,000 m/2π = 6,366,000m. Thus, the speed you need to stay up indefinitely is v=√(6,366,000 x 9.8) = 7900 m/s = 17,800 mph. That’s roughly Mach 35, or 35 times the speed of sound. You need some altitude too, just to keep air friction from killing you, but for most missions, the main thing you need is velocity, kinetic energy, not potential energy, as i’ll show below. If you achieve more speed than 17,800 m/s, you circle the earth higher up; this makes docking space-ships tricky, as I’ll explain also.

It turns out that kinetic energy is quite a lot more important than potential energy for sending an object into orbit, and that rockets are the only way practical to reach orbital speed; no current cannon or gun can reach Mach 35. To get a sense of the energy balance involved in rocketry, consider a one kg mass at orbital speed, 7900 m/s, and 200 km altitude. You can calculate that the kinetic energy is 31,205 kJ, while the potential energy, mgh, is only 1,960 kJ. For this orbital height, 200 km, the kinetic energy is about 16 times the potential energy. Not that it’s easy to reach 200 miles altitude, but you can do it with a sophisticated cannon, or a “simple”, one stage, V2-style rocket, you need multiple stages to reach 7,900 m/s. As a way to see this, consider that the energy content of gasoline + oxygen is about 10.5 MJ/kg (10,500 kJ/kg); this is only 1/3 of the kinetic energy of the orbital rocket, but it’s 5 times the potential energy. A fairly efficient gasoline + oxygen powered cannon could not provide orbital kinetic energy since the bullet can move no faster than the explosive vapor. In a rocket this is not a constraint since most of the mass is ejected. I’ll explain further below.

A shell fired at a 45° angle that reaches 200 km altitude would go about 800 km — the distance between North Korea and Japan, or between Iran and Israel. That would require twice as much energy as a shell fired straight up, about 4000 kJ/kg. This is a value still within the range for a (very large) cannon or a single-stage rocket. For Russia or China to hit the US would take much more velocity, and orbital, or near orbital rocketry. To reach the moon, you need more total energy, but less kinetic energy. Moon rockets have taken the approach of first going into orbit, and only later going on. While most of the kinetic energy isn’t lost, I’m still not sure it’s the best trajectory.

The force produced by a rocket is equal to the rate of mass shot out times its velocity. F = ∆(mv). To get a lot of force for each bit of fuel, you want the gas exit velocity to be as fast as possible. A typical maximum is about 2,500 m/s. Mach 10, for a gasoline – oxygen engine. The acceleration of the rocket itself is this ∆mv force divided by the total remaining mass in the rocket (rocket shell plus remaining fuel) minus 1 (gravity). Thus, if the exhaust from a rocket leaves at 2,500 m/s, and you want the rocket to accelerate upward at 9 G, you must exhaust fast enough to develop 10 G, 98 m/s^{2}. The rate of mass exhaust is the mass of the rocket times 98/2500 = .0392/second. That is, about 3.9% of the rocket mass must be ejected each second. Assuming that the fuel for your first stage engine is less than 80% of the total mass, the first stage will flare-out in about 20 seconds at this rate. Your acceleration at the end of the 20 seconds will be greater than 9G, by the way, since the rocket gets lighter as fuel is burnt. When half the weight is gone, it will be accelerating at 19 G.

If you have a good math background, you can develop a differential equation for the relation between fuel consumption and altitude or final speed. This is readily done if you know calculous, or reasonably done if you use differential methods. By either method, it turns out that, for no air friction or gravity resistance, you will reach the same speed as the exhaust when 64% of the rocket mass is exhausted. In the real world, your rocket will have to exhaust 75 or 80% of its mass as first stage fuel to reach a final speed of 2,500 m/s. This is less than 1/3 orbital speed, and reaching it requires that the rest of your rocket mass: the engine, 2nd stage, payload, and any spare fuel to handle descent (Elon Musk’s approach) must weigh less than 20-25% of the original weight of the rocket on the launch pad. This gasoline and oxygen is expensive, but not horribly so if you can reuse the rocket; that’s the motivation for NASA’s and SpaceX’s work on reusable rockets. Most orbital rocket designs require at least three stages to accelerate to the 7900 m/s calculated above, and the second stage is almost invariably lost. If you can set-up and solve the differential equation above, a career in science may be for you.

Now, you might wonder about the exhaust speed I’ve been using, 2500 m/s. If you can achieve higher speeds, the rocket design will be a lot easier, but doing so is not easy for a gasoline/ oxygen engine like Russia and the US uses currently. The heat of combustion of gasoline is 42 MJ/kg, but burning a kg of gasoline requires roughly 2.5 kg of oxygen. Thus, for a rocket fueled by gasoline + oxygen, the heat of combustion is about 10.5 MJ/kg. Now assume that the rocket engine is 30% efficient. Per unit of fuel+ oxygen mass, 1/2 v^{2} = .3 x 10,500,000; v =√6,300,000 = 2500 m/s. Higher exhaust speeds have been achieved, e.g. with hydrogen-fueled rockets. The sources of inefficiency are many including incomplete combustion in the engine, gas flow off the center-line, and friction flow in the engine and between the atmosphere and gases leaving the rocket nozzle. If you can make a reliable, higher efficiency engine, a career in engineering may be for you.

At an average acceleration of 10 G = 98 m/s^{2} and a first stage that reaches 2500 m/s you find that the first stage will burn out after 25.5 seconds. If the rocket were going straight up (bad idea), you’d find you are at an altitude of about 28.7 km. A better idea would be an average trajectory of 30°, leaving you at an altitude of 14 km or so. At that altitude you can expect to have far less air friction, and you can expect the second stage engine to be more efficient. It seems to me, you may want to wait 15 seconds or so before firing the second stage: you’ll be another few km up and it seems to me that the benefit of this altitude will be worthwhile. I guess that’s why most space launches wait a few seconds before firing the second stage.

As a final bit, I’d mentioned that docking a rocket with a space station is difficult, in part, because docking requires an increase in angular speed, w, but this generally goes along with a decrease in altitude; a counter-intuitive behavior. Setting the acceleration due to gravity equal to the angular acceleration, we find GM/r^{2} = w^{2}r, where G is the gravitational constant, and M is the mass or the earth. Rearranging, we find that w^{2} = GM/r^{3}. For high angular speed, you need small r: a low altitude. When we first went to dock a space-ship, in the early 60s, we had not realized this. When the astronauts fired the engines to dock, they found that they’d accelerate in velocity, but not in angular speed: v = wr. The faster they went, the higher up they went, but the lower the angular speed got: the fewer the orbits per day. Eventually they realized that, to dock with another ship or a space-station that is in front of you, you do not accelerate, but decelerate. When you decelerate you lose altitude and gain angular speed: you catch up with the station, but at a lower altitude. Your next step is to angle your ship near-radially to the earth, and accelerate by firing engines to the side till you dock. Like much of orbital rocketry, it’s simple, but not intuitive or easy.

Robert Buxbaum, August 12, 2015. A cannon that could reach from North Korea to Japan, say, would have to be on the order of 10 km long, running along the slope of a mountain. Even at that length, the shell would have to fire at 450 G, or so, and reach a speed about 3000 m/s, or 1/3 orbital.

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