Current mood:

accomplished

Fusion to Mars
I'm not sure exactly how I got onto the subject, but sometime last week I started thinking about intra-solar-system travel. Sticking to reaction-mass thrusters and not breaking any laws of physics, how fast would we be able to move around if we could improve the efficiency of various systems up to their theoretical limits?

The quick answer: An Earth-Mars trip could be made comfortably in 2-5 days (depending on the relative positions of Earth and Mars at the time). That's faster than current trans-oceanic shipping.

Travel Time at 1 g

I started out not worrying about how to generate the thrust and just looking at how long the trip would take at a constant 1 g of acceleration, which would make for a nice, comfortable trip for any human passengers. The distance between Earth and Mars can be anywhere from 56M to 400M km. Assuming a straight shot, accelerating at 1 g half way there, turning over at the midpoint, and decelerating at 1 g for the other half of the trip, how long does it take to cover those distances?

x = ¹/₂ a t²
(56e9m / 2) = ¹/₂ 9.81m/s² t²
t = 75554s

That's just for accelerating to the midpoint -- the other half of the trip would take equally long. So the whole trip at 1 g when Earth and Mars are closest to each other would take 151109s, or 1.75 days. For the max distance, we get

x = ¹/₂ a t²
(400e9m / 2) = ¹/₂ 9.81m/s² t²
t = 201928s

So, when Earth and Mars are farthest apart, the trip would take 403855s, or 4.67 days. For the rest of the calculations, looking at how the thrust could be generated, I assumed a trip time of 4 days, which would be enough to get to Mars or back nearly all the time. It probably wouldn't be a good idea to fly straight through the sun to get to Mars anyway...

One more note: The maximum speed achieved during those flights (at the midpoint) would be 741km/s for the 1.75-day trip, or 1981km/s for the 4.67-day trip.

Required Reaction Mass

Now, is it possible to maintain 1 g of acceleration constantly for 4 days straight? My initial guess was that it would take so much reaction mass as to be completely infeasible, but I ran the math anyway.

The best possible case is that you're ejecting your reaction mass at the speed of light, so I started there. For each kg of the mass of the ship, how much reaction mass needs to be ejected to accelerate it by 9.81m/s? Based on conservation of momentum:

mc = 1kg 9.81m/s
m = 3.27e-8kg

So it takes 32.7 micrograms of speed-of-light reaction mass to accelerate 1kg by 9.81m/s. If we use that much reaction mass every second, it will maintain a constant 1 g of acceleration.

32.7mcg/s * 4 days = 11.3g

If we can generate speed-of-light thrust, we'd only need 11.3g of reaction mass per kg of the ship for the entire 4-day trip to Mars. In other words, only about 1% of the weight of the ship would have to be devoted to reaction mass.

Of course, ejecting that mass at the speed of light would take infinite energy, so we'll have to slow it down a little bit. To keep away from any significant relativistic effects, I decided to set the thrust velocity at 0.1c. Since that's 1/10th the speed, we'd need 10x as much reaction mass to maintain the same thrust. That puts the required reaction mass at 113g per kg to be pushed -- about 11% of the ship's weight would have to be devoted to reaction mass.

Required Energy

Now, how much energy would it take to eject all that reaction mass, over the course of the entire trip, at 0.1c?

E = ¹/₂mv²
E = ¹/₂ 113g (0.1c)²
E = 5.08e13J

The most efficient energy storage possible would be as matter + antimatter, so I decided to see how much antimatter it would take to generate that much energy.

E = mc²
5.08e13J = mc²
m = 0.565g

About half a gram per kg of ship weight. And only half of that has to be antimatter -- the other half is matter. Antimatter does indeed store a fuckload of energy.

The required antimatter mass is completely insignificant. So if we had a 100% efficient engine that could take antimatter as fuel and use it to accelerate reaction mass to 0.1c, a ship would only have to devote about 11% of its weight to fuel + reaction mass for that 4-day trip to Mars.

Switching to Fusion

Antimatter can be... a bit awkward to work with. So I decided to see how well the next-best energy source would work. Each fusion reaction of ²H + ³H -> ⁴He + n releases 17.6MeV. One deuterium atom plus one tritium atom weigh a total of 5.03amu. That means that one kg of ²H + ³H fusion fuel could provide

1kg / 5.03amu * 17.6MeV = 3.38e14J

From above, we needed 5.08e13J per kg of ship weight for the 4-day trip. With a 100% efficient fusion engine, that means we'd a mass of fuel per kg of the ship equal to

5.08e13J / 3.38e14J/kg = 151g

So, in addition to the 113g of reaction mass, we'd need 151g of fusion fuel. Now the ship is devoting 264g per kg to fuel + reaction mass, or about 23% of its total weight (not 26%, since the fuel and reaction mass will be used up over the course of the flight). Still not bad at all.

Then I realized that the spent fusion fuel (⁴He + neutrons) could be used as the reaction mass, rather than carrying separate reaction mass. If we had a 100% efficient fusion engine that could take ²H + ³H as fuel and spit out the spent fuel as reaction mass, the velocity of that thrust would be

E = ¹/₂mv²
17.6MeV = ¹/₂ 5.03amu v²
v = 0.0867c

At that thrust velocity, the required mass/second to maintain 1 g of acceleration for one second would be

m * 0.0867c = 1kg 9.81m/s
m = 338mcg

Over the entire 4-day trip, that adds up to 130g of combination fuel/reaction mass. So with that 100% efficient fusion engine, you could take a 4-day trip to Mars, even when it's mostly on the other side of the sun from the Earth, and your ship would only have to devote about 12% of its mass to fuel.

That leaves a lot of room for less-than-perfect components. For example, say that the fusion engine is only 20% efficient at turning fuel into thrust. Instead of having a thrust velocity of 0.0867c, it would be 0.0388c, which would require devoting 25% of the ship's mass to fuel.

Also, a continuous burn for the entire trip is not that efficient. A lot of fuel could be saved by burning at the start and end of the trip and coasting in the middle (like modern spacecraft), though of course the trip would take longer that way.

Some Perspective

Mars is pretty far away. For comparison, with the same 100% efficient fusion drive technology, you could take your 10-ton space car to the moon in 6.6 minutes, using only 1.49kg of fuel -- about the weight of half a gallon of gas.