This project will examine the possibility of using a fusion reactor to propel a spacecraft. Although fusion reactors are still being experimentally developed, they are a technology that shows promise for the future. We will examine a Tokamak fusion reactor, as these have had the greatest success. For the purposes of this project, we will assume that the reactor design has been perfected and is able to sustain fusion and produce the predicted energy. This is not supposed to be realistic. It is merely a thought experiment designed to explore a new application of fusion reactors.
How Current Tokamak Reactors Work
The Reaction
Tokamak reactors produce energy through nuclear fusion. The necessary ingredients for a Tokamak reactor are Deuterium and Tritium. These are isotope of Hydrogen with 1 and 2 neutrons, respectively. Deuterium is plentiful in our oceans. Tritium must be created by bombarding lithium, also widely available, with energetic neutrons. In order to fuse, the repulsive Coulombic forces of the deuterium and tritium nuclei must be overcome. To do so, they must be heated up to about 1.5 million ºC. A gas of this temperature is ionized and is referred to as a plasma. When Deuterium and Tritium are heated to this temperature, they will fuse, producing an αparticle (He nucleus), a neutron, and 17.6 MeV of energy, as seen in the equation below.
D + T → He + n + 17.6MeV The 17.6MeV manifests itself as kinetic energy in both the αparticle and neutron, as shown below. Note that although the neutron gets most of the energy (80%), conservation of momentum dictates that each gets half of the momentum.
Figure 1. Fusion of deuterium and tritium to produce 4He and a neutron with their corresponding energies [3].
The Reactor
In order to produce the temperatures necessary to initiate and sustain fusion, the plasma must be confined. There are currently two methods of doing this: inertial confinement and magnetic confinement. Inertial confinement uses laser or ion beams to heat the plasma. Tokamak reactors use magnetic confinement which uses a torus of magnetic and electric fields to compress and heat the plasma, lasers and ion beams. This shape allows for stabilization of magnetic fields while simultaneously keeping the plasma confined.
There are two magnetic fields which constrain the plasma in the torus. The first is a toroidal field which coils around the tokamak to keep the hot plasma away from the walls which could not withstand the heat. The second is a poloidal field created by a current running through the plasma which causes the plasma to spiral on itself reducing instabilities in the torus.
There is a third magnetic field which induces current in the plasma. The combination of all three fields helps to stabilize and confine the plasma.
Figure 2. Cutaway of torus within a tokamak reactor. [4]
Figure 3. Addition of the three magnetic fields within a tokamak reactor. [12]
ITER
ITER is an experimental tokamak fusion reactor currently being built in Cadarach, France. (ITER originally stood for International Thermonuclear Experimental Reactor, but the acronym has been dropped due to uneasiness with the words "experimental" and "nuclear" in the title.) ITER is the most advanced tokamak reactor on earth, so it was a useful resource to gain information on what is currently possible. Some statistics [7]:
Weight: 23,000 tons (3x the Eiffel Tower)
Cost: €12.8B ≈ 18.5 USD
Energy: 50MW in, 500 MW out
Burn fraction: ≈ 1% [9]
The burn fraction represents the fraction of DT particles that actually fuse.
How Ours Will Work
In a traditional tokamak design, the plasma is contained within a vacuum vessel. This vessel is lined with a shielding blanket (Figure 4) that protects the rest of the reactor from the heat and neutrons of the fusion reaction. Our design will essentially involve strapping a tokamak reactor to the back of a spacecraft so that the plane of the torus is perpendicular to the direction of motion. The back half of the tokamak will be opened up. That is, there will be no vacuum vessel or shielding plates  the plasma will be open to the vacuum of space. This is a great advantage as it not only simplifies the design by eliminating The need for pumps!, but it allows much of the heat of fusion to be immediately radiated away. Figure 4. Shielding blanket for the ITER Tokamak reactor. Composed of beryllium, copper, and stainless steel [1].
Thrust
Normally, the heat generated within these blanket panels is used to produce electricity. However, our design will generate thrust through two means: the expulsion of the hot products of fusion (alpha particles) out the back, through diverter plates, and the capture of neutronsat the front Or expulsion of half the neutrons out the back. The alpha particles that result from the fusion reaction are no longer of any use, as they are more difficult to fuse and get in the way of DT fusion. Thus, it is beneficial to expel them. Fortunately, they are positive charge allows them to be manipulated by the magnetic fields. When properly constructed, the magnetic fields can be used to channel the waste alpha particles out through diverter plates at the bottom of the reactor (Figures 5 & 6). The spacecraft gains thust through the expulsion of these high speed particles.
The neutrons are more difficult to control, as they have no charge. Thus, we will only gain thrust by capturing those with forward momentum. To do so, we will leave the front of shielding plates in place. This has the benefit of both protecting the rest of the spacecraft from high speed neutrons and allowing the creation of tritium. The shielding plates contain lithium, which, when bombarded with high speed neutrons, Transmutes tobreaks down into tritium. Tritium has a halflife of only 12 years, so it is beneficial to store it as lithium, which is much more stable, and only break it into tritium as needed.
Figure 5. Crosssection of the plasma within a tokamak reactor. Notice that the spent plasma can escape out the bottom. [10]
Figure 6. Closeup crosssection of a diverter plate. The alpha particles are flung to the outside of the torus where they are ejected through the diverter plates. [11]
Assumptions
100% fusion (Burn fraction = 100%)
Long term sustainable fusion
Spacecraft can cool itself
No momentum loss from captured protons I don't understand. Capturing neutrons that are moving forward should provide momentum in the forward direction.
100% momentum gain from all alpha particles
Currently, tokamak reactors only achieve a burn fraction of about 1% (This is what ITER is projected to achieve). In order to optimize our system, simplify calculations, and demonstrate the best possible case for fusion, we are assuming a burn fraction of 100%. Additionally, as of now fusion has only been sustained on the order of seconds, and even ITER only projects to sustain fusion for around 15 minutes. Cooling would be a significant engineering problem for the project. The problem is magnified by the fact that because the ship is in space, everything must be cooled radiatively. This is a problem we have not tackled but have assumed can be accomplished. In order to model the bestcase scenario for the hydrogen scoop, we have assumed that captured protons don’t slow us down at all. Finally, we have assumed that all alpha particles ejected out the back contribute 100% of their momentum in our direction of motion. Realistically, there are two streams of alpha particles ejected at roughly 90° to each other, so not all of their momentum is in the direction of travel. However, this could be accomplished through a sort or magnetic magic?.
Spacecraft Parameters
Weight
We have used a spacecraft weight of 2,000,000 kg. This is based two other proposed fusion spacecrafts. The first is Daedalus which has a a dry weight of 1,690,000kg. Daedalus was designed to use inertial confinement fusion, so it doesn’t have to deal with the weight of the huge magnets necessary in a tokamak. The second is Discovery II which was designed to use a tokamak reactor and weighs 883,000 kg. This is significantly lighter than ITER’s 23,000,000 kg. We assume that this is likely due to the fact that no effort was made to minimize ITER’s weight. They may also have assumed a smaller reactor. Thus, based on these other designs, we feel our estimate of 2,000,000 kg is fairly conservative. As a reference point, the space shuttle weighs only 110,000 kg.
The mass of the propellant required to reach 10% the speed of light is 24,000,000 kg. Calculations shown in Performance section.
Temperature In order to obtain fusion in a vacuum, we must achieve a temperature of at least 100 million Kelvin. ITER is designed to achieve 150 million Kelvin. However, this is with only a 1% burn fraction. With our 100% burn fraction, we expect to achieve temperatures in excess of 1010 K ?????.
Velocity
We have chosen a desired velocity of 0.1c. This seemed like a reasonable speed for a deep space mission. For example, Daedalus is designed to reach a top speed of .12c What is Daedalus?. As we accelerate to higher speeds, we get decreasing returns for the amount of fuel we must burn. We also run into relativistic effects. At 0.1c, we can reach the nearest star, Alpha Centauri, within a human lifetime (42 years).
Size We did not concern ourselves with the size of our spacecraft, since it was unnecessary for most physics calculations. However, we imagine that the size of the tokamak reactor and its fuel will be insignificant compared to that of the radiative panels necessary to cool the reactor. As a means of reference, a the dimensions for Discovery II What's Discovery II?can be seen below. (Note that Discovery II was designed to refuel its DT tanks partway through its mission, such as on Titan, one of Saturn’s moons. Our fuels tanks would likely be much larger than those pictured.)
Figure 7. Proposed design for Discovery II [6].
Performance
Our mission objective is to accelerate to a speed of 10% the speed of light; 30,000,000 m/s. The Tsiolkovsky Rocket Equation relates the maximum change in velocity to the exhaust velocity, the initial mass and the final mass.
Δv = ve*ln(m0/m1)
The ratio between the initial mass and the final mass is known as the mass ratio, which relates the empty weight of the spacecraft to the fuel weight.
Using the rocket equation we can calculate how much fuel is required to achieve the desired increase in velocity. Note that it does not matter how fast the rocket accelerates, higher accelerations required a higher mass flow rate but not more fuel. The exhaust velocity can easily be calculated by knowing the temperature of the plasma in the tokamak or by knowing the energy of the alpha particles. Assuming 100% fusion rate means there is no diffusion of heat because every particle undergoes fusion and generates 3.5 MeV of energy. Having a burn rate lower than 100% would mean that the energy produced by one fusion will diffuse to the particles which did not fuse, lowering the temperature and exhaust velocity.
T = ⅔ *E*n / k
Where k is the Bolzman constant, T is temperature, E is energy and n is the burn fraction. For a single fusion reaction the energy of an alpha particle is 3.5MeV. For plasma with 100% fusion each particle has 3.5 MeV of energy and the temperature is uniform. For plasma with less than 100% fusion the energy must be multiples by a percentage factor to account for diffusion of energy.
The momentum and velocity of each particle can easily be calculated once the energy/temperature is known.
E = m*v^2
p = m*v
Where m is the mass of the alpha particle, v is the velocity and p is the momentum. The exhaust velocity can the be put in terms of the energy of the plasma.
ve= sqrt(3*k*T/m)
The exhaust velocity for the tokamak plasma engine is in the order of 10 million meters per second. This is four orders of magnitude higher than liquid propellant rockets.
The thrust is determined by the size of the engine and how much DT it can fuse and expell. The thrust is a parameter we picked based on the acceleration desired for the rocket from Newton’s second law. Using the following equation will yield the mass flow rate of the engine.
F=mdot*ve
Where mdot is the mass flow rate and F is the thrust force.
There were two design conditions studied for this rocket engine, one where the average acceleration was 1g (9.8 m/s^2) and one where the maximum acceleration was 1g. The table below shows all the parameters calculated from the equations above.
Table 1: Performance of Plasma Engine
Acceleration Avg
1 G
0.22 G
Max Acceleration
5.1 G
1 G
Thrust Needed
90,000,000 N
20,000,000 N
Time to 0.1c
37 days
164 days
Distance 1st yr
940 billion km 0.09 light years
660 billion km 0.06 light years
Mass Flow Rate of DT
7.7 kg/s
1.7 kg/s
Mass of Fuel
24,000,000 kg
24,000,000 kg
With these values it was possible to calculate the mass, acceleration, velocity, and distance as a function of time using basic kinematic equations. The results are shown in the images below. It can be seen that in both cases the velocity of 0.1 c is reached in the first 6 months. It can be seen how the acceleration increases even though the thrust remains constant because the mass is decreasing considerably as the propellant is ejected. Also to note the order of magnitude of the distance traveled, Neptune is only 4 trillion meters from the sun (4e12 m), which means that in both configurations the space craft leaves our solar system within the first weeks. Figure 8: Mission Analysis for 1G average engine.
Figure 9: Mission Analysis for 1G max engine.
Comparison with Conventional Rockets
The performance numbers for this fusion rocket are mesmerizing and at first they make you wonder how realistic they are. The best way is to compare the performance of this thrust engine with the most common and tested rocket engines in use today, those of the space shuttle. Below is a table comparing the tokamak plasma engine with the Space Shuttle Main Engines (SSME), three of which are present on the shuttle, and the Space Shuttle Solid Rocket Boosters (SSSRB), two of which are used. Remember that the overall performance characteristics of the space shuttle is the sum of the two columns since, at take off, all 3 SSME and all 2 SSSRB are ignited.
Table 2: Comparison between fusion, liquid and solid rockets
Tokamak Fusion Engine
Space Shuttle Main Engine (x3)
Space Shuttle Solid Rocket Boosters (x2)
Isp
1,300,000 s
455 s
360 s
Thrust
90,000,000 N20,000,000 N
5,500,000 N
24,000,000 N
Exit Velocity
10,000,000 m/s
2,200 m/s

m dot
7.7 kg/s1.7 kg/s
1,500 kg/s
8,100 kg/s
Propellant Mass
24,000,000 kg
720,000 kg
1,000,000 kg
The first thing to note is the thrust, which is in the same order of magnitude as the space shuttle. The main difference is the incredible superiority of the specific impulse for the fusion engine which is due to the significant increase of the exit velocity. The specific impulse, Isp, is a key parameter in rocket design which describes the efficiency of a rocket to transform propellant weight into momentum. A higher Isp means that more momentum is generated per unit mass of propellant.
Isp = p/(m*g)
Where p is the momentum of the particle, m is the mass of the propellant and g is the acceleration due to gravity. For a fusion engine, p is the momentum of the postfusion alpha particle, and m is the mass of the prefusion DT because the mass changes and a neutron is lost during the nuclear fusion. In regular rockets, where mass is conserved, the mass in the numerator and denominator cancel simplifying the equation to Isp = ve/g, where ve is the exit velocity and g is the gravitational pull on earth.
Because of the considerably higher exit velocity and Isp, the tokamak plasma engine needs to burn much less fuel to obtain the same thrust. This can be seen by comparing the mass flow rates, which are three orders of magnitude lower. This will allow the fusion rocket to burn for much longer periods of time without having to bring around excessive amounts of propellant. Finally we can see that the propellant needed to reach 0.1 c with fusion is just one order of magnitude larger than that used by the space shuttle to reach lower earth orbit (LEO).
Burn rate comparison
Our tokamak model assumes 100% of the DT propellant fuses which is a reasonable assumption for a fully matured fusion technology. Currently, ITER expects to achieve 1% burn rate being the first self sustained tokamak. We ran some numbers to understand what impact having only 1% burn rate would have on the performance.
With 1% fusion, it was found to be impossible to reach velocities of 10% the speed of light. If maintaining the same thrust of 90 million Newtons and reducing the burn rate we came up with the following numbers. The fuel needed to reach such speed would be 3x10118 kg, the mass of the Milky Way is 1.4x1042 kg. Because of this inconceivable mass it would take forever to accelerate, in fact it would travel 1093 meters in the first year, that is less than the size of an electron by multiple orders of magnitude. It was then assumed that we would only accelerate to 1% the speed of light, and adjusted the thrust accordingly, which gave more reasonable results. The mass flow rate was 8.6 kg/s and the mass of the propellant was 20 million kg. These numbers are similar to those obtained before but the space craft will be traveling ten times slower traveling only 0.01 lightyears every year.
Hydrogen Scoop
One of the considerations of this project is to examine the possibility of scooping up hydrogen as our craft travels through space. The formula to calculate the rate at which we'll pick up hydrogen is fairly simple: ScoopRate = ρ * A * v * m[mass/s] = [protons/m3] [Area] [velocity[ [mass/proton]ρ = density of protons in space
A = area of the scoop
v = velocity of our spacecraft (m/s)
m = mass of a proton = 1.67e27 kg For proton density, we'll use average values for different areas of space, as seen in Table 1 below.
Table 3: Hydrogen Density in Universe
Density (protons/m3)
Interplanetary space
5,000,00010,000,000
Intergalactic space
10010,000
Average universe
10100
With these considerations we can say that the only place where it is reasonable to scoop up Hydrogen is within the Solar System. With the current design the plasma engine will blast out of this region within the first few days or weeks of ignition. Thus the idea of scooping up hydrogen is unrealistic for a spacecraft with such thrust and speed because it will only pick up hydrogen for a very short portion of it’s flight.
The analysis to determine the size of the scoop was done assuming a constant hydrogen density equal to that within the solar system to obtain some tangible numbers. We compared scoops with three frontal areas; the size of Manhattan, the size of California, the crosssectional area of the Moon. These three size are very large and most likely impossible to obtain. Creating solid scoops of this diameter would be a feat in itself, requiring vasts amount of metal considerably increasing the weight of the space craft by more than the propellant weight. Another option is to create a magnetic scoop, but even this option would require massive magnets and a lot of electricity to power them putting a big strain on the propulsion system. Another consideration to make is that for every hydrogen we capture it will hit us at 3x107 m/s creating a considerable loss in momentum and effectively creating a lot of drag. Figure 10: Scoop performance for scoop sizes: Moon (red), California (green), Manhattan (blue) Below we are three graphs showing how much hydrogen the three scoops would be able to pick up. Remember that the tokamak burns between 110 kg/s of hydrogen, it is thus expected for the scoop to collect an equal amount of hydrogen. The xaxis is the velocity of the space craft, at higher velocities the scoop will collect more hydrogen but will also be farther away from hydrogen rich areas of the universe. It can be seen that the only scoop capable of reaching 1 kg/s is that the size of the Moon, but does so only after several months. At the max speed of 0.1 c, all three scoops collect very little hydrogen compared to what is burned by the 90 million N thruster. These graphs show that enormous scoops are required to pick up considerable amounts of hydrogen in space. If we then consider that the density of hydrogen drops by a factor of one thousand outside the solar system, the idea of scooping hydrogen becomes even more unrealistic. The size, weight and power requirements for these scoops would negate any advantage of not having propellant tanks on board.
Cost The cost for such an engine can not be accurately estimated. The technology required is decades away and millions of dollars worth of research must be spent. As of right now, its worth priceless. The best estimate we can do is based on “a long time ago in a galaxy far, far away...” approximately 1000 years before the battle of Yavin. I would say this engine could cost about 1,000,000,000 Republic Dataries, comparable in cost with the Millennium Falcon.
Appendix  MATLAB Code
MATLAB code used for calculations and graphs.
close all
clear all
clc
c = 3e8;
mDT = 8.3e27; %Mass of D+T
k=1.38e23;
n = 1
Temp = 5.6e13*2/3/k*n
E = 3/2*k*Temp;
p = sqrt(2*mDT*E);
Ve = p/mDT;
Msc = 2000000; %Mass of spacecraft
delta_v = .1*c;
m_ratio = exp(delta_v/Ve);
Mf = m_ratio*MscMsc
Mtot = Mf+Msc;
T = 2e7; % Thrust
m_dot = T/Ve %kg of D+T per second
%% Time calculations
t = linspace(0, 365*24*3600, 1001);
i = 1;
m_t = Mtot;
for i = 1:length(t)
if m_t > Msc
m_t(i) = Mtot  m_dot.*t(i);
a_t(i) = T./m_t(i);
g_avg = sum(a_t)/i/9.8;
k = i;
else
m_t(i) = m_t(i1);
a_t(i) = 0;
end
end
Tokamak Propulsion Engine
Project Overview
This project will examine the possibility of using a fusion reactor to propel a spacecraft. Although fusion reactors are still being experimentally developed, they are a technology that shows promise for the future. We will examine a Tokamak fusion reactor, as these have had the greatest success. For the purposes of this project, we will assume that the reactor design has been perfected and is able to sustain fusion and produce the predicted energy. This is not supposed to be realistic. It is merely a thought experiment designed to explore a new application of fusion reactors.
How Current Tokamak Reactors Work
The Reaction
Tokamak reactors produce energy through nuclear fusion. The necessary ingredients for a Tokamak reactor are Deuterium and Tritium. These are isotope of Hydrogen with 1 and 2 neutrons, respectively. Deuterium is plentiful in our oceans. Tritium must be created by bombarding lithium, also widely available, with energetic neutrons. In order to fuse, the repulsive Coulombic forces of the deuterium and tritium nuclei must be overcome. To do so, they must be heated up to about 1.5 million ºC. A gas of this temperature is ionized and is referred to as a plasma. When Deuterium and Tritium are heated to this temperature, they will fuse, producing an αparticle (He nucleus), a neutron, and 17.6 MeV of energy, as seen in the equation below.D + T → He + n + 17.6MeV
The 17.6MeV manifests itself as kinetic energy in both the αparticle and neutron, as shown below. Note that although the neutron gets most of the energy (80%), conservation of momentum dictates that each gets half of the momentum.
Figure 1. Fusion of deuterium and tritium to produce 4He and a neutron with their corresponding energies [3].
The Reactor
In order to produce the temperatures necessary to initiate and sustain fusion, the plasma must be confined. There are currently two methods of doing this: inertial confinement and magnetic confinement. Inertial confinement uses laser or ion beams to heat the plasma. Tokamak reactors use magnetic confinement which uses a torus of magnetic and electric fields to compress and heat the plasma, lasers and ion beams. This shape allows for stabilization of magnetic fields while simultaneously keeping the plasma confined.There are two magnetic fields which constrain the plasma in the torus. The first is a toroidal field which coils around the tokamak to keep the hot plasma away from the walls which could not withstand the heat. The second is a poloidal field created by a current running through the plasma which causes the plasma to spiral on itself reducing instabilities in the torus.
There is a third magnetic field which induces current in the plasma. The combination of all three fields helps to stabilize and confine the plasma.
Figure 2. Cutaway of torus within a tokamak reactor. [4]
Figure 3. Addition of the three magnetic fields within a tokamak reactor. [12]
ITER
ITER is an experimental tokamak fusion reactor currently being built in Cadarach, France. (ITER originally stood for International Thermonuclear Experimental Reactor, but the acronym has been dropped due to uneasiness with the words "experimental" and "nuclear" in the title.) ITER is the most advanced tokamak reactor on earth, so it was a useful resource to gain information on what is currently possible.
Some statistics [7]:
The burn fraction represents the fraction of DT particles that actually fuse.
How Ours Will Work
In a traditional tokamak design, the plasma is contained within a vacuum vessel. This vessel is lined with a shielding blanket (Figure 4) that protects the rest of the reactor from the heat and neutrons of the fusion reaction. Our design will essentially involve strapping a tokamak reactor to the back of a spacecraft so that the plane of the torus is perpendicular to the direction of motion. The back half of the tokamak will be opened up. That is, there will be no vacuum vessel or shielding plates  the plasma will be open to the vacuum of space. This is a great advantage as it not only simplifies the design by eliminating The need for pumps!, but it allows much of the heat of fusion to be immediately radiated away.
Figure 4. Shielding blanket for the ITER Tokamak reactor. Composed of beryllium, copper, and stainless steel [1].
Thrust
Normally, the heat generated within these blanket panels is used to produce electricity. However, our design will generate thrust through two means: the expulsion of the hot products of fusion (alpha particles) out the back, through diverter plates, and the capture of neutrons at the front Or expulsion of half the neutrons out the back. The alpha particles that result from the fusion reaction are no longer of any use, as they are more difficult to fuse and get in the way of DT fusion. Thus, it is beneficial to expel them. Fortunately, they are positive charge allows them to be manipulated by the magnetic fields. When properly constructed, the magnetic fields can be used to channel the waste alpha particles out through diverter plates at the bottom of the reactor (Figures 5 & 6). The spacecraft gains thust through the expulsion of these high speed particles.The neutrons are more difficult to control, as they have no charge. Thus, we will only gain thrust by capturing those with forward momentum. To do so, we will leave the front of shielding plates in place. This has the benefit of both protecting the rest of the spacecraft from high speed neutrons and allowing the creation of tritium. The shielding plates contain lithium, which, when bombarded with high speed neutrons, Transmutes to breaks down into tritium. Tritium has a halflife of only 12 years, so it is beneficial to store it as lithium, which is much more stable, and only break it into tritium as needed.
Figure 5. Crosssection of the plasma within a tokamak reactor. Notice that the spent plasma can escape out the bottom. [10]
Figure 6. Closeup crosssection of a diverter plate. The alpha particles are flung to the outside of the torus where they are ejected through the diverter plates. [11]
Assumptions
Currently, tokamak reactors only achieve a burn fraction of about 1% (This is what ITER is projected to achieve). In order to optimize our system, simplify calculations, and demonstrate the best possible case for fusion, we are assuming a burn fraction of 100%. Additionally, as of now fusion has only been sustained on the order of seconds, and even ITER only projects to sustain fusion for around 15 minutes. Cooling would be a significant engineering problem for the project. The problem is magnified by the fact that because the ship is in space, everything must be cooled radiatively. This is a problem we have not tackled but have assumed can be accomplished. In order to model the bestcase scenario for the hydrogen scoop, we have assumed that captured protons don’t slow us down at all. Finally, we have assumed that all alpha particles ejected out the back contribute 100% of their momentum in our direction of motion. Realistically, there are two streams of alpha particles ejected at roughly 90° to each other, so not all of their momentum is in the direction of travel. However, this could be accomplished through a sort or magnetic magic?.
Spacecraft Parameters
Weight
We have used a spacecraft weight of 2,000,000 kg. This is based two other proposed fusion spacecrafts. The first is Daedalus which has a a dry weight of 1,690,000kg. Daedalus was designed to use inertial confinement fusion, so it doesn’t have to deal with the weight of the huge magnets necessary in a tokamak. The second is Discovery II which was designed to use a tokamak reactor and weighs 883,000 kg. This is significantly lighter than ITER’s 23,000,000 kg. We assume that this is likely due to the fact that no effort was made to minimize ITER’s weight. They may also have assumed a smaller reactor. Thus, based on these other designs, we feel our estimate of 2,000,000 kg is fairly conservative. As a reference point, the space shuttle weighs only 110,000 kg.The mass of the propellant required to reach 10% the speed of light is 24,000,000 kg. Calculations shown in Performance section.
Temperature
In order to obtain fusion in a vacuum, we must achieve a temperature of at least 100 million Kelvin. ITER is designed to achieve 150 million Kelvin. However, this is with only a 1% burn fraction. With our 100% burn fraction, we expect to achieve temperatures in excess of 1010 K ?????.
Velocity
We have chosen a desired velocity of 0.1c. This seemed like a reasonable speed for a deep space mission. For example, Daedalus is designed to reach a top speed of .12c What is Daedalus?. As we accelerate to higher speeds, we get decreasing returns for the amount of fuel we must burn. We also run into relativistic effects. At 0.1c, we can reach the nearest star, Alpha Centauri, within a human lifetime (42 years).Size
We did not concern ourselves with the size of our spacecraft, since it was unnecessary for most physics calculations. However, we imagine that the size of the tokamak reactor and its fuel will be insignificant compared to that of the radiative panels necessary to cool the reactor. As a means of reference, a the dimensions for Discovery II What's Discovery II? can be seen below. (Note that Discovery II was designed to refuel its DT tanks partway through its mission, such as on Titan, one of Saturn’s moons. Our fuels tanks would likely be much larger than those pictured.)
Figure 7. Proposed design for Discovery II [6].
Performance
Our mission objective is to accelerate to a speed of 10% the speed of light; 30,000,000 m/s. The Tsiolkovsky Rocket Equation relates the maximum change in velocity to the exhaust velocity, the initial mass and the final mass.
Δv = ve*ln(m0/m1)
The ratio between the initial mass and the final mass is known as the mass ratio, which relates the empty weight of the spacecraft to the fuel weight.
m0/m1 = (empty weight + propellant weight) / (empty weight)
Using the rocket equation we can calculate how much fuel is required to achieve the desired increase in velocity. Note that it does not matter how fast the rocket accelerates, higher accelerations required a higher mass flow rate but not more fuel. The exhaust velocity can easily be calculated by knowing the temperature of the plasma in the tokamak or by knowing the energy of the alpha particles. Assuming 100% fusion rate means there is no diffusion of heat because every particle undergoes fusion and generates 3.5 MeV of energy. Having a burn rate lower than 100% would mean that the energy produced by one fusion will diffuse to the particles which did not fuse, lowering the temperature and exhaust velocity.
T = ⅔ *E*n / k
Where k is the Bolzman constant, T is temperature, E is energy and n is the burn fraction. For a single fusion reaction the energy of an alpha particle is 3.5MeV. For plasma with 100% fusion each particle has 3.5 MeV of energy and the temperature is uniform. For plasma with less than 100% fusion the energy must be multiples by a percentage factor to account for diffusion of energy.
The momentum and velocity of each particle can easily be calculated once the energy/temperature is known.
E = m*v^2
p = m*v
Where m is the mass of the alpha particle, v is the velocity and p is the momentum. The exhaust velocity can the be put in terms of the energy of the plasma.
ve= sqrt(3*k*T/m)
The exhaust velocity for the tokamak plasma engine is in the order of 10 million meters per second. This is four orders of magnitude higher than liquid propellant rockets.
The thrust is determined by the size of the engine and how much DT it can fuse and expell. The thrust is a parameter we picked based on the acceleration desired for the rocket from Newton’s second law. Using the following equation will yield the mass flow rate of the engine.
F=mdot*ve
Where mdot is the mass flow rate and F is the thrust force.
There were two design conditions studied for this rocket engine, one where the average acceleration was 1g (9.8 m/s^2) and one where the maximum acceleration was 1g. The table below shows all the parameters calculated from the equations above.
Table 1: Performance of Plasma Engine
0.09 light years
0.06 light years
With these values it was possible to calculate the mass, acceleration, velocity, and distance as a function of time using basic kinematic equations. The results are shown in the images below. It can be seen that in both cases the velocity of 0.1 c is reached in the first 6 months. It can be seen how the acceleration increases even though the thrust remains constant because the mass is decreasing considerably as the propellant is ejected. Also to note the order of magnitude of the distance traveled, Neptune is only 4 trillion meters from the sun (4e12 m), which means that in both configurations the space craft leaves our solar system within the first weeks.
Figure 8: Mission Analysis for 1G average engine.
Figure 9: Mission Analysis for 1G max engine.
Comparison with Conventional Rockets
The performance numbers for this fusion rocket are mesmerizing and at first they make you wonder how realistic they are. The best way is to compare the performance of this thrust engine with the most common and tested rocket engines in use today, those of the space shuttle. Below is a table comparing the tokamak plasma engine with the Space Shuttle Main Engines (SSME), three of which are present on the shuttle, and the Space Shuttle Solid Rocket Boosters (SSSRB), two of which are used. Remember that the overall performance characteristics of the space shuttle is the sum of the two columns since, at take off, all 3 SSME and all 2 SSSRB are ignited.
Table 2: Comparison between fusion, liquid and solid rockets
The first thing to note is the thrust, which is in the same order of magnitude as the space shuttle. The main difference is the incredible superiority of the specific impulse for the fusion engine which is due to the significant increase of the exit velocity. The specific impulse, Isp, is a key parameter in rocket design which describes the efficiency of a rocket to transform propellant weight into momentum. A higher Isp means that more momentum is generated per unit mass of propellant.
Isp = p/(m*g)
Where p is the momentum of the particle, m is the mass of the propellant and g is the acceleration due to gravity. For a fusion engine, p is the momentum of the postfusion alpha particle, and m is the mass of the prefusion DT because the mass changes and a neutron is lost during the nuclear fusion. In regular rockets, where mass is conserved, the mass in the numerator and denominator cancel simplifying the equation to Isp = ve/g, where ve is the exit velocity and g is the gravitational pull on earth.
Because of the considerably higher exit velocity and Isp, the tokamak plasma engine needs to burn much less fuel to obtain the same thrust. This can be seen by comparing the mass flow rates, which are three orders of magnitude lower. This will allow the fusion rocket to burn for much longer periods of time without having to bring around excessive amounts of propellant. Finally we can see that the propellant needed to reach 0.1 c with fusion is just one order of magnitude larger than that used by the space shuttle to reach lower earth orbit (LEO).
Burn rate comparison
Our tokamak model assumes 100% of the DT propellant fuses which is a reasonable assumption for a fully matured fusion technology. Currently, ITER expects to achieve 1% burn rate being the first self sustained tokamak. We ran some numbers to understand what impact having only 1% burn rate would have on the performance.
With 1% fusion, it was found to be impossible to reach velocities of 10% the speed of light. If maintaining the same thrust of 90 million Newtons and reducing the burn rate we came up with the following numbers. The fuel needed to reach such speed would be 3x10118 kg, the mass of the Milky Way is 1.4x1042 kg. Because of this inconceivable mass it would take forever to accelerate, in fact it would travel 1093 meters in the first year, that is less than the size of an electron by multiple orders of magnitude.
It was then assumed that we would only accelerate to 1% the speed of light, and adjusted the thrust accordingly, which gave more reasonable results. The mass flow rate was 8.6 kg/s and the mass of the propellant was 20 million kg. These numbers are similar to those obtained before but the space craft will be traveling ten times slower traveling only 0.01 lightyears every year.
Hydrogen Scoop
One of the considerations of this project is to examine the possibility of scooping up hydrogen as our craft travels through space. The formula to calculate the rate at which we'll pick up hydrogen is fairly simple:
ScoopRate = ρ * A * v * m[mass/s] = [protons/m3] [Area] [velocity[ [mass/proton]ρ = density of protons in space
A = area of the scoop
v = velocity of our spacecraft (m/s)
m = mass of a proton = 1.67e27 kg
For proton density, we'll use average values for different areas of space, as seen in Table 1 below.
Table 3: Hydrogen Density in Universe
With these considerations we can say that the only place where it is reasonable to scoop up Hydrogen is within the Solar System. With the current design the plasma engine will blast out of this region within the first few days or weeks of ignition. Thus the idea of scooping up hydrogen is unrealistic for a spacecraft with such thrust and speed because it will only pick up hydrogen for a very short portion of it’s flight.
The analysis to determine the size of the scoop was done assuming a constant hydrogen density equal to that within the solar system to obtain some tangible numbers. We compared scoops with three frontal areas; the size of Manhattan, the size of California, the crosssectional area of the Moon. These three size are very large and most likely impossible to obtain. Creating solid scoops of this diameter would be a feat in itself, requiring vasts amount of metal considerably increasing the weight of the space craft by more than the propellant weight. Another option is to create a magnetic scoop, but even this option would require massive magnets and a lot of electricity to power them putting a big strain on the propulsion system. Another consideration to make is that for every hydrogen we capture it will hit us at 3x107 m/s creating a considerable loss in momentum and effectively creating a lot of drag.
Figure 10: Scoop performance for scoop sizes: Moon (red), California (green), Manhattan (blue)
Below we are three graphs showing how much hydrogen the three scoops would be able to pick up. Remember that the tokamak burns between 110 kg/s of hydrogen, it is thus expected for the scoop to collect an equal amount of hydrogen. The xaxis is the velocity of the space craft, at higher velocities the scoop will collect more hydrogen but will also be farther away from hydrogen rich areas of the universe. It can be seen that the only scoop capable of reaching 1 kg/s is that the size of the Moon, but does so only after several months. At the max speed of 0.1 c, all three scoops collect very little hydrogen compared to what is burned by the 90 million N thruster.
These graphs show that enormous scoops are required to pick up considerable amounts of hydrogen in space. If we then consider that the density of hydrogen drops by a factor of one thousand outside the solar system, the idea of scooping hydrogen becomes even more unrealistic. The size, weight and power requirements for these scoops would negate any advantage of not having propellant tanks on board.
Cost
The cost for such an engine can not be accurately estimated. The technology required is decades away and millions of dollars worth of research must be spent. As of right now, its worth priceless.
The best estimate we can do is based on “a long time ago in a galaxy far, far away...” approximately 1000 years before the battle of Yavin. I would say this engine could cost about 1,000,000,000 Republic Dataries, comparable in cost with the Millennium Falcon.
Appendix  MATLAB Code
MATLAB code used for calculations and graphs.
close all
clear all
clc
c = 3e8;
mDT = 8.3e27; %Mass of D+T
k=1.38e23;
n = 1
Temp = 5.6e13*2/3/k*n
E = 3/2*k*Temp;
p = sqrt(2*mDT*E);
Ve = p/mDT;
Msc = 2000000; %Mass of spacecraft
delta_v = .1*c;
m_ratio = exp(delta_v/Ve);
Mf = m_ratio*MscMsc
Mtot = Mf+Msc;
T = 2e7; % Thrust
m_dot = T/Ve %kg of D+T per second
%% Time calculations
t = linspace(0, 365*24*3600, 1001);
i = 1;
m_t = Mtot;
for i = 1:length(t)
if m_t > Msc
m_t(i) = Mtot  m_dot.*t(i);
a_t(i) = T./m_t(i);
g_avg = sum(a_t)/i/9.8;
k = i;
else
m_t(i) = m_t(i1);
a_t(i) = 0;
end
end
v_t = cumtrapz(t, a_t);
x_t = cumtrapz(t, v_t);
time = t(k)/3600/24
dist = max(x_t)
dist_ly = dist/10e15
g_avg
g_max = max(a_t)/9.8
%% Plot
close all
figure()
subplot(4,1,1)
plot(t/3600/24, m_t)
ylabel('mass (kg)')
% title('1G Avg')
subplot(4,1,2)
plot(t/3600/24,a_t/9.8)
ylabel('acceleration (G)')
subplot(4,1,3)
plot(t/3600/24,v_t/c)
ylabel('velocity (c)')
subplot(4,1,4)
[AX,H1,H2] = plotyy(t/3600/24,x_t, t/3600/24,x_t/10e15);
set(get(AX(1),'Ylabel'),'String','distance(m)')
set(get(AX(2),'Ylabel'),'String','distance (ly)')
xlabel('time (days)')
%% Scoop Calculations
rho = 10000; %atoms of H per m^3
A = [100 5400 59e6 4.2e11 9.5e12]; %Area (10x10, football field, manhattan, cali, moon)
scooprate = rho*A'*v_t*1.7e27;
figure(10)
plot(v_t,scooprate)
figure(11)
plot(v_t,scooprate)
figure(12)
loglog(v_t,scooprate)
axis([0,3e6,0,1])
figure(13)
plot(t,scooprate)
figure(14)
subplot(3,1,1)
semilogx(x_t,scooprate)
axis([0,2e14,0,1])
subplot(3,1,2)
semilogx(x_t,scooprate)
axis([0,2e14,0,.05])
subplot(3,1,3)
semilogx(x_t,scooprate)
axis([0,2e14,0,.000005])
References