Much of the struggle with terrestrial fusion power seems to be with keeping the ultra-hot plasma contained. In a classic Tokamak configuration, magnets using massive amounts of power, suspend the plasma in a torus that contains the incredible heat and prevents it from melting the containment apparatus. So far, this appears to be one of the great challenges and a reason why we do not currently have cheap, sustainable fusion power.

If you had a fusion reactor in geosynchronous orbit, it would seem like many of these problems of containment would be greatly simplified by weightlessness. The magnetic containment would just be needed to hold the plasma in place, but not need to hold it against gravity.

Another aspect to this might be if space elevators (aka know as 'beanstalks') are used as connecting conduits for bringing generated power back to earth.

What is the practicality of orbital fusion reactors as a source of energy and would weightlessness simplify containment challenges? Current terrestrial test fusion reactors need to be incredibly heavy and massive. Could they be lighter and simpler in space? What are the advantages to orbital fusion power plants?

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    $\begingroup$ (1) The confinement of the plasma is well understood, with multiple solutions. (2) At this stage, the major engineering problems are the heat exchangers and the evacuation of the helium produced by the reaction. (3) Gravity is not a significant force in the design of fusion reactors. (4) When designing a fusion reactor for use in space, a new major engineering problem would be cooling; the only known way to do it would be with gigantic radiators. (5) See the design of the ITER reactor for the current state of the art in fusion reactor engineering. $\endgroup$
    – AlexP
    Jan 12, 2019 at 21:10
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    $\begingroup$ It's not gravity the magnets are fighting (as WillK mentioned, there's only a few grams of gas), the magnets are fighting the extreme forces (i.e. tonnes and tonnes) that want to make the gas expand and cool down (and vaporise a millimeter or two of reactor wall) $\endgroup$
    – Samwise
    Jan 12, 2019 at 21:59
  • $\begingroup$ The Gravity force on the plasma is many,many,many,many,many millions of times weaker than their own electrostatic, thermal and centrifugal effects. many $\endgroup$
    – user79911
    Nov 4, 2020 at 15:25

3 Answers 3


I think that as regards containing hot plasma, gravity is the least of their worries. Think how much mass in in that hot plasma. Probably hardly any because the less mass there is, the easier it is to heat it to fusion.

But lets figure it out with the hard hardness of hard science! Here are stats for the EUs fusion project.


The plasma volume is 830 cubic meters. Considering that volume of hydrogen gas at 1 atmospheres and 0 C I got 66 kg. I can lift that on a good day.


But maybe this plasma is at high pressure? It looks like pressures are not super high. From 2016: http://news.mit.edu/2016/alcator-c-mod-tokamak-nuclear-fusion-world-record-1014

The team set a new world record for plasma pressure in the Institute’s Alcator C-Mod tokamak nuclear fusion reactor. Plasma pressure is the key ingredient to producing energy from nuclear fusion, and MIT’s new result achieves over 2 atmospheres of pressure for the first time.

So 830 cubic meters of plasma at 2 atmospheres. That would be double the weight or 132 kg. I would need help to lift it.

But the whole thing about a Tokumak is that it is hot; from first source, 15 million C. I am proud of the linked calculator; it would accept that value of 15,000,000 C. It gave me

= 3.2757271682719E-6kilogram/meter^3 or 0.0000032 kg / m^3. *830 m^3 that would be 0.0026 kg or 2600 mg. That is 10 grains of rice, which I can lift.

I conclude the force of gravity on the contained plasma is not much of a consideration. Really the consideration is keeping something that is that hot in a place where you can heat it up more.

I have been known to misplace a decimal here and there. Anyone feeling an itch to duplicate my math, I would like to know if I screwed something up.

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    $\begingroup$ It's actually even hotter than that! So the plasma would probably be even lighter. The Earth-bound units have to push temps right past that of the Sun's core towards more like 150 million degrees (Celsius or Fahrenheit, doesn't really matter) in order to get meaningful reaction rates, the sun can run cooler because a) it's got a bajillion tonnes weighing down on it (not sure what that works out to in elephants though...) and b) it doesn't matter if it takes 5 billion years for a couple of hydrogen atoms to fuse (afterall, it's not running on taxpayer-funded company time) $\endgroup$
    – Samwise
    Jan 12, 2019 at 21:53
  • $\begingroup$ I wouldn't necessarily trust that calculator for temperatures in the millions of degrees. Although theoretically hydrogen at high temperatures behaves most similarly to an ideal gas, 15(0) million C is likely past most experiments. Your answer still holds, though. $\endgroup$ Jan 13, 2019 at 6:52
  • $\begingroup$ I seem to recall that the JET reactor (plasma volume of 100 m^3) uses O(10 mg) of deuterium/tritium per "shot", so I guess for ITER it's probably O(100 mg). $\endgroup$ Jan 13, 2019 at 10:33

Unfortunately, it's not at all practical. The basic issue is that the reactor (at least any reactor built with foreseeable technology -- Mr. Fusion is on the far side of Clarke's Law) is very, very heavy (making orbit a Bad Place to put it since costs are still around $3000/lb to put things into low Earth orbit and several times that to GEO), while the plasma itself weighs very, very little.

The ITER plasma volume is on the order of 2000 cubic meters and the plasma density is 0.6x1020 atoms/cubic meter, so there is 1.2x1023 atoms total, which (if it's using a deuterium-tritium mixture, which is likely for the first reactors, at least) is right around 1 gram.

Besides that, waste heat dissipation in space is very difficult, and a fusion reactor will produce a lot of waste heat. (Foreseeable designs produce more waste heat than usable energy.) The only practical method for getting rid of waste heat in space is through radiators, and that would be a significant chunk of additional weight...all of which must be moved to GEO at high cost.

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    $\begingroup$ Add a sentence or two to explain what makes the rest of the reactor so heavy and expensive to put into orbit and we have a great answer. (Add a bit more about the need for heat dissipation in a vacuum and it's IMHO perfect.) $\endgroup$
    – JBH
    Jan 12, 2019 at 22:27
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    $\begingroup$ I didn't think that the high cost of putting things into orbit was needed, but the heat dissipation issue is a very good point. (I'll add both.) $\endgroup$
    – Mark Olson
    Jan 12, 2019 at 22:33

I second Willk's answer: gravity doesn't really matter at all for plasma containment, so trying to build a tokamak in orbit would be a huge complication for basically no gain. However, I just took a class on plasma physics so I would be remiss if I didn't cram a bunch more math down peoples' throats.

Now, the main equation that will be governing plasma movement in a tokomak on Earth is $$m_j n_j \frac{D\mathbf{v_j}}{Dt}=q_j n_j (\mathbf{E+v_j \times B})-\nabla p + m_j n_j \mathbf{g}$$

where $m$ is the mass per particle, $n$ the number density, $q$ the particle charge, $\mathbf{v}$ the fluid velocity, $p$ the pressure, $\mathbf{E}$ the electric field, $\mathbf{B}$ the magnetic field, $\mathbf{g}$ the gravitational field, and the subscript denoting which species we are talking about (normally ion vs electron). Now I know that was a whole bunch to dump at once, but I have a very simple goal here: to show you that that the term involving $\mathbf{g}$ (the gravitational force term) is much smaller than the other forces at play.

You see, the monstrous equation I gave is really nothing more than a dressed up version of Newton's second law: $\mathbf{F} = m \mathbf{a}$. The left hand side is called the convective derivative and decribes how the plasma is being pushed around (analogous to $\mathbf{a}$), while the right hand side lists the forces acting on the plasma. So, let's get a rough sense of the orders of magnitude that we have for the forces.

First off, we will ignore the electric field, since that tends to be approximately zero in steady state plasmas due to a phenomenon called Debye shielding. I'm also going to ignore the term involving the magnetic field because that's the thing we want to adjust.

So, we want to analyze the approximate magnitude of the term $$\nabla p$$ which for thermodynamics reasons is equivalent to $$\gamma kT \nabla n$$ The plasma recombines at the walls of the vessel, so $n=0$ there. Meanwhile, at the center of a typical fusion plasma we have a typical value of $n=10^{19} m^{-3}$, and a cross sectional radius of maybe $1 m$, giving us an approximate gradient of $10^{19} m^{-4}$. Using the approximation of an isothermal plasma with $\gamma = 1$, and ITER's projected temperature of $kT = 8 keV$, we obtain $$\nabla p \approx 13 \times 10^3 N/m^3$$

Now, to compare this to the gravitational term. Using the heaviest particle mass (that of tritium ions in the case of ITER) and $n=10^{19} m^{-3}$, we get $$m n \mathbf{g} \approx 5 \times 10^{-7}N/m^3$$ which is over 10 orders of magnitude less than the force felt due to pressure gradients! So, when you're designing the magnetic field topology, you can pretty safely ignore gravity.

As for an intuitive reason: fusion plasmas are hot. This means particles are bouncing around incredibly quickly, and they bounce off each other so frequently that gravity has basically no time to alter their trajectory in any noticeable way. This is much like how you don't really need to worry about gravity when you're shooting at a target 10 feet away-- the bullet moves so fast that it doesn't really matter.


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