# What would be the specific impulse of a continuous nuclear fusion drive?

Let's assume problems of running sustained nuclear fusion are overcome (be it by making the mechanism highly energy-positive, or just supplying all the energy deficit externally, like beamed power.) Let's also assume we managed to direct all products (and none of the substrates) of the fusion in one direction (through the nozzle). Plus matters of cooling, safety etc, all the engineering trivia.

Essentially, the drive reacts Deuterium and Tritium, converting them into Helium through nuclear fusion, and the newly created extremely hot helium is ejected through the nozzle, as reaction mass.

What would be the exhaust speed of such a drive - speed at which the atoms of helium would be ejected; the specific impulse of such a drive?

I tried to ballpark typical temperatures of fusion plasma into average speeds of particles, and got nowhere, as the gas equations don't really work with plasma. Could you give me a ballpark value?

• That's not how fusion works. You are confusing released energy (photons) with the randomly-directed motion of 'hot' atoms. You can't focus those particles any more than you could focus the emissions from the sun. At least, not without typical Sci-Fi use of unobtanium containers and uncompilium software. Commented Oct 19, 2018 at 13:53
• Many such questions are addressed at Project Rho: projectrho.com/public_html/rocket/enginelist.php#modes Commented Oct 19, 2018 at 14:10
• @CarlWitthoft: Ionized particles can be guided quite well using magnetic field - that's how a stellator works. Momentum from the photons will be much smaller than from the ions due to mass difference, so I'm not so worried about reflecting them. And Project Orion somehow deals with them...
– SF.
Commented Oct 19, 2018 at 15:37

The easiest way is just to think in terms of energy. Using numbers from wikipedia, the mass of a deuterium nucleus is 2.014 daltons, that of a tritium nucleus is 3.016, helium 4 is 4.0026 and a neutron is 1.0087

Thus the net energy production is about 0.019 or very roughly 1/250 of the mass of the products, and in the perfect engine you describe, all of this ends up as kinetic energy in the exhaust.

So we get that $$1/2 mv^2 = 1/250 mc^2$$ From which we can quickly find $$v$$ to be about $$c/11$$.

Specific impulse is exhaust velocity divided by $$g$$, so we get about 2.7 million seconds as the $$I_{sp}$$.

• A month of specific impulse... That's damn respectable!
– SF.
Commented Oct 19, 2018 at 12:02
• @SteveLinton Nice work, thanks for posting it. Reading the link posted by Russell Borogrove in the question, projectrho.com/public_html/rocket/enginelist.php#modes. I see an answer of about 1M seconds for a fusion plasma rocket. Is their lower value the "real world expectation" because of expected losses (neutrons and such)? Commented Oct 19, 2018 at 23:44
• @RandyHill That seem's plausible. That link deals with a Deuterium-3He reaction which produces relatively few neutrons, in fact, so in this case the losses would likely be greater. Commented Oct 20, 2018 at 7:47