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How many joules of electric energy can you draw out of an RTG battery over its lifetime - and what's the mass of that battery? (whole; shell, thermocouples, fuel and all.)

(realizing, that the power drops off asymptotically towards zero over time, pick any reasonable cut-off point you like - say, 10% of initial output or any other you prefer.)

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  • $\begingroup$ Power ratings are available here - note they have been made in a lot of sizes. Might be useful for anyone who wants to have a go. $\endgroup$
    – Andy
    Jul 11, 2016 at 14:38
  • $\begingroup$ @Andy: it lacks any data poins on energy drop-off over time. Is it a direct function of Pu-238 decay (half the output after half-life period) or does it depend on other factors? $\endgroup$
    – SF.
    Jul 11, 2016 at 16:21
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    $\begingroup$ Okay, you've got all the pieces -- show us the numbers for an RTG-laser drive interstellar craft. $\endgroup$
    – Russell Borogove
    Jul 12, 2016 at 3:06
  • $\begingroup$ @RussellBorogove: It's somewhere in the back log of the Pod Bay, and doesn't look promising, I got something of order of 1m/s per 11 days. That's a measly 9km/s over lifetime of the craft if I use the 291 years figure and disregard the exponential drop-off. Locally emitted photons make a really lousy reaction mass. $\endgroup$
    – SF.
    Jul 12, 2016 at 5:00
  • $\begingroup$ (and it's not a laser propulsion (laser being a wasteful device) but a photon propulsion, taking all of RTG radiation, thermal, gamma, and so on, and reflecting it "backwards", for least losses and maximum thrust. (actually minus the 5% or so made into electricity, which would be used to run the probe systems. The "waste heat" would be the propulsive force.) ) $\endgroup$
    – SF.
    Jul 12, 2016 at 6:29

2 Answers 2

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The lifespan of a Radioisotope Thermoelectric Generator (RTG) is entirely dependent on how big you want it to be. A larger RTG will live longer, but will, obviously, be more massive. Curiosity's RTG weighs 45 kg, including all the exterior cladding.

All of the RTGs in use by NASA use Plutonium-238 as a fuel source. According to Wikipedia, Pu-238 has a half-life of 87.7 years, meaning the power output will decay by about 0.787% per year. Curiosity's RTG was designed to provide 125 watts of power when it was first built. Assuming this constant decay rate, it will take 114.35 years for the power delivered by Curiosity's RTG to decay to 10% Of course, Curiosity requires a great deal more than 12.5 watts of power to run, so it will have failed long before this.

Doing some rough calculations, over the course of 114.35 years, Curiosity's RTG will supply a total of 248,000,000,000 Joules. To calculate this, I simply integrated: $$ E = 125watts - (0.00787) \frac{1 year}{3.154*10^7 seconds} (125 watts)*t $$ (I don't have experience with MathJax, so this is the best I could do with the equation formatting. I am pretty sure this math is correct, though someone might want to check it for me.).

More information about the RTGs used on specific missions can be found here, as given by Andy in the comment to the original question..

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  • $\begingroup$ Nice, so 10 gigajoules per kg. With a magical drive that weighs nothing, works at 100% efficiency and converts all input electrical energy into kinetic energy (speed), the RTG would still barely exceed 140km/s (10GJ of kinetic energy per kilogram.) Nuclear drive won't be a relativistic drive. $\endgroup$
    – SF.
    Jul 11, 2016 at 16:35
  • $\begingroup$ @SF. note RTGs are under 10% efficient by the way. I gather typical ones may produce 100 W of electricity but actually output more than 1 kW as thermal energy (waste heat.) $\endgroup$
    – Andy
    Jul 11, 2016 at 16:39
  • $\begingroup$ @Andy: but that "magical drive" is not entirely an unobtainium: a ion drive in form of a linear accelerator, exploiting void and vastness of space (simply, being longer than possible on Earth, and simultaneously very lightweight) could very well give fuel ISp of relativistic proportions, and might even run on interplanetary medium - but it would run on electricity, not on thermal energy. $\endgroup$
    – SF.
    Jul 11, 2016 at 16:52
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    $\begingroup$ This is a reasonable assumption, however actual RTG's don't decay in such an orderly fashion. There was a link to a presentation in the last RTG question that had a very interesting graph with output-over-time vs Pu238 decay on page 27. The best result was for Voyager's power system, which still decayed at twice the speed one would expect from just decay. $\endgroup$
    – Ordous
    Jul 11, 2016 at 19:19
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    $\begingroup$ Here is the link to that presentation. $\endgroup$
    – Phiteros
    Jul 11, 2016 at 19:24
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If you allow the time to go to infinity, on the assumption that power continues to be produced at arbitrarily low heat output (which it doesn't), and that the conversion system maintains the same efficiency (it doesn't), then you get simply:

$$E_\infty=P_0{t_{1/2}\over\log 2}$$

where $P_0$ is the initial power and $t_{1/2}$ is the half-life. This assumes that only one decay is contributing to the energy output. It gets more complicated for a chain of daughter products, but that won't matter in this case if you only go to a few hundred years. To cut it off at time $T$, the energy up to that point is:

$$E_T=P_0{t_{1/2}\over\log 2}\left(1-e^{-{T\log 2\over t_{1/2}}}\right)$$

Since you asked about going down to some fraction of the original power, let's call that $f$, this simplifies to:

$$E_f=P_0{t_{1/2}\over\log 2}\left(1-f\right)$$

So for $f=0.1$, $t_{1/2}=87.7\,\mathrm{y}$, and $P_0=125\,\mathrm{W}$, we get $E_f=449\,\mathrm{GJ}$. That's a while though, since $f=0.1$ at $T=291\,\mathrm{y}$.

The above significantly underestimates the reduction in power of a real RTG. The degradation of the thermoelectric converters can equal or exceed the decay rate of the Plutonium. See this presentation for more details. The half-life can however be used as shown above to estimate the total heat output of the RTG, if $P_0$ is the initial heat flux.

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