# Are any electrically propelled missions to the outer solar system being planned? If not why not?

In answering this question I discover that a probe capable of $10^{-4}$ G continuous acceleration can get to Europa (starting from LEO) in about 16 months. Assuming I haven't made a mistake in my calculations, this seems a very appealing approach, yet I'm not aware of any planned missions using it. Following the success fo Dawn, are there any? If not, why not?

According to this, the NSTAR Ion Thruster that powered Deep Space 1 (and Dawn as well?$^2$)

operates over a 0.5 kW to 2.3 kW input power range providing thrust from 19 mN to 92 mN

So it would be roughly capable of operating almost up to Jupiter (4% of 10kW = 0.4 kW - see PearsonArtPhoto answer - and assuming that all energy would go to the engine). However, I think it does not ever provide the required $10^{-4}g$ of thrust...

Max thrust is 91mN, Dawn dry mass (generously excluding propellant mass) is 747 Kg, 0.091 / 747 = 0.00012 m/s $\approx$ 0.00001 g ($10^{-5}g$)

(Deep Space 1 was lighter, 373 Kg of dry mass - but still not light enough)

No, there are no planned missions using an Ion drive to the outer solar system. The reason is something that you haven't taken in to account. Sunlight drops significantly as one goes further from the Sun. One might be able to get the continuous acceleration you indicated to get to the Asteroid belt, but to go much further then that would require large solar panels, or else an RTG. The RTG used for Curiosity provides 110 W of power. Dawn has the capacity to use 10 KW of power near Earth. At Ceres, it's furthest distance, the power usage is roughly 15% of that, but still 1.5 kW. At Jupiter, the power would be more like 4%.

There was one mission that proposed something similar, but using a full nuclear reactor. The mission was called JIMO, using a 200 kW nuclear engine. Even at that, the mission was proposed to take 4-5 years to get to Jupiter.

Bottom line, one would need to power the solar panels that far out in order to achieve anything, and that becomes a very difficult problem to manage.

• ...and while packing more, bigger RTGs (and even designing more lightweight versions - bulk of mass is the protective casing) technically would solve the problem, the entire West (America, Europe) has nearly completely depleted their supply of Pu-238. Producing more would necessitate reactivating the cold war nuclear weapons production facilities (it was a byproduct of making bomb-grade PU-239). If you're thinking about RTG powered missions, look towards China. – SF. Apr 26 '18 at 12:45
• @SF Interesting subject! I googled a bit and it seems USA is restarting their Pu-238 production, see en.wikipedia.org/wiki/Plutonium-238 and spacenews.com/… – BlueCoder Apr 26 '18 at 13:52
• Added a blurb about JIMO, which was a proposed mission to use a nuclear reactor and ion engines to explore Jupiter. I had forgotten about JIMO... – PearsonArtPhoto Apr 26 '18 at 13:57
• Although it hasn't been done before, you could also beam the energy to the probe's solar panels in the form of a large orbital or lunar laser to counteract the sunlight problem, getting the generating the power and getting the laser to hit would be difficult though – Dragongeek Apr 26 '18 at 15:19
• @Dragongeek: Putting a laser capable of sending this sort of energy at that sort of distance in space is always a political nightmare. After all, who stops you from turning it towards Earth? And even if it wouldn't have enough power to make a difference through the atmosphere, that's just you claiming it wouldn't... – SF. Apr 26 '18 at 15:28

Explorations of Psyche and Callisto enabled by Ion propulsion

It is not being seriously planned as such, but according to this, it could be done, and on a discovery mission budget.

• Dawn is a Discovery mission. – Tom Spilker Apr 28 '18 at 1:46

The kinetic power in a rocket engine's exhaust plume (in the EP community it's called the jet) — and this is for any rocket engine, not only EP engines — is Pjet = F Ve / 2 , where F is the thrust and Ve is the exhaust velocity. In an electric thruster this kinetic power is the same as the electric power that must be supplied to accelerate the ions to their velocity in the jet. Since Ve = Isp g, where Isp is specific impulse and g is Earth's surface gravitational acceleration, Pjet = F Isp g / 2. If you keep F constant, then if Isp goes up, so does Pjet, so for higher Isp it takes more power to maintain that constant force. This Pjet is only that part of the electric power supplied to the system that actually shows up as usable propulsive power. Because EP systems aren't 100% efficient, a fair chunk of the power you actually have to supply to the system doesn't show up as propulsive power. EP engines themselves have efficiencies typically in the 50-70% range, though there are examples slightly higher and decidedly lower (NASA's NEXT thruster runs around 71-72%; the mu10 ECR used on Hayabusa is less than 40%). Some of the electric power goes to ionizing the propellant, and that energy doesn't show up as propulsive power, it just makes a nice, pretty blue exhaust plume as electrons from the neutralizer recombine with the ions. Power processing units, the devices that take the electric power from the power source (like a solar array and its power conditioners) and generate the voltages and currents that the EP components need, are 90-95% efficient. Other accessory components, like controllers, propellant feed systems, etc. can pull tens of Watts (~40 for NEXT), and this counts against the system's overall efficiency. So even though Dawn can generate 10 kW of electric power at 1 AU, you can't turn all that power into propulsive power. If you know a system's efficiency, then you can calculate the power required from the power source to get a given thrust level: P = F Ve /(2E), where E is the system overall efficiency: E = Pjet / Pe , where Pe is the electric power supplied to the EP system, not just the engine. With these equations, and knowledge of a spacecraft's dry mass, EP system specs, and power production capability (or you can postulate those, if you want to assume technology advances give you better performance than existing systems), you can have all kinds of fun with back-of-the-envelope calculations. You can calculate propellant flow rates, and then model the resulting variation of acceleration as the spacecraft's propellant load decreases, and use use all this to model a trajectory. Note that for solar power sources, the power capacity is a strong function of heliocentric distance, which adds a twist to the modeling. One ramification: if a mission requires a really large ∆V, such as a mission to get out to Neptune in a short time, then the Isp needs to be large as well, and the larger-than-average thrust needed to accelerate to the high ∆V in a reasonable period of time combines with that to yield a very high electric power demand. This has been an issue when we've examined EP systems for Uranus or Neptune missions.

To synthesise the various helpful answers and add one more bit on information I've come across: Solar power is out in the outer system, and RTGs don't produce enough power, so we're left with fission reactors which are not yet available.

Even looking ahead, according to this study Kilopower expects to produce around 2W/kg. Based on the numbers for Dawn in one of the answers above, You need about 25 kW of power per Newton of thrust, so the top acceleration using these technologies (and no payload) would be about $8\times 10^{-5} m s^{-2}$ or a little less than $10^{-5} g$, so the transit times I calculated were wildly unrealistic.

• Power per thrust varies with system efficiency but also with exhaust velocity (and thus specific impulse). For the NSTAR system on Dawn you get about 40 milli-N thrust per kW of power. The newer NEXT engine and support system, in every respect giving better performance than the NSTAR system, yields only ~34 mN per kW. This is because the NSTAR engine has a peak exhaust velocity of 31 km/s (Isp 3160 s) while the NEXT engine's is 41.1 km/s (Isp 4190 s). I'll explain in my next answer to this question. – Tom Spilker Apr 28 '18 at 21:32