One interesting result is a rule of thumb to find required thrust given average acceleration. Google failed me on finding an exact solution, but it looks like there is a simple approach that is within 1% of the target value.

I eventually settled on a design massing 33.4 tons, 1.6 MW solar-electric, Isp 6,000 and 40 N thrust using PIT thrusters with water propellant.

More after the break.

Let's look at an electric tug with payload comparable to my reference tug, both a solar PV and a nuclear version. The main routes for this vehicle will be between LEO, GEO, EML1/2 and Mars orbit. Unlike the chemical tug we can't get much out of the Oberth effect, so the delta-V requirements are higher. Just like the chemical tug, the LEO to EML1 leg has the highest dV requirements (about 7km/s), so if we design for that case then the other trips will be faster, carry more cargo or burn less propellant.

A key design factor here is trip time. If we throw enough power at the problem we can get to EML1 in the same amount of time as a chemical rocket, but that is a poor use of the mass. We need to decide how long we are willing to wait for the cargo and design enough thrust into the ship to make the trip in that span. I'm going to suggest four weeks to EML1 as a reasonable compromise, so let's see the consequences of that choice.

**Estimating thrust requirements for average acceleration**

To apply 7km/s of delta-V in 28 days we need to make an average acceleration of 2.89 mm/s. To allow some wiggle room let's assume we can only thrust 90% of the time, meaning now we need 3.22 mm/s. Since this is our average acceleration, we need to find either the initial or final acceleration to find the thrust of the propulsion system. To do that we will first need to know the vehicle's propellant mass fraction, so let's take a few test cases at Isp of 3000, 6000 and 10,000.

Propellant mass fraction (Mf) is equal to 1 - e ^ (- dV / Ve), where dV in this case is 7000 m/s and Ve is Isp * g. See the rocket equation page for more details.

Isp 3000 -> Mf of 0.21175

Isp 6000 -> Mf of 0.11217

Isp 10000 -> Mf of 0.06890

It would be nice if the average acceleration also matches up with the midpoint of fuel consumption, but somehow I doubt it. Let's find out.

Given a dry mass of, say, 10 tons and an Isp of 3000, the fueled mass is ( 1 / ( 1 - Mf ) ) * dry mass, or 12.686 tons. When half of the fuel is burned the craft masses 11.343 tons. The target acceleration is 0.00322 m/s, so the required thrust is 36.52 newtons. Thrust is mass-flow (mdot) times exhaust velocity, so mdot is 1.2414 grams per second. That rate of propellant consumption would require 25 days to empty the tank, or 27.825 days after accounting for our 90% duty cycle.

That surprises me. It's not exact but it is close enough for exploratory work. The case of 10,000 Isp works out to 27.947 days, so it looks like this rule of thumb is valid across a fair range of Isp values. I also spot-checked some different mission dV values and found similar agreement, always within 1%. If anyone out there knows of an exact solution I would love to hear it.

To calculate this yourself you need your mission dV, Isp, thrust duration and a test mass. If you set the test mass to 1kg (or 1t) then you can find a multiplier to use for different dry masses. The relationships are linear.

The required acceleration a will be dV in meters per second divided by thrust duration in seconds.

First, find fuel mass fraction, which is 1 - e ^ (- dV / Ve).

Convert to dry mass fraction Md, which is -Mf + 1

Convert to the 'gear ratio', which is simply 1 / Md

Multiply by dry mass M1 to get fueled mass M0 and note this value.

Find the fuel mass by taking M0 - M1 and note this value.

Find the 'halfway point', which is half the fuel mass plus M1; let's call this Mh.

Find the thrust F, which is Mh * a. This is the value you are looking for.

Find mdot, which is thrust divided by exhaust velocity, or F / (g * Isp ).

Find the real thrust duration, which is fuel mass divided by mdot. This should be within 1% of your stated thrust duration; if it is then the average acceleration value is accurate enough to use.

If you have a known spacecraft (known dry mass and fuel mass, known thrust), you can use thrust divided by (dry mass plus half the fuel mass).

**Electric tug design**

**To align with the chemical tug, let's target a payload of 40 tons from LEO to EML1. Note that EML2 is a better target, but for purposes of comparison I'm using the LEO to EML1 trip as the most costly trip in the set. As mentioned above, we need to deliver in 28 days or provide an average acceleration of 3.22 mm/s. I don't have an exact solution, so I can't solve the problem in a single step. That's fine; spacecraft design is an iterative process.**

Let's assume an electric thruster at Isp = 6000 and mission dV of 7000 m/s. Also assume a one-way trip (meaning fuel is available at both endpoints). Power alpha is assumed to be 18 kg/kW, whether that be nuclear or long-life solar. Thrusters will be the NuPIT design shown in the last post, using the design values for the 5 N, 200 kW unit at 2.75 kg/kW (550 kg per thruster).

As a first guess let's try eight thrusters, 1.6 MW. That's 33.2 tons, for a dry mass without tanks of 73.2 tons. We will need approximately 10 tons of liquid water propellant; using a tankage fraction of 2% would be reasonable in this case, so tack on 200kg for tanks for a total dry mass of 73.4 tons. Actual propellant load is 9,273 kg, so tankage is sufficient. The half-fueled mass is 78,037 kg and approximate average acceleration is 0.513 mm/s. We're not even close. Trip time would be 157.9 days, or 2.3 one-way trips per year.

Maybe 20 thusters / 4 MW? Power alpha would improve to about 15, yielding 71 tons of power and propulsion. 40 tons of payload and perhaps 0.4 tons of tankage gives a dry mass of 111.4 tons, fuel mass of 14.1 t and average acceleration of 0.844 mm/s. This is clearly not going our way. Trip time would be 106.3 days, or 3.4 one-way trips per year.

Let's aim much higher, 50 thrusters / 10 MW. Power alpha would continue to improve to about 14, yielding 167.5 t of power and propulsion. 40 tons of payload and 0.6 tons of tankage gives a dry mass of 208.1 tons, 26.29 tons of fuel and an average acceleration of 1.13 mm/s. Trip time would be 71.7 days or 5.1 one-way trips per year.

Clearly, short trip times require increasingly absurd power levels. Matching the payload size of a chemical thruster with the 1.6 MW version means only making one round-trip per year. In fact, looking at that version of the ship, if we eliminate the payload entirely the highest acceleration the ship can make is 1.2 mm/s on its last gasp of propellant. Since fuel mass, dry mass, power and thrust are all linear relationships* that means no matter how we scale up the ship it can never get better than this. (The power system alpha does actually get better as we scale up, but moving a ship that masses several times your payload is inefficient and extremely expensive.)

One thing we can do is increase the thrust of each propulsion unit, which usually means decreasing the Isp significantly. Let's look at a VASIMR thruster for comparison, since I have some data on performance at different Isp levels handy. A VASIMR thruster at 200 kW and 6000 Isp produces about 4.75 N of thrust, a fairly close match to the NuPIT. We need about six times that thrust (28.5 N), which occurs right at an Isp of 1000. That would bring the 8-thruster 1.6 MW vessel up to about 230 N of thrust. However, dropping the Isp so dramatically brings the fuel fraction just over 50%. That pushes our dry mass up to 75t, fuel mass to 78.1t and nets us only 2.0 mm/s average acceleration. It's a 40.5 day trip or 9 trips per year, but now we are burning more fuel than the chemical tug thanks to our drastically higher dry mass. Still no net benefit to be had.

**Putting the tug to work**

Let's look at what an electric tug actually saves: propellant. In a fully functional ecosystem of cis-lunar services propellant is fairly plentiful. The speed, convenience and throughput of chemical vehicles far outweighs the efficiency of ion vehicles in this environment. Where an electric tug shines is in the buildup phase, where all of the propellant is coming from Earth. The tug would save money during a critical part of the project. What that means is we do not need to survive dozens of Van Allen belt transits over two decades, we just need to make a reasonable number of trips over two or three years. We also don't need to standardize on the same payload sizes as the chemical tug, nor do we need to make trips in 1 month. I would say that using the same power system alpha for the solar version as I do for the nuclear version is very pessimistic; these vessels would not need to function at Mars orbit, though they do need significantly thicker front-glass shielding on the panels than other craft.

So, a lunar ISRU plan would still start with a single chemical tug / lander as described in part 1. Using performance for the detailed reference tug, a 15-ton package can be delivered from LEO direct to the lunar surface. This will be 12.4 tons of ISRU equipment and 2.6 tons of spares (2.1 year supply). Refilling tug 1 will take 6 months, after which it can deliver 33 tons to EML1.

In the meantime, an electric tug (call it tug A) will deliver a 9-ton fuel depot (135 ton capacity) to EML1. Let's use our 40 N / 1.6 MW / 33.4t / 6000 Isp vehicle from above. It does the job in about 92 days, which means there is a window of three months after the launch of the first ISRU package to get the 33.4t tug, 5.36t propellant and 9t payload into LEO.

At the first lunar launch, 33 tons are delivered to the EML1 depot. Tug A will collect this and head to LEO, taking 127 days and consuming 7.45t of water. During this trip a LEO depot is launched, identical to the one at EML1. The tug turns back around and heads for EML1, taking 4.22t of water for the return trip and leaving 21.33t of cargo in LEO. This could be a mix of surface samples and water as desired. Let's assume five tons are samples and the rest is fuel.

The return trip takes 72 days, during which tug 1 will have delivered another 33 tons to the EML1 depot. Tug A repeats its performance, returning to LEO with a full load of 23.78t water and another 5t of samples. At this point we are at 598 days elapsed since start of ISRU operations, which should be enough time to settle on and construct additional hardware to expand the lunar surface capacity.

This is significantly longer than the all-chemical scenario and has an IMLEO of 129.71 tons, within a few tons of all-chemical. Hardware costs are higher since more of the mass is spacecraft and much less of it is fuel. The main benefit is that schedule pressures are greatly reduced; final design, construction and testing of the second round of ISRU plant is allowed more than a year and a half of time rather than two months. More operational data is available and the tolerance for mistakes or inefficiencies is higher. Another benefit is that this profile includes depots in LEO and at EML1; even if things do not progress beyond the first ISRU package the infrastructure is still useful for this and future projects.

This baseline hardware could continue to deliver 21 tons of cargo to LEO every ~200 days for about a decade, eventually reaching 426 tons over 20 trips at a cost of 141 tons of Earth mass or a leverage of about 3 to 1. Things improve if we continue to expand, since about 44% of that mass was fuel to get the first ISRU plant in position; additional ISRU hardware is delivered using lunar propellant.

The next phase would be to send more ISRU hardware. Tug A can pick up 33 tons at EML1, deliver 19.18t of net payload to LEO over 127 days, pick up a 17-ton package and head to EML1 in 109 days. All of the required propellant is lunar and picked up at EML1. Round trip time is 236 days (a bit under 8 months). The harvesting process run by tug 1 has a shorter turnover time of 6 months, so on average an extra 19 tons is accumulated at the depot. That's not quite enough to provide for a cargo landing, so tug A may not always be bringing a full load of cargo to LEO (meaning shorter round trips in practice).

An alternative might be to use 12-ton packages that will fit into a Falcon 9 for cheaper launch costs; the delivery time for that is 98 days. If less cargo is returned to LEO then that trip time can be shortened as well; for example, 6 tons of return cargo plus round-trip fuel would make each leg of the trip take 98 days, or 196 days round-trip. Each 6.5-month trip would deliver another 10 tons of ISRU with two years of spares. Two electric tugs could deliver 80 tons of ISRU capacity in 26 months, roughly a single Mars synodic period. That would place 95 tons of ISRU with expected output of 950 tons of propellant annually at a cost of 9.5 tons of spares. Net propellant delivered to EML1 would be 505 tons annually, or could be 168 tons to LEO annually with chemical tugs. The annual demand for spares (both ISRU and depots) can be met in a single hardware run with minimal fuel costs, leaving 3/4 of the electric tug schedule open for assignments like delivering new chemical tugs or GEO debris retrieval (a mission that avoids the majority of the radiation belts and prolongs the tug's useful life).

The total phase 2 IMLEO would be 134.4 tons, all of it hardware. Lunar mass to LEO during this period would only be another 48 tons since capacity is focused on buildup. This phase would run for 26 months, or a total of 46 months since first launch.

Ongoing maintenance would require approximately 12 tons per year. The initial depots would be insufficient, so we need another 27t of hardware for fuel storage. If we rate all flight and depot hardware with a 10-year lifespan and pro-rate the replacement mass then we need an additional 12.2 tons annually (24.2t total). Depending on how the output is allocated, this could be considered an ongoing leverage of 23.5 tons in EML1 per ton IMLEO or 7.8 tons in LEO per ton IMLEO. Another way to look at it would be as a fuel supply for three manned Mars missions covering four synodic periods (104 months), or a full ISRU program length of 150 months (12.5 years). Overall Earth mass to LEO is then 511.15t to harvest 5,501.7 tons of gross lunar propellant, yielding 2,924.6 tons net lunar propellant at EML1. That's a gear ratio of about 5.7 to 1. If you are only interested in delivering fuel to LEO then you can net 972.9 tons, still a favorable 1.9 to 1 mass ratio. 511 tons is a lot of mass to launch, but only three payloads require a heavy lift vehicle: the initial chemical tug stack (62t fuel and 22t hardware, split across two Falcon H) and the two electric tugs (33.4t each, also requiring a Falcon H unless they can be built in parts and flown on two Vulcan launches). The remaining 360 tons would be delivered by 30 Falcon 9 launches, or by some combination of any price-competitive launchers with at least 12 tons of payload.

Launch costs would be roughly $2.1 billion. Hardware would run another $6.7 billion (at $15m per ton). Operations might cost $125-$250 million. Call it a total of $9 billion over about 15 years (12.5 years of operation plus 2.5 years of r&d, manufacturing and testing). Overall cost of fuel at EML1 would be $3,094.44 per kg, about $3.1 million per ton and expected to decline to $0.9 million per ton in the long run. Savings are about the same as the all-chemical approach, a bit over $4.5 billion vs. NASA baseline. Additional savings could be realized by using the chemical tugs as cargo haulers to and from Mars as described in part 2, resulting in excess capacity that could be sold or used for other purposes. One of those purposes might be ISS reboost and water supply for life support. Another might be developing a significant water supply on the Moon for growing food, in support of manned missions.

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For this post I think the mass could be reduced somewhat by advances in large solar panel array performance. For example, wing-level alpha as high as 300 W/kg is commercially available (3.33 kg/kW). That would yield a power generation mass of only 5.33 tons rather than 28.8 tons for a nuclear system, though there are structural and power conversion masses that have to be added back in as those numbers are included in the nuclear figure but not the solar figure. I suspect savings could be realized, leading to improved payload and performance.