A lot of satellites use hydrazine propellant but the Prospector 1 is using water. What are the benefits and drawbacks of that decision?


1 Answer 1


Unlike chemical rockets such as hydrazine (a monopropellant, which doesn't require separate oxidizer and fuel to burn), water is just the reaction mass — not the energy source. The Comet-1 thruster that the DSi Prospector series is using is electrothermal, so it requires a substantial electrical power supply in order to energize the water (presumably by turning it to steam). According to their data sheet,

Comet-1 thrusters are the ideal balance of cost and performance, occupying a place in the market between low-cost, low-performance cold gas and resistojets, and high-cost, high-performance monopropellant and electric systems.

In other words, the tradeoff is that in exchange for lower cost (and In-Situ Resource Utilization, ), $I_{SP}$ and probably specific power are lower than hydrazine, and $I_{SP}$ is much lower than ion engines. Since the Prospectors are intended to refuel as they go, and therefore they don't need to carry anywhere near all their fuel at launch, low $I_{SP}$ (150–175 s) isn't quite such a problem, but it's still a limiting factor.

Suppose the mission is to harvest two asteroids close enough that only 250 m/s $\Delta V$ is necessary to reach them from a suitable parking orbit. Each trip, the miner retrieves 10% of its dry mass in payload and leaves it in the same parking orbit. (These are made-up numbers, but they shouldn't be too far off, and things get more dramatic with higher $\Delta V$ or more payload.) With a 450 s $I_{SP}$ from a $LH_2 / LO_2$ chemical rocket (the best you're going to get), that requires a fuel fraction of about 21%, which is not too bad. But with the Comet-1, you can launch with only enough reaction mass to get to one of them and a tank large enough to contain a full round trip, then refuel at every asteroid harvested. The result is launching with only about a 14% remass fraction, which (even with the larger tank) is likely a cost improvement. And that stays the same no matter how many missions you go on, whereas with ten asteroids, $H_2 / LO_2$ requires a starting fuel fraction of about 69%, which is 2½ times the wet mass (starting mass) of the two-asteroid mission. (Doubling the asteroids to twenty increases the wet mass by almost another factor of 3, to 8 times the mass for two.)

Of course, the eventual plan at Planetary Resources is to electrolyze the water brought back and store $LH_2$ and $LO_2$ in orbital fuel depots for further use. This has all the advantage of easy refueling between trips and the much better $I_{SP}$ of hydrolox without the larger solar panels on the craft, but requires more complicated engines, routine fully automated docking, and, crucially, long-term storage of cryogenic fuel in space, which has not yet been done and probably relies on cryogenic selective surfaces to be practical.

I used this rocket equation calculator for the initial figures, then automated calculating $\Delta V$ for ten- and twenty-asteroid missions with this PowerShell script:

$dry = 1000
$pay = 100
$dv = 250
$v_e = 450*9.81
$asteroids = 10        # or 20
$mass = $dry
1..(2*$asteroids) | ForEach-Object {
    if ($_ % 2 -eq 0) {
        # Outgoing trip, no payload
        $mass -= $pay
    else {
        # Incoming trip, payload
        $mass += $pay
    $mass *= [Math]::Exp($dv / $v_e)
  • 2
    $\begingroup$ I honestly don't know anything about this topic, but reading "the Prospectors are intended to refuel as they go" sounds like a pretty good reason to use water. A lot easier than finding some hydrazine. $\endgroup$
    – Steve
    Commented Aug 15, 2016 at 20:52
  • $\begingroup$ Where'd you get the idea that LH2/LO2 engines are only 340s Isp? The SSME has a better Isp than that at sea level, and much better (452s, approximately 33% more) in vacuum. Even much smaller and less-complicated H2/O2 rockets like the RL10 used on the Centaur (Atlas V upper stage) manage at least 450s of Isp. $\endgroup$
    – CBHacking
    Commented Dec 1, 2016 at 8:14
  • $\begingroup$ @CBHacking: I'm not sure, tbh; I was working off memory and, obviously, was mistaken. I've reworked the figures; the main point stands, although it's less impressive. $\endgroup$ Commented Dec 1, 2016 at 8:21

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