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I'm trying to see how much thrust in needed to move 6,000 kg (1 surplus LK lander) from low earth orbit to lunar orbit? Even better would be to know what is currently available (off the shelf) to do the job knowing that there will have to be extensive modification to attach it in orbit. I'm thinking of one or two surplus stage "D" that are available but don't know if they would have enough thrust.

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    $\begingroup$ How long do you want it to take? $\endgroup$ – Organic Marble Oct 16 '17 at 0:25
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What is needed is not thrust (above a certain basic amount), but delta-v, a function of the type of engine and the ratio between fueled mass and dry mass, according to the Tsiolkovsky equation.

The required delta-v from low earth orbit to low lunar orbit is about 4040 meters per second. If I remember rightly, the plan for LK was to have the Soyuz 7K-LOK spacecraft carry it into a lunar-suborbital descent trajectory, release the lander, then return to a safe orbit; the LK doesn't have enough fuel to descend from and ascend back to low lunar orbit on its own -- so you'd need something more than 4040 m/s to do the job; I'm not sure exactly how much, possibly 4500 m/s.

We can evaluate whether the Blok-D would have this capability. Dry mass would be 2.5 tons (Blok D) + 6 tons (LK) = 8.5 tons; propellant is ~15 tons for a total of 23.5 tons loaded. Exhaust velocity for the RD-58 engine is ~3460 m/s.

So the equation is: $$\Delta v = 3460\ \mathrm{m/s} \ln \frac {23.5} {8.5} = 3518\ \mathrm{m/s}$$

This is enough to reach the moon, but not to insert into lunar orbit; using a Blok D for such a mission would raise another problem anyway: the stage uses liquid oxygen as an oxidizer, and would boil off a substantial amount of it on the 3 day journey.

Ideally, you'd want a larger stage than Blok D, using storable propellants (typically UDMH and NTO in Russian designs), to carry out the mission.

Briz-M comes closer; it carries 20 tons of propellant in a 2.5 ton stage, burning storable UDMH/NTO with an exhaust velocity of 3198 m/s. Adding in the mass of the LK, we come to:

$$\Delta v = 3198\ \mathrm{m/s} \ln \frac {28.5} {8.5} = 3869\ \mathrm{m/s}$$

This is still well short of what's needed; it could insert the LK in a high lunar orbit but not get it into the required trajectory for landing.

What might work would be a Briz-M - Fregat-MT - LK stack. The Briz-M provides 2308 m/s of ∆v, the Fregat provides 2275 m/s, and the total of 4584 m/s may well be enough to get on the proper descent trajectory for the LK to finish the job. Fregat also probably isn't designed for a three day trip, so some payload or propellant mass would be lost to larger batteries, fuel cells, or some other power supply.

You're going to need something to catch it on the ascent, though!

Unlike the Blok D, this configuration wouldn't be available off the shelf prior to about year 2000.

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While Russell Borogove mentioned it I think it deserves more focus: Thrust is the wrong thing to measure here. Thrust is how hard you can push--akin horsepower in your car. Your question is thus akin to asking how much horsepower does it take to drive from New York to Los Angeles.

Rather, you are interested in whether it has the fuel to get there. With your car you measure miles per gallon * gallons = how many miles you can go. In space it doesn't cost any fuel to keep going, what you care about is how much you must speed up or slow down. Here's the map:

(You'll need to expand this to read it decently.) enter image description here

Follow the lines to where you want to go (to/from Earth, it will overstate all other paths) and add up the numbers. That's how much total velocity change you need. See Russell Borogove's post for what the rockets can actually do. Note that his numbers are slightly higher--this chart shows the mathematical minimums, not what you can actually do in the real world. It also does not consider gravity assists. If you're heading beyond our nearest neighbors you can get there for less fuel than this shows at the cost of time. Look at MESSENGER. Spend the 11,580 m/s and it could have gotten there in half a year. I'm not finding how much it used on the ejection burn but after that it used 2,212 m/s--but took 7 years to make the trip.

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Russell Borogove is correct that the parameter more useful to you is likely to be delta-V, which you need a lot of to get to the moon.

To answer the thrust question, instead of the available-stages question: if you have sufficiently sophisticated navigation options, there are very low-thrust trajectories available for lunar insertion because of the nature of the Earth-Moon system. You can keyhole through the Earth-Moon L1 point and into a predominantly lunar trajectory with very small accelerations (as opposed to being on a short timeline from a hyperbolic transfer as you'd see on interplanetary patched-conic trajectories, which would require a minimum acceleration to provide enough delta-V to close the orbit in the appropriate window). My understanding is that the limit would truly be down to your ability to control the spacecraft, and your ability to wait.

There's no shortage of articles to peruse on "low-thrust transfers" to the moon; it's a favorite topic in space-navigation circles.

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  • $\begingroup$ Are you thinking of routes similar to Belbruno's Hiten trajectory? Belbruno type routes from LEO to a loosely bound lunar orbit take at least 3.1 km/s and a few months. Farquhar's 9 day route from LEO to EML2 takes 3.4 km/s. $\endgroup$ – HopDavid Oct 16 '17 at 21:02
  • $\begingroup$ I am not personally familiar enough with the topic to know the names you're talking about. I wouldn't expect any low-thrust trajectory to be short or delta-V optimal, but that's as far as I've gone. $\endgroup$ – Erin Anne Oct 17 '17 at 0:59
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Lifting a spaceship from launchpad to a low orbit is very different to increasing the height of an orbit from LEO to GEO. For a launch from ground, there must be enough thrust to fight gravity and to accelerate to the necessary velocity for a LEO. Increasing the height of an orbit can be done with very low thrust if there is enough time for the maneuver. But for both cases a strategy to use a minimal amount of fuel is essential.

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