# What would be the lightest possible moon launch vehicle?

(Reposting from Physics.Stackexchange, as it is more appropriate here.)

Edit: I calculated it, but the results are so big, I'll write it as a small pop-sci article, then will post a summary and the data here. Lot's of funny findings and ideas.

I tried calculating this, but it gets too complicated.

Assume, we have a Moon orbit station and ISS on Earth orbit. We have a Moon base. We want to send a tourist for a week on the Moon and back. We need to launch only the oxygen/fuel cells and the fuel for trans-lunar and trans-earth injection.

The lunar lander lands and takes off in one piece (we don't need a pile of used stages on the lunar base). It can be lighter than 10 metric tons of Apollo LEM. Fuel cells (200 kg each) may go, because now we have solar panels.

So, how low can we get with newer technologies? And the main question, how small can the launch weight be? Is this doable with conventional Soyuz or Proton rocket?

Assume, we might have a coilgun to do trans-Earth injection right from the lunar surface. Delta-v of 2.7 km/s means 90 km of rail and accelerating @ 5g, if I remember my calculations correctly.

Some data from Wikipedia:

• Apollo Lunar Module
• Ascent stage 4,547 kg of which 2,353 kg is propellant
• Descent stage: 10,149 kg (8,200 kg of propellant for 2,500 m/s delta-v)
• Command and Service module
• command module 5,809 kg
• service module 24,523 kg

all together 46,980 kg

• Saturn V
• 1st stage with fuel: 2,300,000 kg
• 2nd stage: 480,000 kg
• 3rd stage: 120,800 kg

The third stage fired only partially (165 + 335 seconds) to get to LEO, and then was used for TLI.

So, 120 tons were sent to LEO, and only 47 tons left after TLI. How low can the latter get?

The lunar module will be orbiting the Moon and reused. This means several tonnes less for TLI and Lunar orbit injection. But the fuel for it has to fly from Earth.

If CM gets smaller, this can also make the vehicle lighter.

Service module can be reduced by using inflatable materials, and it can be reused and stored at ISS. So we save 25 tons x 9 km/s (launch from Earth), but add 3km/s of delta-v to park it after the way back.

So, all together, we need to launch

• the Earth landing module (CM analog)
• fuel for
• TLI
• LOI
• lunar landing and takeoff
• TEI
• fuel to park SM in LEO

If coilgun is used, we don't need takeoff and TEI fuel. (Hm... we need to launch the SM back too :)

How much does this weigh?

• You might wanna read this answer on What is the main problem of Luna's base? Why is there a 3rd generation of orbital bases and no Luna base?. I would also like to ask of you not to repost same questions on various Stack Exchange sites. If you feel a question of yours should be migrated, please flag it and specify a custom reason with the option "moderator attention needed" and it will be migrated for you. Thanks! Aug 24 '13 at 21:06
• @TildalWave: Thanks. Physics site members told this is the appropriate place. I can remove it from there if this is such a big deal (although I wanted to keep the comments there). Aug 24 '13 at 22:09
• I made these calculations myself. It gradually becomes an article, so I'll write and post it somewhere and will leave a short summary here. Aug 29 '13 at 19:35

This is rather difficult, but let's see what we can do with it. First of all, let's make a few assumptions.

1. Let's use the Dragon capsule as our base platform. The manned version should be able to hold 7 people, and enough supplies for them to launch and land.
2. Let's take the direct approach to the moon.
3. Let's just assume that a Dragon could hold enough supplies for the 2 week trip to the Moon.
4. Let's assume you enter LEO for an orbit. From there, there are 2 stages, one to get to the moon and land, and the second to get home. This violates the no littering principal, but it makes the math a lot easier...
5. Let's just use Merlin 1-D engines. This is because it is the rocket enginue with the highest thrust/weight ratio. It's vacuum thrust is about ~801 kN, and a weight of 500 kg roughly. The ISP is 310.
6. Let's assume a tank/fuel ratio of about 3% (Tank weights 3% of fuel)

Okay, so how much does the dragon itself weigh? Cargo + dry mass is about 8,000 kg. Okay, so what are the delta V requirements? From Wikipedia, here's the stats:

• LEO to Moon- 5.93 km/s
• Moon to Earth- 2.74 km/s

So how do you figure this out? I made a spreadsheet which helps me calculate this, using the following formula: $\text{fuel}=[\text{payload}+\text{engine}+(1/2+0.03)\text{fuel}(\text{iter}-1)]/(I_{SP}*9.8)$. It's approximate, but should be fairly close. To verify it, I plugged in some of the apollo numbers and got something in the same ballpark (But slightly better).

Okay, given this spreadsheet, let's crunch some numbers. Let's start with the return trip. You have a mass of 8,000 kg previously mentioned. Add 500 for the engine, and plugging it in to my magic spreadsheet gives 14700 kg fuel, 440 tank, and 23700 total.

Okay, so how do you get that to land on the moon from LEO? It turns out that the 5.93 km/s delta v is higher than is practical with a single stage. Let's try 2 stages, split down the middle (Unrealistically, but there it is). The second stage is still a single Merlin D engine, 46904 kg fuel, 1400 kg tank, and 71500 kg total. The first stage is then 145600 kg fuel, 4370 kg tank, and 222,000 kg total.

Okay, so if you can get 220,430 kg to LEO, you are on your way to the moon! How much is that? It turns out that is twice as large as the Saturn V was capable of lifting to LEO. Considering we are taking 7 people instead of 3, that isn't bad, but it does mean that technology isn't as far along as we would hope.

Bottom line is, with some careful staging this could be reduced, but not significantly. I don't doubt that some careful physics could reveal that my analysis isn't perfect either, or that if we did an Apollo style mission we couldn't save a few kg anyways. Feel free to create your own spreadsheet to explore this if you want;-)