22
$\begingroup$

The question How useful is placing an asteroid in the Moon's orbit? quotes a news article saying the Obama administration is opposed to a lunar landing mission because of the cost of such a mission:

The Obama administration is opposed to another moon landing, saying such a mission would be too costly. It wants instead to focus on capturing an asteroid and placing it into the Moon's orbit for future exploration.

But placing an asteroid in lunar orbit and going there to pay a visit to it isn't exactly cheap either (even if the delta-v to bring the asteroid into a lunar orbit isn't that large and everything goes perfectly, there's still the cost of first getting to the asteroid in its original orbit, then getting to lunar orbit to do anything with it). Sending robots to do the exploration and possibly mining of the asteroid would almost certainly cut down on the cost, but that same argument also holds for a lunar landing mission: technology has advanced a lot since the 1960s, so we can do a lot with robots today that we couldn't reasonably or even at all do back then. The difference in capability between the early planetary landing probe missions and the recent Mars missions, for example, shows this well.

All this got me wondering. If we assume for a second that we would want to go to the Moon in some manner, then roughly how much higher would the monetary cost be just to land on the Moon and getting back into lunar orbit, compared to only establishing a (reasonably) stable lunar orbit? It's probably safe to say that it would add a fair bit to the cost of the mission, but how much? Let's ignore the matter of human EVA capability; people aren't going to be walking around on the surface of the moon from a lunar orbit either (even though landing people on the Moon without human EVA capability seems a bit silly, to put it mildly; there isn't a lot that can be done by Moon-local remote control that can't be done over an Earth-Moon link, if only you have enough power and receiver sensitivity for a good quality downlink from the Moon), and a lot can be done without sending humans either to lunar orbit or to the lunar surface.

A good answer would explore this both from the angle of manned and unmanned missions. Order of magnitude figures are quite good enough.

Improve this question
$\endgroup$
4
  • 2
    $\begingroup$ Nice question. It might be useful if answers break down the cost into mission phases eg. £1m to lunar orbit, +£3m for manned, +£1m for landing, +£8m for manned landing etc. $\endgroup$
    – ThePlanMan
    Jun 7, 2014 at 0:02
  • $\begingroup$ Some asteroids can be nudged from their heliocentric orbits into a lunar orbit for as little as .17 km/s. Have you read the Keck report? $\endgroup$
    – HopDavid
    Jun 7, 2014 at 0:33
  • $\begingroup$ @HopDavid No, I haven't. I see you linked to it in your answer to the linked question, where it would appear to be more relevant. I just might have a look, but I don't think it changes much for this question. $\endgroup$
    – user
    Jun 7, 2014 at 11:44
  • $\begingroup$ Do you mean people specifically? We just gave birth to a plant on the Moon, does that count? I can't find a price tag on Chang'e 4 though. $\endgroup$
    – Mazura
    Jan 16, 2019 at 10:56

2 Answers 2

Reset to default
22
$\begingroup$

1.87 km/sec Δv between low lunar orbit and the surface. Thus you need 3.74 km/sec plus some reserve because suicide burns are for KSP, not real life.

To make a trip from low Earth orbit to low lunar orbit is 4.04 km/sec Δv and I believe the return is 1.40 km/sec Δv.

Thus your total Δv is 5.44 for the trip to low lunar orbit (and less if you go to a higher orbit, I don't know how high you can go and yet remain in orbit) vs 9.18+ km/sec Δv for the landing. The tyranny of the rocket equation says the latter trip is MUCH harder.

Edit: FraserOfSmeg provided a link giving \$1.2M/kg to the lunar surface. There's an even more interesting piece of data there--the same company is charging \$198K/kg to lunar orbit. (Edit: They no longer list the price to orbit, only to the surface.) Thus the cost of landing from orbit increases the price sixfold. Furthermore, looking at the ascent stages for the Apollo lunar landers they were more than 50% fuel. Even if we can reuse the descent engine that means a lot more weight involved. Using the Apollo numbers combined with the Astrobotic ones I get \$2.2M/kg returned to lunar orbit. Thus the price of a landing/return mission is elevenfold higher than an orbital mission.

Improve this answer
$\endgroup$
8
  • $\begingroup$ @HopDavid I'm aware of the mascon problem, it's just that those were the only numbers I could find and my math isn't up to calculating it. A higher orbit makes the difference even greater. $\endgroup$
    – Loren Pechtel
    Jun 7, 2014 at 0:35
  • $\begingroup$ That's the delta-v budget, but like FraserOfSmeg's comment on the question indicates, I was hoping for the cost in terms of money, not delta-v. This is still useful information, however, so I'd suggest you do not delete it outright. I'll edit the question to clarify this as well. $\endgroup$
    – user
    Jun 7, 2014 at 11:51
  • 2
    $\begingroup$ Budget for such a project is speculative. Loren's observation is relevant though. An additional 4 km/s added to the delta V more than doubles Gross Lift Off Weight (GLOW). $\endgroup$
    – HopDavid
    Jun 7, 2014 at 13:58
  • 9
    $\begingroup$ +1 for "suicide burns are for KSP, not real life" $\endgroup$
    – Shane
    Jul 7, 2016 at 19:17
  • 12
    $\begingroup$ @Shane Except since I wrote that we find SpaceX routinely doing suicide burns with boosters worth tens of millions of dollars. $\endgroup$
    – Loren Pechtel
    Sep 22, 2018 at 2:38
6
$\begingroup$

Finding a nice equation for this is next to impossible. The cost per kilogram might steadily decrease from 1-10kg, but then increase thereafter since you need a different launcher/modular system to get the mass to the surface (speculative reasoning).

This all being said, Astrobotic give you a cost per kilogram to the surface of the moon at \$1.2m. That's not specifically for a manned payload, but the majority of the cost difference would be design/build side (and the fact the you need to launch many kilos for a manned mission). Typical estimates for the cost/kilogram to LEO are in the range of \$10,000. As for the cost in the interim of lunar orbit but not landing I would imagine it would be something very similar to the extra amount of propellant needed to get you to the orbit. If that ends up being 20 kilograms of propellant, then you're talking an extra \$200,000.

You can't give exact figures because the mass of propellant required is based on the dry mass of the system.

This is how I would suggest costing a mission like this:

1) Calculate mass required on for operations in orbit around the moon

2) Add mass required for surface operations (if applicable)

3) Add mass for earth re-entry (if applicable)

4) Add propellant mass for decent and landing onto the moon (if applicable)

5) Add propellant mass for take off from moon and orbit manoeuvres leading to Earth re-entry (if applicable)

6) Add propellant mass for lunar orbit insertion

7) Add Calculate mass required for launch

8) Total launch cost x \$10,000 is flight cost

9) Add design\ build\ test costs

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.