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For a one-way robotic or cargo mission, with current technology, how much of a lander that touches down on the Moon wouldn't be propellant or rocket parts? Modern engines like the CECE by Aerojet Rocketdyne could be used, or any other that could realistically be ready for a mission in the next few years.

After asking how much a Saturn V today could soft-land for such a mission, eventually I realized I'm still unclear how much of that is rocket. This paper by Spudis and Lavoie says this about their lunar cargo lander parameters:

The lander has a dry mass of 8300 kg, a propellant mass of 22000 kg and a payload capacity of 12000 kg.

I had penciled in a much higher payload fraction than that. One CECE would be enough to land the mass a Saturn V equivalent could launch, and masses only 250 kg. I looked up that stages for chemical rockets are roughly 8% inert mass and went with a 4 ton lander with 21 tons of payload for that Saturn V lander. Am I way off?

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    $\begingroup$ I suppose Spudis and Lavoie's cargo lander is overweight because it's a variant of a manned ship, and hardware commonality is a priority? en.wikipedia.org/wiki/Apollo_Lunar_Module#Specifications has some mass breakdown of the LM; besides engine, propellants, and payload, you need thermal control, RCS, power systems, radars and other sensors, guidance/navigation hardware, communications, landing gear, and the structure that holds it all together. You could try figuring out how much those items mass, and see if you can manage 8%. $\endgroup$ – Russell Borogove Dec 21 '15 at 23:23
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The Apollo LM fueled ascent stage, which could be considered payload, was about 30% of the total fueled mass of the LM. However it departed from lunar orbit, so you'd first have to insert into orbit, which takes another stage. That's about 900 m/s. For storable propellants you can assume an Isp of 325 s. Let's say you can make a 15% dry mass stage. (The Apollo service module was 25%, but it did a lot of other things.) Then overall you would get 21% of your TLI mass to the surface as payload. 25% would be an optimistic number, perhaps getting rid of some things on the descent stage. However you need landing gear and a platform to support the payload, or a skycrane with a cable and gear on the payload. Whatever it is, you need more mass than just a stage to land.

If you could keep cryogens from boiling off for long enough, you could do better on the Isp.

Your numbers are about 28%. In the same ballpark.

Reference: Lunar Module Quick Reference Data.

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Let's consider Chang-e 3 as an example, where we'll assume that the lander is our payload and our spacecraft is designed to start in lunar orbit (at 100 km altitude) with enough fuel to make the descent. According to this Spaceflight 101 reference, the specifications are as follows:

  • Dry mass of lander: 1200 kg
  • Rover: 120 kg
  • Wet mass of entire probe: 3780 kg

That wet mass includes the lander, rover, and all the fuel required to transfer from the final launch vehicle trajectory, insert into lunar orbit, and perform the landing. If we consider that to be our spacecraft design, the payload fraction is obviously 3.2%. However, let's do a bit of math and assume that another stage has done the work of inserting us into a lunar orbit -- so how much fuel do we need for descent and landing?

To do this, we'll simplify the problem as being composed of a burn to de-orbit and descend (with apolune at the initial altitude and perilune at the surface), followed by a burn to bring the speed down to null.

In orbit at 100 km (at a radius of 1837.4 km), the speed of the spacecraft is 1.63 km/s from (where $\mu_{Moon} = 4904.9 \frac{km^3}{s^2}$):

$v_{orbit} = \sqrt{\frac{\mu_{Moon}}{r_{Orbit}}}$

Our descent trajectory is an elliptical orbit with semi-major axis $a = 1787.4$ km from:

$a = \frac{1}{2}\left( r_{Orbit} + R_{Moon} \right)$

From that we can compute our speed at the start and end of descent, which gives us $v_{start} = 1.61$ km/s and $v_{end} = 1.70$ km/s:

$v_{start} = \sqrt{\mu_{Moon} \left(\frac{2}{r_{Orbit}} - \frac{1}{a}\right)}$

$v_{end} = \sqrt{\mu_{Moon} \left(\frac{2}{R_{Moon}} - \frac{1}{a}\right)}$

So we'll need an initial $\Delta v$ of about 0.02 km/s to insert into our descent and then a landing $\Delta v$ of about 1.70 km/s to stop the vehicle. That gives a total of 1.72 km/s, and then assuming a specific impulse of $I_{SP} = 300$ s, we can compute our mass fraction. Note we'll assume the final mass is the lander (1200 kg), the rover (120 kg), and some fuel left over that totals 1400 kg.

$\frac{m_{initial}}{m_{final}} = e^{\frac{\Delta v}{g_0 I_{SP}}}$

Substituting those values, we get a mass fraction of about 1.8, and that yields an initial mass of 2520 kg. So for that spacecraft you would have a payload fraction of about 4.8%.

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    $\begingroup$ The Chang-e 3 lander is a functional science station, not just a carrier for the rover, so the 3.2% isn't valid. $\endgroup$ – Russell Borogove Dec 21 '15 at 23:39
  • $\begingroup$ Yes, upvoted! I recognize that but forgot to address it in my answer -- mainly since I couldn't find a mass value for those payloads. This is really just an example of how we can manipulate some of the sample data to reduce it to a common fraction. $\endgroup$ – Brian Lynch Dec 21 '15 at 23:41
  • $\begingroup$ The Chang'e rover is much smaller than what that lander is capable of, so it is a rather poor example to use for a mass fraction. That lander was designed for a future lunar sample return mission, carrying a return rocket. $\endgroup$ – Mark Adler Dec 22 '15 at 0:01
  • $\begingroup$ Yes, I do recognize that, I will modify it with one or two more examples when I have a chance. Shouldn't be hard to do the same for the Apollo LEM. $\endgroup$ – Brian Lynch Dec 22 '15 at 0:03
  • $\begingroup$ It's good to be walked through the math. It indicates how i could have determined this myself with more examples. I did also discard Chang'e 3 for this, for the reasons mentioned. But i could go through it for the LEM, and maybe a Luna or Surveyor mission. $\endgroup$ – kim holder Dec 22 '15 at 15:37
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In addition to the above answers presenting realistic examples, there is one more interesting consideration. Would it be beneficial to re-use the cargo lander? In other words, when is the mass of the propellant needed to return the lander to orbit, and the propellant needed to bring that propellant to the surface of the Moon, smaller than the dry mass of the lander?

We must also consider the ratio of total mass to dry mass (excluding payload), because a lander only capable of landing on the Moon and return is of no use. Most of modern upper stages has a mass ratio between 10 and 15, and because a lander is a bit more complex than an upper stage, I choose a ratio of 10.

Plotting this, we see that re-using the lander is better than single use when the ISP is larger than 680s.

enter image description here

X-axis is ISP of lander, y-axis is proportion of mass in lunar orbit delivered to lunar surface.

Re-using the lander is not economical with chemical rockets, but it gives a slight advantage for a hydrogen NTR.

But we must also consider the cost of propellant relative to building an entire new lander. For that matter, the number of possible reuses is also important.

The amount of cargo in lunar orbit delivered to the surface is for a single use lander

$$\mathbf{\frac{\frac{r}{ℯ^{\frac{\Delta v}{I_{sp}g}}} - 1}{r}}$$

Where $r$ is the landers ratio of dry mass to total mass, $I_{sp}$ is the specific impulse of the lander and $\Delta v$ is one way delta-v.

For a re-usable lander, the formula is

$$\mathbf{\frac{r- \left(ℯ^{\frac{\Delta v}{I_{sp} g}} \right)^{3}}{r \; ℯ^{\frac{\Delta v}{I_{sp} g}} - ℯ^{\frac{\Delta v}{I_{sp} g}}}}$$

Substituting $ℯ^{\frac{\Delta v}{I_{sp} g}}$ with $\rho$, we get

$$\mathbf{\frac{r- \rho^{3}}{\rho \left(r-1 \right)}}$$

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    $\begingroup$ Those slopes are quite close to each other, so your answer will be sensitive to small changes in your assumptions. Also "beneficial" depends on cost, for which delivered mass is only a portion of the cost. When taking into account the cost of building the lander, the point where it is worthwhile is pushed lower, and will depend on how many reuses you can get out of it. $\endgroup$ – Mark Adler Dec 22 '15 at 18:51
  • $\begingroup$ @MarkAdler Sure, I was only pointing out the theoretical possibility. How practical it is in the real world, or the cost of it, are completely different questions. Do you have a better ratio than the 10 I used? I can update my graph. $\endgroup$ – Hohmannfan Dec 22 '15 at 18:54
  • $\begingroup$ @MarkAdler It is not that sensitive anyway, increasing the ratio to 30 does only lower the point to an ISP of 640s. For an infinite ratio, the threshold is 624s $\endgroup$ – Hohmannfan Dec 22 '15 at 18:59
  • $\begingroup$ The mass ratio of the lunar lander descent stage was about 5. $\endgroup$ – Mark Adler Dec 22 '15 at 19:10
  • $\begingroup$ @MarkAdler Note that what I ask for is not the mass ratio, it is the ratio of (lander dry mass)/(lander dry mass + propellant + maximal cargo) $\endgroup$ – Hohmannfan Dec 22 '15 at 19:13

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