Dr Philip J. Stooke of The University of Western Ontario has compiled a map of planned landing sites of Google Lunar X Prize contestants.
Archive for January, 2011
Contrary to what was said in the previous post, our Moon does have an atmosphere… sort of.
When orbiting an airless body, one would expect no light scattering to occur. But when the Apollo astronauts were doing circles above the lunar surface, when approaching the terminator they saw this:
What was that?
Universe Today has a story on this, which quotes a scientist investigating this phenomenon:
“For the first set of experiments, imagine just a piece of surface with dust particles on it, and we shine light on this surface,” he said, “so that half is illuminated, half is not, pretending that there is a terminator region, that the sun is set on one side and is still shining light on the other. When you shine light on the surface with properties that are appropriate, you can emit photo electrons, but you only emit electrons from the lit side, and some of those electrons land on the dark side, — you have a positive charge surplus on the lit and a negative charge pile-up on the night side. Across a couple of millimeters you can easily generate a potential difference of maybe a
wattvolt, or a handful of wattsvolts, which translates actually as a small-scale, but incredibly strong electric fields. This could be like a kilowattkilovolt over a meter. But of course, it only exists over a sharp boundary, and that sharp boundary may be the key to understanding how you get dust moving to begin with.”
(I took liberty of fixing obviously mistaken physical units).
Okay. Since someone has already done the hard part (i.e. calculated electric field near the lunar terminator), we can now apply high school physics, to calculate how fast can dust be ejected from the terminator zone. The result is below (click to enlarge).
The analysis assumes that a constant electric field exists across the lunar terminator. The first graph assumes that the acceleration happens with the potential difference of V=1V along the distance of s=1mm. The second graph assumes a potential difference of V=1000V along the distance of s=1m. In both cases the electric field is 1kV/m (see the quote above), but the result is not the same. Since it’s known that the potential difference between the light and dark side of the Moon is several hundred volts, then probably the second version is closer to the truth.
Dust grains made of different elements are investigated. For each element, two curves are plotted. The bottom one assumes that only a single atom in the grain has been ionized once (i.e. the electric charge of the grain is equal to the elementary charge). The top one assumes that 50% of surface atoms have been ionized once (the illuminated side of the grain becomes ionized).
Please be aware, that this calculation is based on some highly speculative assumptions. The model is highly sensitive to the parameters of electric field near the lunar terminator.
I hope we will know more on this fascinating subject when LADEE flies.
- A collection of links on electrostatic dust transport (BAUT)
- And the complete BAUT thread which prompted me to do the calculation
MATLAB/Octave code is below.
function moondust() A_Au = 197; rho_Au = 19300; A_Al = 27; rho_Al = 2700; A_Fe = 55.8; rho_Fe = 7874; A_Ca = 40; rho_Ca = 1550; figure; plot_speed(A_Au, rho_Au, 'y;Au;'); plot_speed(A_Al, rho_Al, 'k;Al;'); plot_speed(A_Fe, rho_Fe, 'b;Fe;'); plot_speed(A_Ca, rho_Ca, 'g;Ca;'); print dust_velocity.png; endfunction function v = dust_velocity(q, m) % V and s estimates from: http://www.universetoday.com/81727/lunar-dust-transport-still-a-mystery/#more-81727 V = 1; % Volts s = 1e-3; % meters E = V/s; a = q/m*E; t = sqrt(2*s /a); v = a*t; endfunction function plot_speed(A, rho, style) amu = 1.66e-27; % atomic mass unit [kg] q = 1.6e-19; % elementary charge [C] %A = 197; % atomic mass (Au) %rho = 19300; % density (Au) [kg/m^3] v_esc = 2380; % escape velocity [m/s] NN = logspace(0, 10, 100); % number of atoms in the grain vv1 = ; vv2 = ; mm = A*amu*NN; for N=NN m = A*amu*N; % Worst case: single atom in the grain ionized vv1 = [vv1; dust_velocity(q, m) ]; % Best case: half of grain surface completely ionized qx = N^(2/3)/2; qx = floor(qx); if qx == 0 v = nan(); else v = dust_velocity(qx*q, m); end vv2 = [vv2; v ]; end % calculate grain size [nm] grain = (mm/rho).^(1/3) / 1e-9; if ishold() else loglog(grain, v_esc*ones(1, length(grain)), 'r;escape velocity;'); end hold on; loglog(grain, vv1, style); loglog(grain, vv2, style(1)); xlabel('grain size [nm]'); ylabel('velocity [m/s]'); title('Electrostatic transport of particles on lunar surface'); endfunction
Readers not interested in cultural musings used as introduction to today’s post, are kindly requested to skip the part written in italic below and proceed directly to the next paragraph.
With the Epiphany behind us, we can say that the period of Christmas festivities is now over. It’s interesting to note that in my part of the world, each year we revert more from using Christian symbolic of, well, Christmas, to pagan Germanic tradition of Yule. (Or, to be more specific, we seem to be doing away with the Christian part of the symbolics, so only the pagan part remains). Now, I am not going to get into the Yule-vs-Christmas discussion, because first, the subject has been widely covered already, and second, this is, after all, a blog about astronomy, and not culture or religion. But I can’t help noting one very ironic twist. Although one can say that Christmas is not a native holiday in my country, so shifting focus back to Yule symbolics is coming back to one’s cultural roots, the joke is that Yule isn’t native here either. Depending on the exact location, actual Slavic holiday celebrated around winter solstice would be Święto Godowe (here’s a Wikipedia article on it, readable although mutilated by Google Translate), Kračún or Koleda. That’s not like these customs were completly lost, though, as some parts of it were simply incorporated into the Christian rituals. But as we are moving away from Christian character of midwinter celebrations, and the major deity nowadays is an overweight, bearded dude in a red coat, affiliated with the Coca-Cola company, there’s a good chance that this part of my cultural legacy will die out completely in my lifetime. So if you want to know what the world will be missing by the time humans return to the Moon, you can have a look here and here.
Regardless of what particular religious (or commercial) holiday someone celebrates at this point of the year (or spends at work, like me), one must remember that midwinter festivities are in fact astronomical in nature. After the winter solstice, northern hemisphere days become longer and so the world is again moving from darkness into light. And that transition from darkness to light (and back) brings me to the topic of today’s post.
Illumination data are very important from the perspective of future surface operations. Since the Moon has no atmosphere, there is no ambient (scattered) light. A given point on the surface is either directly illuminated by Sun (i.e. you can see the Sun if you are there), or it is not. An illuminated region will be hot (+107°C); a dark region will be cold (-153°C). In the equatorial regions (where Apollo landed) the Sun is high above the horizon during the day, so everything is illuminated (and hot); at night, nothing is illuminated (the Sun is below the horizon), so the temperature drops.
Polar regions are different. There, the Sun grazes the horizon all the time. (Since the tilt of the Moon’s axis is minuscule, there is no polar day / polar night effect we have on Earth). So, if you stand on a mountain peak you can see the Sun all the time, going around you. This situation is known as a peak of eternal light. On the other hand, if you dive into a deep crater, you never see the light. Of course, the whole thing is a bit tricky: remember, there is no ambient light. So if you happen to be high up, but a mountain obstructs your view of the horizon, you will be in darkness when the Sun happens to be behind the mountain.
Now, assume that you want to build a base. What would be a good spot to do so? The equatorial regions are not a good choice; the 250°C temperature difference between day and night makes engineers nervous. On the other hand, a peak of eternal light would be a nice place. Permanent illumination ensures that the thermal environment would be stable — around -50°C. Also, it provides access to abundant solar energy. (Although in practice you have to erect your solar panels vertically and rotate them following the Sun, which complicates matters a bit).
But wait a moment. We also know that the permanently shadowed regions harbor water and other interesting volatiles. Wouldn’t it be better to set up a base there? Not really, because at ca. -200°C the conditions are not really likable. So what does one need? Well, of course: a well illuminated region near a permanently shadowed region!
Meet the Shackleton crater:
(for a larger view, click the image or download a full resolution 300dpi PDF with description).
The above map combines the illumination data from LRO (linked above), expressed as shades of gray with the Kaguya laser altimetry data (red isolines) and some annotations. I have drawn isolines only every 250m, to avoid too much clutter inside the crater.
Letters A, B and D mark the areas which are illuminated at least 80% of time. These locations have been identified by Bussey et. al. (see the paper here) by creating a relief model of the terrain and performing a computer simulation of solar illumination; see the paper for details. Also, Paul Spudis’ site contains an iconic image of these interesting locations marked over the Kaguya “Earthset” photography. (The “C” location is too far from Shackleton to be included on my map.)
I must note that a major discrepancy exists between the LRO image and the Bussey et.al. paper. The LRO image is 8-bit grayscale image (i.e. values between 0 and 255) with actual pixel values between 2 and 254. So, logically,the value of 254 would correspond to 100% illumination (peak of eternal light) and 2 to 0% illumination (eternal darkness). At the same time, Bussey et.al. say that the best illuminated point (D) receives around 86% of illumination on average. Since there is no additional information to resolve this, I have chosen to render the LRO data in 0-100% range anyway.
One can now easily see the attractiveness of the “A” spot for mission planners. The following image, taken from a BBC article about the ill-fated Constellation program, confirms this expectation:
As for getting the volatiles from the crater floor, the matter is a bit tricky. A quick look at my map tells you that the crater floor has the elevation around -2500m and the “A” spot is at +1500m. This produces a 4000m difference in elevation over a horizontal distance of about 8km. That means an average slope of 26 degrees or 46%. The upside is, we have engineered such systems on Earth already. This is Pilatus railway in Switzerland, world’s steepest cog railway, climbing a track with 38% average and 48% maximum slope.
The downside is, building something like that on the Moon is pure madness. It would operate in hard vacuum, low gravity and have to be able to survive 150K temperature difference between the crater bottom and its rim. And there will be no industry around to supply the materials.
I have no idea what will be used to transport the water out of the lunar cold traps, but I am sure that it will require some brilliant engineering.
- Spudis et.al., Geology of the South Pole of the Moon and the age of the Shackleton crater
- Spudis et.al., Geology of Shackleton Crater and the south pole of the Moon
- Shackleton area imaged by SMART-1 (includes potential landing sites)