TerraForming Mars

More
17 years 4 months ago #18575 by Joe Keller
Reply from was created by Joe Keller
Let's terraform Mars now. Maybe Pres. Bush could get this done before the end of his term.

Bury an H-bomb beneath the trailing surface of Phobos, at the depth maximizing the total kinetic energy of the ejecta falling onto Mars. This fall would multiply the energy of the H-bomb many-fold. Also, the ejecta mostly would hit Mars obliquely.

Mars' permafrost would be exposed and melted by these hot meteoric plowshares. Water frost would form on the night side. Frost's high albedo insulates against radiative night-time heat loss. The resulting planetary temperature rise would melt more permafrost.

The meteors also would stir up dust, which would heat the air though cooling the surface. Cooling the surface is counterproductive, but hopefully this effect would be small.

Meteor impacts might have caused past wet eras on Mars. It would be most effective to do this when Mars is nearest the sun.

Please Log in or Create an account to join the conversation.

More
17 years 4 months ago #19387 by Joe Keller
Replied by Joe Keller on topic Reply from
Plan for an Earth-like Mars


Summary. Previous authors have described plans for warming Mars, with greenhouse gases, or with nuclear explosions on Mars. After liquid water is freed from permafrost, microbes will use solar energy to free oxygen from Martian minerals. Assuming 2% efficiency of algae using solar energy for various chemical reactions, in 610 yrs. Mars could have the same oxygen pressure as Earth.
The explosion of a 20-megaton subsurface hydrogen bomb on Phobos, could deliver 48 times its energy, to the surface of Mars, in the form of meteors near Mars' equator ("tropical shot"). Alternatively ("temperate shot"), it could deliver 9 times its energy, in the form of meteors landing at permafrost-rich latitude between 39S and 70S, and roughly another 18 times its energy, between 39S & the equator. (The walls of a crater on Phobos, would be used to direct the debris somewhat, increasing the immediately delivered meteoric debris by 9%, and reducing by 60%, the nuisance debris entering from early-decaying low orbits.) For any such Phobos-to-Mars shot, radioactive isotope levels on Mars would be less than 4% those of a shallow undergound burst on Mars; thus the ratio of energy to radioactivity would be more than 500-1000x better.
The amount of atmospheric dust from these meteor impacts (proportionally 30-60x Krakatoa), would be comparable to previously observed planet-wide dust storms, and temporarily would reduce atmospheric convection, increasing the half-life of the immediately freed lower atmospheric water. The water in Mars' atmosphere is slightly less than needed to provide thick enough frost to stop infrared radiative cooling at night. Meteoric debris from Phobos might be 8-18% water, and further, would heat and expose permafrost and hydrated minerals on Mars. This small increase in atmospheric water, could give night-time frost which would warm the planet, sublimate the carbon dioxide ice cap (greenhouse CO2), and melt more permafrost.

Please Log in or Create an account to join the conversation.

More
17 years 4 months ago #18638 by Joe Keller
Replied by Joe Keller on topic Reply from
(cont., part 2)


The Immediate Prospect. Phobos is expected to crash into Mars in 40 million yrs. (and to disintegrate into rings, when it crosses the Roche limit millions of years before that). This will be a 4000x larger event than would result from any proposed 20 megaton shot. Such crashes might have caused riverine eras on Mars. The most recent riverine era was about 1 million yrs. ago, according to a published, refereed estimate. (Another published, refereed estimate says that the largest river ever on Mars carried half the water of Earth's Amazon.) Phobos & Deimos morphologically greatly resemble a common type of asteroid. It is unlikely that Phobos has been in Mars orbit 4 billion years and that we just happen to be observing the last 1% of its life. So, asteroids relatively recently have been, and will be, captured by Mars, and crash into Mars. (Orbital regularization occurs so quickly that it must be from unknown forces.) What is planned herein, happens naturally every few million years, sometimes on a much larger scale.
Atom bombs now can be made much cleaner (i.e., less fallout), perhaps by using a smaller fission trigger or by choosing a tamper that produces less of the harmful isotopes. Gases from the explosion on Phobos would disperse into interplanetary space. Some solids near the explosion would remain on Phobos. Only 4% of solid fallout would hit Mars immediately; another 1.4% would have perigee less than 200mi and thus would hit within a few yrs., as the orbit degraded (calculated for the "tropical shot", but the "temperate shot" isn't much different this way). Fallout radiation generally decays according to the "Rule of 7 and 10" (Effects of Nuclear Weapons, U. S. Defense Dept., 1962), i.e., if the radiation level is called 100% at one day post explosion, it's 10% at one week, 1% at seven weeks, 0.1% at one year, etc. A 20 megaton bomb might weigh no more than a typical interplanetary probe. A 2000 megaton (buried 100^(1/3)=4.6x deeper) bomb would require 2 metric tons of deuterium, assuming 4% mass-energy conversion. Bombs which attempt to spread the chain reaction to surrounding local material are contraindicated because Phobos is large enough that a runaway nuclear reaction there would severely damage Mars and maybe even Earth.
When the pressure of Mars' atmosphere equals 47mmHg (the vapor pressure of water, or for practical purposes medical "normal saline", at 37C; the pressure at about 60,000ft on Earth) human body fluids won't spontaneously boil; survival without a pressure suit will be possible. Because of Mars' lower gravity, the mass of atmosphere overhead then would be the same as at about 40,000ft on Earth. Radiation protection still would be needed because Mars has only 1/800 of Earth's protective magnetic field, and at first, the atmosphere would contain little ozone. A perfect unpressurized oxygen delivery system (real oxygen masks fall short of this) needs an ambient pressure of 92mmHg (that found at about 46,000ft on Earth) to deliver the partial pressure of oxygen found at 11,000ft on Earth. Some pressure will be achieved quickly by carbon dioxide (already 4mmHg, which might quickly double due to simultaneous sublimation of both polar CO2 caps), water vapor (at saturation, 17.5mmHg at 20C but only 4.6mmHg at 0C and practically zero below that), and the release of dissolved gases from the melted permafrost. If the permafrost froze at 0C, at saturation, under 1 atmosphere of CO2, then the melting of a 500m depth of water over Mars' entire surface, would release 93% of the CO2, adding another 44mmHg. Hydrogen sulfide is more than twice as soluble in water as CO2 is, and might also be present in the permafrost.
Independently living cells need certain molecular machinery, which seems, on Earth, to require a volume of a few cubic microns; thus globules found in meteors originating from impacts on Mars, are thought to be too small to be fossilized cells. Lack of life on Mars, could be why past meteor impacts did not completely or permanently "terraform" Mars. Algae from Earth would be parachuted onto Mars from the "mother ship", as soon as surface water appeared. Before the detonation, Earth radio telescopes, and then also the mother ship, would transmit easily decodable television images to Mars, and send a lander to the Martian surface with posters, a movie projector, a loudspeaker (acoustic code) and a Very Low Frequency (subterranean penetrating) transmitter, warning what is to happen and how to stop us (send a radio signal, make a noise, shine a light, make dust, make a cross with rocks). (The Mars lander also would monitor the Phobos debris impact.) The argument about intelligent artifacts on Mars (e.g., the "King face") is not yet resolved, mainly because of the "Cydonia face" with its high degree of symmetry and alleged persistence under varied lighting angles.

Please Log in or Create an account to join the conversation.

More
17 years 4 months ago #19388 by Joe Keller
Replied by Joe Keller on topic Reply from
(cont., pt. 3)


Craters on Phobos. Everywhere in the solar system (e.g., Arizona), meteor impact craters are about 6 times as wide as deep, with curved bottoms (OR Norton, “Cambridge Enc. of Meteorites”, 2002, Fig. 12.6, p. 280; p. 285). J Blunck, “Mars and Its Satellites”, 2nd ed., 1982, has photos of the solid model of Phobos which RJ Turner made in 1975 using Mariner 9 images. Blunck (Figs. 15&16, pp. 154 & 155) also contains the map which Turner made using his model. (A preliminary map version by TC Duxbury, 1974, is in Icarus 23:217,296 & 26:95.) Phobos has many deep, curved craters. Like our moon, Phobos keeps the same side always toward the planet, with about 4 degrees libration (3 degrees “optical libration” = (2/3) x the orbital eccentricity; plus 1 degree tidally forced)(HH Kieffer, ed., “Mars”, 1992, ch. 36, RM Batson et al). Phobos & Deimos have nearly circular orbits nearly in Mars’ equatorial plane.
For a ”tropical shot”, a crater on Phobos’ equator at 90E longitude (i.e., the trailing edge) is ideal. I suppose that the maps of Phobos choose the prime meridian so that the libration correction is zero at apogee & perigee.
The ideal “temperate shot” would be angled 35 degrees southward; ideally, this is also 35S in “planetographic” coordinates (like the “geographic coordinates” used for Earth maps, which determine latitude from the normal to the idealized sea level, or “reference ellipsoid”, surface) but the Phobos maps use “planetocentric” coordinates because of Phobos’ 20x23x28km ellipsoidal shape (the long axis is radial to Mars, the short axis is NS), so ideally the crater should be 31 1/3 degrees S planetocentric latitude (and 90E longitude), assuming that the crater is level with the surrounding reference ellipsoidal surface.
Impact craters roughly approximate hyperboloids of revolution with perpendicular asymptotes (see U.S. Geological Survey contour map in HH Niniger, “Arizona Meteorite Crater”, 1956, Fig. 1, p. 17, reprinted from GP Merrill, Smithsonian Misc. Collections vol. L, 1908). The best location for a bomb, is beneath the center of the crater, at a depth equal to the radius of curvature at the center. If the equation of the hyperbola is x^2-y^2=a^2, this location is the origin, i.e., the intersection of the asymptotes. A shallower bomb would waste energy, accelerating matter into the crater wall. A deeper bomb would waste energy, accelerating matter that is outside the asymptotes of the crater. If there is no large hard meteor core buried in the way, then drilling a hole for the bomb, and the explosive excavation, are easier because of the fragmented crater material (regolith, i.e. fragmented rock, is generally 10 to 100m deep on Phobos & Deimos: WF Bottke et al, “Asteroids III”, 2002, p. 9). Regolith is estimated to be 300m deep in Phobos’ “grooves” (Batson, op. cit., ch 37, p. 1269)_. The negligible (1 cm/sec/sec) gravitational acceleration on Phobos also helps. A mole-like hole digger could move debris from its front to its rear as it descends.
The “tropical shot” (with a 20 megaton, i.e., 2*10^16 calorie, bomb) maximizes its debris payload to Mars (which is practically the same as maximizing the impact energy on Mars) when buried 655m deep, assuming Phobos’ specific gravity is 1.95. The “temperate shot”, for which anything hitting Mars poleward of 39S is counted as payload, is buried 371m deep. Assuming the typical width:depth=6:1 ratio, the craters chosen (“mortar tubes” by virtue of the inertia of their walls) would be 786 & 444 wide, resp. The minimum successful ejecta speeds, projected onto the vertical crater axis, would be 580m/s & 1360m/s, resp. For both, this would occur at 22 degrees from the vertical axis; the maximum ejecta speed, occurring along the vertical axis, would be 1.75x greater. The use of craters as “mortar tubes” increases the payload of the “tropical shot” by 9% and reduces its near-miss nuisance debris ( e.g., 0 < perigee < 200mi ) by 60%; similarly for the “temperate shot”.
The above includes small corrections for the eccentricity of Phobos ( e = 0.0135 or so; major axis = 9380km ) and its orbit’s inclination to Mars’ equator (1.1 degree, according to AT Sinclair, Astronomy & Astrophysics 220:321+, 1989). The tropical shot” should be done at apogee; for it, the orbital inclination is unimportant. The “temperate shot” should be done at apogee, when apogee is approximately simultaneous with the descending node. Phobos’ node and apsides move in opposite directions, each with period 2.25 Earth yr., so optimal “Phobos” windows occur every 1.13 Earth yr., vs. optimal “Mars” windows every Martian yr. of 1.88 Earth yr (Sinclair op. cit.; also JA Burns, Reviews of Geophysics & Space Physics 10:463+, 1972).
Satisfactory craters are visible on the Phobos map, made from Viking images, of DP Simonelli et al, Icarus 131:52-77, 1998, Fig. 6(a), p. 65. We’re lucky, because craters larger than the size we need, are far fewer. Centered at 269W(91E)32.5S is a dark faint crater measuring about 3 degrees latitude across & 2 degrees longitude across, for an average diameter of about 420m. The reflectivity near this crater is about 0.075 (Ibid., Fig. 2(b), p. 62). At 272W(88E)10N is a bright-rimmed crater measuring about 3x4 degrees = 580m; there might be a more subtle crater of comparable size, closer to 270W0N.

Please Log in or Create an account to join the conversation.

More
17 years 4 months ago #19226 by Joe Keller
Replied by Joe Keller on topic Reply from
(cont., pt. 4)


Position of Mars. Mars’ perihelion precedes its southern hemisphere summer solstice by about 30 degrees of orbit. Our shot should be made sometime between these two points, because warmth will sublimate exposed permafrost. Theoretically, permafrost exists poleward of about 40S or 45N (PL Read, “The Martian Climate Revisited”, Springer 2004, Fig. 8.21, p. 242), and the upper surface of permafrost lies no deeper than 5m in this latitude range. The Viking landers found 1% H2O near the Martian surface, from hydrated minerals (P Cattermole, “Mars”, 1992, p. 18). Type CI & CM asteroids (Phobos & Deimos might be such) are common; these are “carbonaceous chondrites” containing 6 to 18% H2O (JS Kargel, “Mars: a warmer, wetter planet”, 2004, p. 189) and thought to be, when the size of Phobos or Deimos, able to retain their permafrost for billions of years if never closer to the sun than Mars is. So, the main water source is shallow permafrost, which might not exist near the equator. Hydrated surface minerals, and the pieces of Phobos, will be backup sources.
A 20 kiloton bomb exploded 50ft deep makes a crater 100ft deep and 800ft wide (U. S. Defense Dept., “The Effects of Atomic Weapons”, McGraw-Hill 1950, sec. 5.92, p. 170); this implies that the debris of the 20 megaton “tropical shot” would excavate craters totaling 36 cubic km on Mars; and the debris of the “temperate shot”, 6.5 cu km poleward of 39S and roughly 13 cu km to the equator side of 39S. Thus a “temperate shot” is the surer bet because it deposits about half as much material anyway as a “tropical shot”. Regolith probably approaches 100m deep in crater bottoms (maybe 300m; see above), so the “temperate shot” is likelier to achieve nearly the proper drill depth for detonation.
Let the sub-Phobos point be the prime meridian. The debris trail of a “temperate shot” begins at 0S180E, curves southwest to 39S90E, then southeast to 63S180E on a curve with maximum S latitude 70S. From detonation to impact is about 3.5 hr (i.e., Mars rotation through 51 degrees longitude) for all of the track between 39S & 70S.

Please Log in or Create an account to join the conversation.

More
17 years 4 months ago #18639 by Joe Keller
Replied by Joe Keller on topic Reply from
(cont., pt. 5)


Rings. For a “tropical” (resp. “temperate”) 20 megaton shot, there initially would be about 3.5 (resp. 2) billion cu meters of debris in the rings, mostly very near the orbit of Phobos. Phobos will acquire a slow rotation about its (dynamically stable) NS axis; there is an almost 50% chance that the trailing and leading edges will permanently trade places. The crater from either 20 megaton shot would be about 4km diam. The 10km diam. “Stickney” crater on Phobos looks like the expected result of a 20*(10/4)^3 = 300 megaton shot; it’s equatorial, like a “tropical shot” (though now 60W longitude), and would have had about double the efficiency, for making rings, because it was big enough to blow off a curved end of Phobos.
A significant amount of debris would have orbital inclination up to 5 (resp. 20) degrees for the “tropical” (resp. “temperate”) shot. For comparison, in 1883 Krakatoa “blew up most of the island” which was 813m high & 13 sq km (horizontal) area (Columbia Enc., 1975). Assuming a conical shape, half the island, would have been 1.8 billion cu m of debris. Mars has about 30% of Earth’s surface area, but the ring would be shading Mars only about 30% as much of the time. So initially the shading effect, of a 20 megaton “temperate” shot, on Mars’ climate, might be about that of Krakatoa on Earth’s.
For Krakatoa, uncharged spherical rock (2g/ml) particles 10 microns in diameter, would have settled out from 10,000m altitude in 45 hrs. by gravity alone (Stokes’ viscosity formula). For the planned Phobos explosion, the larger particles would remain longest in orbit, so the main effect might be determined by albedo. The visual albedo of Phobos (Deimos is about the same) is 6.5% (Burns, op. cit.); at the surface near the planned “temperate shot” crater, it is 7.5% (see above). The rings of Jupiter, Uranus & Neptune are “as dark as charcoal” (L Lovett et al, “Saturn”, 2006, p. 77). The Handbook of Chemistry & Physics, 44th ed. (1963) table of “Diffuse Reflecting Power”, pp. 3091-2092, shows that materials with low visual (0.54 or, usually given, 0. 60 micron) albedo, usually are much less black at 8.8 microns. “Slate (dark clay)” has albedo 0.067 at 0.60 mu but 0.20 at 8.8 mu: if energy comes at the former wavelength and leaves at the latter, this amounts to 10 degrees greater equilibrium temperature at 250K, if the ring sector totally shades Mars. If all particles were the near-optimal 100 micron size, the shaded area from 2 billion cu m of dust, 1/10 of which casts its shadow on Mars, would be only 1/15 of Mars’ cross-section, for 10/15 = 0.7 deg C.

Please Log in or Create an account to join the conversation.

Time to create page: 0.345 seconds
Powered by Kunena Forum