How ESA-NASA’s Solar Orbiter Beats the Heat

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From NASA

Feb. 4, 2020

When Solar Orbiter launches on its journey to the Sun, there’s one key piece of engineering making this ESA-NASA mission possible: the heat shield.

Seeking a view of the Sun’s north and south poles, Solar Orbiter will journey out of the ecliptic plane — the belt of space, roughly in line with the Sun’s equator, through which the planets orbit. Slinging repeatedly past Venus in order to draw near the Sun and climb higher above the ecliptic, the spacecraft bounds from the Sun and back toward the orbit of Earth throughout its mission.

“Although Solar Orbiter goes quite close to the Sun, it also goes quite far away,” said Anne Pacros, the payload manager at the European Space Agency’s, or ESA’s, European Space Research and Technology Centre in the Netherlands. “We have to survive both high heat and extreme cold.” In the dark of space, Solar Orbiter faces temperatures of minus 300 degrees Fahrenheit. At closest approach, 26 million miles from the Sun, it will encounter intense heat and radiation.

But Solar Orbiter’s 324-pound heat shield reflects and guides heat away from the spacecraft and can withstand up to 970 F.

Trailer for the ESA-NASA Solar Orbiter mission

Credits: NASA’s Goddard Space Flight Center/Genna Duberstein. Animation by ESA/ATG Medialab.

Download this video in HD formats from NASA Goddard’s Scientific Visualization Studio

The Solar Orbiter spacecraft is prepared for encapsulation in the Atlas V payload fairing.

The Solar Orbiter spacecraft is prepared for encapsulation in the Atlas V payload fairing. In this image, the front layer of thin titanium foil and star-shaped brackets are visible. The front layer reflects heat, while the brackets provide support. Credits: NASA/Ben Smegelsky

The heat shield is built like a 10-foot-by-8-foot sandwich. The front layer — wafer-thin sheets of titanium foil — strongly reflects heat. A honeycomb-patterned aluminum base, covered in more foil insulation, forms the inner slice closest to the spacecraft and provides support.

Star-shaped titanium brackets keep the layers in place, like a toothpick tasked to hold bread together, but notably, this sandwich is missing its filling. The nearly 10-inch gap in the shield funnels heat out to space. A smaller, second gap lies between the inner slice and the spacecraft. Overall, the shield is 15 inches thick. It also has several eyes: peepholes for five of the spacecraft’s remote-sensing instruments to peer through.

Solar Orbiter’s heat shield is coated with a thin, black layer of calcium phosphate, a charcoal-like powder much like pigments used in cave paintings thousands of years ago.

“It’s funny that something as technologically advanced as this is actually very old,” Pacros said. But the coating resists degradation under the blow of intense solar ultraviolet radiation. Although the black powder does absorb some heat, it’s excellent at shedding that heat back out to space.

Solar Orbiter also has to contend with its own heat. Its instruments work up a sweat while they’re running; panels of radiators on the side of the spacecraft eject heat and ensure the instruments don’t get too hot.

Tight control of the spacecraft’s position and tilt is key to protecting the instruments. Once the spacecraft is past the 88-million-mile mark in its solar flybys — that’s 95% of the distance between the Sun and Earth — the heat shield must be pointed straight at the Sun. That means Solar Orbiter will crab walk through space, keeping the spacecraft and instruments tucked in the heat shield’s shadow.

Solar Orbiter is an international cooperative mission between the European Space Agency and NASA. ESA’s European Space Research and Technology Centre (ESTEC) in the Netherlands manages the development effort. The European Space Operations Center (ESOC) in Germany will operate Solar Orbiter after launch. Solar Orbiter was built by Airbus Defence and Space, and contains 10 instruments: nine provided by ESA member states and ESA. NASA provided one instrument (SoloHI) and an additional sensor, the Heavy Ion Sensor, which is part of the Solar Wind Analyzer (SWA) instrument suite.

For more information on Solar Orbiter, please visit:
nasa.gov/solarorbiter
sci.esa.int/solar-orbiter


Banner image: An animation of Solar Orbiter peering at the Sun through peepholes in its heat shield. Credits: ESA/ATG medialab

43 thoughts on “How ESA-NASA’s Solar Orbiter Beats the Heat

  1. Space – the Hotter Frontier

    One of the heated issues underlying greenhouse theory is whether space is hot or cold.

    Greenhouse theory says that without an atmosphere the earth would be exposed to a near zero outer space and become a frozen ice ball at -430 F, 17 K.

    Geoengineering techniques that increase the albedo, the ISS’s ammonia refrigerant air conditioners, an air conditioner in the manned maneuvering unit, space suits including thermal underwear with chilled water tubing, UCLA Diviner lunar data and Kramm’s models (Univ of AK) all provide substantial evidence that outer space is relatively hot.

    But outer space is neither hot nor cold.

    By definition and application temperature is a relative measurement of the molecular kinetic energy in a substance, i.e. solid, liquid, gas. No molecules (vacuum), no temperature. No kinetic energy (absolute zero), no temperature. In the void & vacuum of outer space the terms temperature, hot, cold are meaningless, like dividing by zero, undefined. Same reason there is no sound in space – no molecules.

    However, any substance capable of molecular kinetic energy (ISS, space walker, satellite, moon, earth) placed in the path of the spherical expanding solar photon gas at the earth’s average orbital distance will be heated per the S-B equation to an equilibrium temperature of: 1,368 W/m^2 = 394 K, 121 C, 250 F.

    Like a blanket held up between a camper and campfire the atmosphere reduces the amount of solar energy heating the terrestrial system and cools the earth compared to no atmosphere.

    This intuitively obvious as well as calculated and measured scientific reality refutes the greenhouse theory which postulates the exact opposite even incorrectly claiming the naked earth would be a -430 F ice ball.

    Zero greenhouse effect, Zero CO2 global warming and Zero man caused climate change.

    • The Moon is pretty much the same distance from the Sun as the Earth is and there is no atmosphere on the Moon. Would the Earth, without atmosphere, not undergo same temperature swings between day and night as the Moon?

      When sunlight hits the moon’s surface, the temperature can reach 260 degrees Fahrenheit (127 degrees Celsius). When the sun goes down, temperatures can dip to minus 280 F (minus 173 C).

      From https://www.space.com/18175-moon-temperature.html

      Averaging -23°C or -9.4°F, although I would only like to live on Earth then during dusk and dawn.

      • Would the Earth, without atmosphere, not undergo same temperature swings between day and night as the Moon?

        No, because on Earth night-time is only 12 hours or so, not much time to cool off. But on Moon, night-time is two weeks. Also albedo is only 0.12, about one third of Earth albedo (0.33). So more heat is absorbed by the Moon.

        So the lunar temperature swings are much greater than terrestrial swings: average 107C daytime with maximum 123C. Average lunar night-time temperature -153C, with minimum -233 °C.

        The “average” lunar temperature, in a statistical sense, is -23C. But surprising perhaps, because lunar albedo is lower, more is absorbed. But virtually all of that heat is emitted back into space, thanks to Kirchoff’s law of radiation. There is no atmospheric “blanket” to retain the warmth, as on Earth.

        • Take away high albedo of clouds, ice, and snow due to no atmosphere and you get close to that of the moon.

          • I have a problem with “albedo”. It is reasonably well defined for the day, but what is it at night?

          • “defined for the day”
            Actually albedo is defined as the ratio of reflected over incident light, so at night it is really not “well defined” because the incident light is relatively close to zero.

            Emissivity is term you should use for the ratio of emitted radiation compared to expected black body radiation.

          • Curious G

            Much against common expectation, the albedo of a surface does not change (much at all) with temperature. Suppose an iron bar has an albedo of 0.9. Heating that bar until it is bright yellow does not change the albedo. It is still ~0.9 and radiating furiously.

          • “I have a problem with “albedo”. It is reasonably well defined for the day, but what is it at night?”
            Curious George ,
            With that simple question you reach right into the centre of the dark heart of climate science.
            MarkW is correct, it is the same, and the reason for that is because the sun shines at night.
            Really? Yes, in the fake model world of climate science the post-albedo solar intensity is simultaneously spread across the full surface area of the Earth. Hence the reason for the divide by 4 illumination divisor that makes the power of sunlight too weak to heat the surface of our planet above the freezing point of water.
            All of this of course is total nonsense. We have built and applied a dynamic atmosphere climate model for the Earth that fully honours the reality of a single lit hemisphere that collects solar energy, linked to a full dual hemisphere globe that emits radiant thermal energy back out to space both day and night.
            See here for more details
            https://www.researchgate.net/project/Dynamic-Atmosphere-Energy-Transport-Climate-Model

          • Could anyone tell me the reason why a flat earth shadow is used in the seemingly fundamental greenhouse equation cited in Hansens papers that creates a 33K GHG effect?

            My searching yields some answers, which I’m not quite satisfied with.
            1) Fourier simplified it because the maths was complicated
            2) it’s got something to do with the rotational speed of planets
            3) the (pi r squared) assumption is roughly similar to a calculation compensating for the change of incident angled light – so they use the similar version

            if any one could provide a sum/paper/citation on where this assumption of a flat Earth originated, I would be very grateful!

          • pompeydano

            You need to look for the Vacuum Planet equation of Astronomy for the origins of this. concept.
            Sagan, C. and Chyba, C., 1997. The early faint sun paradox: Organic shielding of ultraviolet-labile greenhouse gases. Science, 276 (5316), pp.1217-1221. http://science.sciencemag.org/content/276/5316/1217

            See my short essay here for more details:
            Comparative Planetology: Establishing the Role of Meteorology in the Science of Climate.
            https://www.linkedin.com/pulse/comparative-planetology-establishing-role-meteorology-mulholland/

          • Climate Forward Modelling.
            The process of Forward Modelling creates a numerical prediction, that must be matched and verified against external data for the model to be both valid and useful. The modelling process starts with the identification of the set of fundamental principles, that contain the irreducible minimum set of axioms, from which the actions of a system are designed and constructed. With the set of first principles established and measured, then the mathematical algorithm that combines these elements can be created.

            With forward modelling studies of a planet’s energy budget, the first and overarching assumption is that the only way that a planet can lose energy is by thermal radiation from the planetary body to space. This assumption is not in dispute, and it leads to the adoption of the Stefan-Boltzmann (S-B) equation of thermal radiation, which is used to establish the direct relationship between power intensity flux in Watts per square metre (W/m^2) and the absolute thermal temperature of the emission surface in Kelvin (K).

            The second critical assumption made in the analysis of a planet’s energy budget, is that it receives incoming thermal energy in the form of insolation from a single central star. Solar system planets orbit around this central source of light, and consequently all planets have both a lit (day) and a dark (night) hemisphere.

            A technique for establishing the energy budget of a planet, and hence how the power being consumed is distributed within its climate system, is a technical challenge that has already been addressed by astronomy. An equation was required that could be used to compute the average surface temperature of any planet, by establishing its thermal emission temperature under a given insolation loading. To solve this problem, a set of modelling assumptions were made that include the following simplifications: –
            1. That the planet being observed maintained a constant average surface temperature over a suitably long period of time.
            2. To make this assumption valid, the total quantity of solar energy intercepted by the planet is averaged out over its annual orbital year.
            3. This annual averaging therefore removes the effect of distance variation from the Sun, inherent for the trajectory of any planet’s elliptical orbit.

            Next the complex problem of how a planetary orb intercepts solar energy, and how this sunlight energy is distributed over the planet’s surface, was addressed. Planets contain the following geometric features in common:
            1. They are near-spherical globes.
            2. They are only lit on one side from a sun that is located at a focus of their orbit’s ellipse.
            3. They often (but not always) have a daily rotation rate that is significantly faster than their annual orbital period.
            4. They commonly have an obliquity or axial tilt, although each planet’s angle of tilt is unique.

            Given the above list of distinct features, it is clear that the computation for the surface capture of solar energy on an orbiting, rotating, axially tilted planet is a complex mathematical calculation. To address this complexity the following simplification was applied: –
            That all planets intercept solar energy at their orbital distance, as if they are a disk with a cross-sectional area that is equal to the planet’s radius (i.e. π R^2). However, due to daily rotation and seasonal tilt, planets emit radiation from all parts of their surface over the course of each year.
            Therefore, the total surface area of the planet that emits thermal radiation to space is four times the surface area of its intercepting disk (i.e. 4π R^2). It is this geometric fact that is responsible for the “divide by 4” rule that is contained within the calculation of planetary radiative thermal balance.

            Having devised a simplified way of calculating the amount of energy that the total surface of an orbiting, rotating, axially tilted planet would receive during the course of its year, we can now move to the next stage of the calculation. Namely, the computation of the annual average surface temperature associated with this energy flux.

            This is achieved by using the Stefan-Boltzmann law to determine the absolute temperature in Kelvin (K), associated with the average radiative power flux in Watts per square metre (W/m2) of the planet’s emitting surface.
            Equation 1: j* = σT^4
            Where j*is the black body radiant emittance in Watts per square metre; σ is the Stefan-Boltzmann constant of proportionality, and T is the absolute thermodynamic temperature raised to the power of 4.

            The fundamental equation used in astronomy that results from this work is exemplified by the Vacuum Planet radiation balance equation (corrected from the published error pers comm) used by Sagan and Chyba (1997): –

            “The Early Faint Sun Dilemma
            The equilibrium temperature Te of an airless, rapidly rotating planet is: –
            Equation 2: Te ≡ [S π R^2(1-A)/4 π R^2 ε σ]^1/4
            where σ is the Stefan-Boltzmann Constant, ε the effective surface emissivity, A the wavelength-integrated Bond albedo, R the planet’s radius (in metres), and S the solar constant (in Watts/m^2) at the planet’s distance from the sun.”

        • Thanks, got it now.
          How much simpler climate science would be here on on Earth if we didn’t have this complicated atmosphere 🙂

          • Carl,
            As far as the sun is concerned the earth appears as a flat disc. So 1,368 W/m^2 times PI*over 6,371,000^2 = Total W * 3.6 for J/SI h or 3.412 for Btu/Eng h.

            The most popular model of the earth is a ball suspended in and evenly/averagely heated by a bucket of warm poo spreads that energy evenly over the entire spherical ToA.
            A sphere of radius r has four time the area as a disc of radius so divide by four.
            1,368/4 = 342W/m^2 ISR.
            BTW this earth model is REALLY^4 dumb since it is not even close to how the earth heats and cools.

            The lit hemisphere gets all the incoming energy.
            Energy leaves the surface 24/7 in all directions based on Q = U A dT.
            More energy leaves the lit side because it is hotter than the cold dark side.

        • “There is no atmospheric “blanket” to retain the warmth, as on Earth.”
          BINGO!!!
          And a blanket obeys Q = U A (Tsurf – Ttoa) and NOT sigma epsilon A T^4.

          Albedo is the ratio of the energy absorbed by the terrestrial to the amount incident.
          (1-0.3)*1,368 = 957.6 = net QAs net Q increases and decrease with the albedo so does Tsurf. My back-of-envelope suggests that a 1% change in albedo changes Tsurf 1 C.

          For the earth this is NOT the same as emissivity. For the moon WITHOUT an atmosphere they are the same. A surface can radiate BB ONLY into a vacuum.

          Emissivity is the ratio between the energy leaving a surface and the radiation if ALL the energy left as radiation. In a vacuum the ONLLY way is radiation.
          Because of the non-radiative processes of the contiguous atmospheric molecules BB LWIR from the earth’s surface is not possible.
          Per K-T: Earth’s theoretical emissivity = 63/396=0.6, physical emissivity = 63/160=0.39.

          Emissivity & the Heat Balance

          Emissivity is defined as the amount of radiative heat leaving a surface to the theoretical maximum or BB radiation at the surface temperature. The heat balance defines what enters and leaves a system, i.e.

          Incoming = outgoing, W/m^2 = radiative + conductive + convective + latent

          Emissivity = radiative / total W/m^2 = radiative / (radiative + conductive + convective + latent)

          In a vacuum (conductive + convective + latent) = 0 and emissivity equals 1.0.

          In open air full of molecules other transfer modes reduce radiation’s share and emissivity, e.g.:
          conduction = 15%, convection =35%, latent = 30%, radiation & emissivity = 20%

          I demonstrated this with a couple of classical style experiments:
          https://principia-scientific.org/debunking-the-greenhouse-gas-theory-with-a-boiling-water-pot/
          That’s about as sciency as one can get.

          The Instruments & Measurements

          But wait, you say, upwelling LWIR power flux is actually measured.

          Well, no it’s not.

          IR instruments, e.g. pyrheliometers, radiometers, etc. don’t directly measure power flux. They measure a relative temperature compared to heated/chilled/calibration/reference thermistors or thermopiles and INFER a power flux using that comparative temperature and ASSUMING an emissivity of 1.0. The Apogee instrument instruction book actually warns the owner/operator about this potential error noting that ground/surface emissivity can be less than 1.0.

          That this warning went unheeded explains why SURFRAD upwelling LWIR with an assumed and uncorrected emissivity of 1.0 measures TWICE as much upwelling LWIR as incoming ISR, a rather egregious breach of energy conservation.

          This also explains why USCRN data shows that the IR (SUR_TEMP) parallels the 1.5 m air temperature, (T_HR_AVG) and not the actual ground (SOIL_TEMP_5). The actual ground is warmer than the air temperature with few exceptions, contradicting the RGHE notion that the air warms the ground.

          • No Nick, not -430F….without an atmosphere, the Earth would be warmer than the often stated 255 K, because its Albedo would be about .13 instead of .3 Do an SB calc, the answer will surprise you. Also your assumption regarding the equilibrium temperature at 1368 w/SqM is incorrect for a rapidly rotating spherical body like Earth, that shares the 1368 w/SqM over 4 times the radiation receiving area of a non-rotating flat disk of the same diameter.

          • Nick, there is no “333 watt feedback loop”….396 minus 333 is just the SB equation terms for (Tground^4 minus Tsky^4) for an average 63 watts of heat radiated into the sky…..pardon me, that also includes 40 watts going directly to outer space, thereby uninfluenced by Tsky, leaving 23 watts for your (Tground^4 minus Tsky^4) calculation. Also day, night, cloudy, clear, are all much different than the average.
            So if your experiment show the 333 watt feedback loop doesn’t exist, its not even wrong.

      • “Would the Earth, without atmosphere, not undergo same temperature swings between day and night as the Moon?”

        Dr Kramm at univ of AK and the UCLA Diviner Mission seem to think so.

      • 288 K (assumed average) – 255 K (assumed 0.3 albedo) = 33 C (complete garbage)

        Refer to the Dutton/Brune Penn State METEO 300 chapter 7.2: These two professors quite clearly assume/state that the earth’s current 0.3 albedo would remain even if the atmosphere were gone or if the atmosphere were 100 % nitrogen, i.e. at an average 240 W/m^2 OLR and an average S-B temperature of 255 K.

        That is just flat ridiculous.

        NOAA says that without an atmosphere the earth would be a -430 F frozen ice-covered ball.
        https://sos.noaa.gov/Education/script_docs/SCRIPTWhat-makes-Earth-habitable.pdf
        (slide 14)

        That is just flat ridiculous^2.

        Without the atmosphere or with 100% nitrogen there would be no liquid water or water vapor, no vegetation, no clouds, no snow, no ice, no oceans and no longer a 0.3 albedo. The earth would get blasted by the full 394 K, 121 C, 250 F solar wind.

        The sans atmosphere albedo might be similar to the moon’s as listed in NASA’s planetary data lists, a lunarific 0.11, 390 K on the lit side, 100 K on the dark.

        And the naked, barren, zero water w/o atmosphere earth would receive 27% to 43% more kJ/h of solar energy and as a result would be 19 to 33 C hotter not 33 C colder, a direct refutation of the greenhouse effect theory and most certainly NOT a near absolute zero frozen ball of ice.

        30 % albedo: 957.6 W/m^2, 360.5 K, 87.5 C, 189.5 F (by 4, 239.4 W/m^2, 255K)
        11% albedo: 1,217.5 W/m^2 (27.1%), 383.2 K, 109.8 C (22.3), 223.8 F (by 4, 304.4 W/m^2, 270.7 K)
        0% albedo: 1,367.5 W/m^2 (42.8%), 394.0 K, 121.0 C (33.5), 250.0 F (by 4, 341.9 W/m^2, 278.7 K)

    • No Nick, not -430F….without an atmosphere, the Earth would be warmer than the often stated 255 K, because its Albedo would be about .13 instead of .3 Do an SB calc, the answer will surprise you. Also your assumption regarding the equilibrium temperature at 1368 w/SqM is incorrect for a rapidly rotating spherical body like Earth, that shares the 1368 w/SqM over 4 times the radiation receiving area of a non-rotating flat disk of the same diameter.

  2. I hope this NASA Sun mission works out. If their heat shield design works we apparently will soon need it here on earth to try to save ourselves during the doomsday scenario of run-away global warming. Sarcasm aside, I strongly recommend anyone driving anywhere close to Cape Canaveral/Kennedy Space Port to go through the HASA complex. You will never forget it, especially the Saturn5 rocket and the Space Shuttle exhibit.

  3. Although the black powder does absorb some heat, it’s excellent at shedding that heat back out to space.

    This is because of Kirchoff’s law of thermal radiation: “For an arbitrary body emitting and absorbing thermal radiation in thermodynamic equilibrium, the emissivity is equal to the absorptivity.

    This factoid should be considered when studying so-called “urban heat” effects: black asphalt does indeed absorb a lot of heat during the day, but sheds heat to space with equal efficiency at night.

    • “black asphalt does indeed absorb a lot of heat during the day, but sheds heat to space with equal efficiency at night.”

      Sounds technically right to me, but how come the average temperature is higher in cities than in rural areas?

      However, with regards to the satellite it makes sense to use the black stuff on the back side, as you mainly have heat from the inside, which needs to escape to empty space with no solar radiation coming in from the backside.
      I am impressed how big the thing is, looking forward to hear some great discoveries it will contribute.

        • The laws of physics are almost always declared in some “ideal” sense, that is rarely observed in the “real” world. Physics is applied mathematics. “Non-pure” in the sense that we only admit “truths” we expect to see in the real world.

          We have yet to see a “perfect” circle, but we can describe it, mathematically. You say: what about a ball of mercury, unperturbed, floating in orbit around the sun? Wouldn’t that be a perfect sphere? Answer: No, it is close to perfection, but not perfect. Minor perturbations and absorbed energy from local masses and energy will keep it non-perfect. There is really no place in the universe you could go inside our universe, to escape these infinitesimal, mostly “neglible”, non-perfections.

          As for thermodynamic equilibrium, it is also a matter of accepting “almost” as “perfect”. We say a bathtub is in equilibrium if the following conditions apply:
          1) empty bathtub
          2) water in bathtub, but drain is closed and faucet it off.
          3) faucet on but drain is open, or the bathtub is overflowing

          But what about evaporation? or a slow leak (one drop a day)? Is it still in equilibrium?

          So, we should say we are “in charge” of equibrium if we can account for all the sources and sinks and show that they are more-or-less the same.

          The Earth has never been or ever will be in perfect equilibrium.

      • “but how come the average temperature is higher in cities than in rural areas?”

        Because the asphalt does absorb radiation (due to its low albedo) that would have otherwise been reflected to space.

        My only point was that absorption works two ways with equal _efficiency_ , I was not suggesting that we should cover the earth with asphalt, which would turn it into the Moon. (See my comment above). But the mean temp of the Moon is lower than Earth, due to lact of heat-retaining atmosphere.

      • In deserts day temperature is high but due to very low atmospheric vapour content night temperature falls back far more than it would do in a large city due to the vapour’s GHG effect where there is lot of surface evaporation. Space probe is located in an extreme example of a ‘desert’ climate so a fine granulated black surface makes sense. If surface is a thermoelectric layer then heat flux could be converted directly into electrical energy for the on-board use.

        • In deserts day temperature is high but due to very low atmospheric vapour content

          No, desert temperatures are high due to 1) a lack of vegetation & evapotranspiration (which helps cool the boundary layer), 2) predominate sinking air from warm-core High Pressure overhead (adiabatic warming), 3) low-elevation deserts (i.e. Death Valley) will be hotter than high-elevation deserts (again, adiabatic warming – the longer the dry adiabat to the surface, the more warming occurs w/ sinking air). Compare Flagstaff (el. 2192m) to Las Vegas (el. 697m) during a normal Summer heatwave.

          night temperature falls back far more than it would do in a large city due to …

          Prove it. I find desert (or anyplace else without snow on the ground) can only cool about 30F overnight. So, if the daytime high temperature was around 100F, the max it will cool to would be around 70F. Snow on the ground will GREATLY enhance radiational cooling.

          Oh, yeah – anecdotal climatology doesn’t count for ‘proof’ i.e. – “when I was a kid, I remember…”

    • It does “shed” heat, but until it does, it’s temperature is higher than the temperature that other materials that “absorb” less would attain.

    • black asphalt does indeed absorb a lot of heat during the day, but sheds heat to space with equal efficiency at night.

      True, but there is also conduction into such materials (below any surface) which accumulates/stores heat according to its mass & thermal-conductively, and then both conduction & the amount of heat stored has to be considered as to what the overnight temp will follow.

    • Water is required to make steam, and water is scarce in space, and very heavy to launch from earth. Steam-powered generators work on earth due to the pressure gradient between high-pressure steam (generated using a heat source) and atmospheric pressure, and condensed water can be confined by gravity. In space, there is not enough gravity to hold steam near the spacecraft, and not enough pressure to condense it to liquid water, so it would escape into space.

      Solar-powered batteries have their limitations on earth, but are the only feasible power supply for a spacecraft designed to fly closer to the sun than the earth’s orbit. Solar radiation is more intense closer to the sun, and there are no clouds (or night) to interrupt solar radiation in space.

  4. At the point when Solar Orbiter dispatches on its excursion to the Sun, there’s one key bit of designing making this ESA-NASA strategic: the warmth shield.

    Looking for a perspective on the Sun’s north and south shafts, Solar Orbiter will travel out of the ecliptic plane — the belt of room, generally in accordance with the Sun’s equator, through which the planets circle. Throwing over and over past Venus so as to gravitate toward the Sun and move higher over the ecliptic, the rocket limits from the Sun and back toward the circle of Earth all through its crucial.

    “Albeit Solar Orbiter goes very near the Sun, it additionally goes very far away,” said Anne Pacros, the payload supervisor at the European Space Agency’s, or ESA’s, European Space Research and Technology Center in the Netherlands. “We need to endure both high warmth and outrageous virus.” In the corner of room, Solar Orbiter faces temperatures of less 300 degrees Fahrenheit. At nearest approach, 26 million miles from the Sun, it will experience extreme warmth and radiation.

  5. This reminds me of the old joke about a certain country announcing that they were sending astronauts to the sun.

    When asked how they were going to prevent them from a firey death, replied “we’ll send them at night.”

  6. postscript to

    The Insignificance of Greenland’s Ice Mass Loss in Five Easy Charts…

    David Middleton / February 5, 2020

    Guest geological perspective by David Middleton

    ___________________________________

    Steven Mosher February 5, 2020 at 7:27 pm

    Nice chart by McKay

    dave’s conclusion?

    “The inescapable conclusion is that if there ever was a climate “crisis,” it was during the Little Ice Age… It was FRACKING cold! And it got pretty cold again in the 1970’s!”

    what the authors say?

    ‘Finally, it should be noted that whereas these analyses are useful for quantifying some aspects of temperature sensitivity, they are poorly suited to determine the extent to which the records reflect long-term (centuries to millennia) changes in past temperature, or the stability of the modern relation back through time.”
    ____________________________________

    David Hartley February 5, 2020 at 8:44 pm

    ‘Finally, it should be noted that whereas these analyses are useful for quantifying some aspects of temperature sensitivity, they are poorly suited to determine the extent to which the records reflect long-term (centuries to millennia) changes in past temperature, or the stability of the modern relation back through time.”

    Does not the same conclusion impact the ‘warmist’ certainties on predictions?
    ____________________________________

    D’accord. That same conclusion

    “whereas these analyses are useful for quantifying some aspects of temperature sensitivity, they are poorly suited to determine the extent to which the records reflect long-term (centuries to millennia) changes in past temperature, or the stability of the modern relation back through time”

    impacts the ‘warmist’ certainties on predictions.
    ___________________________________

    Steven Mosher. As always decidedly ambiguous.

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