Our neighborhood has a “planet walk” - signposts representing the planets of the solar system, spaced out to scale over 1 mile, which gives a good idea of the relative distance of the various bodies in the solar system. It gave rise to this very silly and frivolous question which nevertheless is nagging at me, and I’m wondering if some of our astronomy experts would know the answer:
According to NASA, the average temperature of the surface of the earth is ~ 60 degrees F. The average temperature outside a commercial jet cruising at 35,000 feet is ~ minus 60 degrees F. The average temperature of the surface of Mercury is 800 degrees F; of the solar corona is in the millions; and of the surface of the sun is ~ 10,000 degrees F. I can’t find a good estimate for the temperature of interplanetary space, but the surface of Pluto is ~ minus 400 deg F, so I would assume that’s in the ballpark.
So, were you to able to travel, like Calvin and Hobbes, in your wagon on a journey from Earth to the sun, you’d pass through a temperature gradient of a balmy spring day at Earth’s surface, through down into the negative-hundreds Farenheit as you left the planet’s atmosphere, to something presumably much colder, to warmer again as you approached Mercury to thousands/millions degrees F as you got within the outskirts of the solar corona.
At what point is the temperature again a balmy 60 - 70 degrees F and you can stop the wagon and hang out? In other words, how far from the edge of the corona is the temperature close to Earth-like? I’m assuming somewhere between Mercury and Venus, but wondering if there is a definitive answer to this.
It’s not a simple question.
In interplanetary space, the temperature of an object depends greatly upon it’s reflectivity. Space itself doesn’t really have a temperature. If you were to put a thermometer in space (say, between Mercury and Venus) it would register widely different temperatures if it was exposed to the sun, or in the shade of another object. As an example, the Space Station surface temperatures vary between +250°F and -250°F as it goes in an out of the Earth’s shadow. Also, a highly reflective surface will not get as hot as a less reflective one.
To add to beowulff’s answer it is really a factor of:
Solar absorptance: reflective materials less, dark materials far higher.
Infrared emittance: Remember there is not atmosphere for convection, so all heat has to be removed by EM radiation.
Note the white colored solar panel shaped objects at 90 degrees to the solar panels, those are radiators.
they need about one square meter for every 100-350W of electrical draw plus the additional cooling for solar radiation.
OK, looking at it from another angle: the corona is extremely hot, and extends thousands of miles from the sun’s surface. I realize that the quality of “temperature” describes the movement of particles, but it seems that as material streams away from the sun in the corona or solar wind, or what have you, you would have a region of particles that become less and less concentrated as they move outward. It seems intuitive that the temperature would drop as the stream becomes more diffuse, and so at some point you’d reach 70 degrees F (or whatever point you wish to name). The solar wind, I thought does continue outward in this way until the heliopause. Not sure about the corona. Is there a distinct edge to it, where the temp on one side is extremely high and on the other extremely cold; or does it gradually become more diffuse and cooler?
Yes temperature is related to speed or more precisely the average kinetic energy of the molecules which for a give mass depends on the speed squared. I don’t think the solar wind loses much speed at all until the particle hits something. Yes the gravitational pull of the sun will slow them down, but that wouldn’t have much of an effect. So the temperature of the solar wind won’t change much as you get farther from the sun.
What will change is the heat. Heat depends on the total energy. And the density of that will fall off with the square of the distance simply because the volume of space falls off with the square of the distance. The temperature could be quite high, but you’d not feel the heat as much because there would be fewer particles to bang into you and transfer their energy to you.
The problem with this line of reasoning is, the corona is extremely diffuse even near the sun. The corona may be a million degrees (K), but if you put a solid object there, conductive heating from the surrounding corona will be negligible because the density is so low. The heat input/output will be dominated by radiation. If the object is very close to the sun’s surface (photosphere), it will come close to the temperature of the photosphere, around 6000 K.
And if you were within the solar corona but shielded from the direct rays from the photosphere (behind some sort of shield, for instance) you’d freeze to death, because you’d be losing heat radiatively to deep space faster than you’d gain it conductively from the corona. On the other hand, if the surface of your ship were somehow a perfect reflector, then it would (very, very slowly) eventually come into equilibrium with the corona, because radiative transfer (either from the Sun or to deep space) would be completely cut off.
Here is an explanation and formula of the thermal equilibrium temperature of planets. The formula under “theoretical model” assumes a spherical planet, but is equally applicable to a spherical spaceship. Note that Teq is dependent on a, the albedo (reflectivity).
If you plug in a=0 (black surface) and D=1 AU, you get Teq=290K, which is pretty close to the 60F you want.
But that assumes there is no other heat input/output. In reality, a spaceship generates its own heat. All the energy used to run the life support system & other machinery turn into heat and add to the heat input from the sun. Which is why real-life spacecraft and space stations have big radiators to get rid of heat, and usually have shiny surfaces (high albedo) to minimize heat input from the sun.
To add to the above, “temperature” doesn’t have the meaning it does on earth. If you or your ship was an inert rock that didn’t generate any heat, what would determine the temperature it reached would be the blackness of the rock facing the sun, the distance from the sun, and the ratio of surface area of the rock facing the sun versus the area facing space.
I guess if the rock were a perfect black body and shaped like a sphere, there is an answer to your quesiton.
This is misleading. How hot is Mercury?
Venus has a higher average temperature than Mercury.
While it’s true to say that the amount of heat received from the Sun falls off with distance, the local effects mentioned in this thread are a better determiner of temperature. Average temperatures don’t have much meaning in space or on a spot on a planet.
However, if you really want to know where it averages 60 degrees, look outside. That place is Earth.
Only because of the “greenhouse effect” of the atmosphere. Without the atmosphere, the average temperature would be about 0 degrees F.
Ok, the answer can be worked out from the solution given to problem set 3 and 4 : http://www.lehman.edu/faculty/anchordoqui/problems303-12-sol.pdf
Am a bit busy right now, but you can use the equations they gave to make a simple python script that you can just type in any distance from the sun, the radius of the sphere, and it’ll tell you the equilibrium temperature. This is as close to “what is the temperature at this distance from the sun” as you can reasonably get unless you have a specific design for a spacecraft.
Here are a few planet location Watt’s per meter values for radiation intensity.
If the surface was 100% efficient in absorbing the solar radiation this is what you would have to dissipate per square meter to maintain the same temperature.
Mercury - 9116
Venus - 2662
Earth - 1337
Mars - 590
Jupiter - 50
Neptune - 1.5
Note that orbits are elliptical, and I am using some round number on the Stefan-Boltzmann’s blackbody equation to calculate the output from the sun so these are just just approximate and will change based on millions of factors.
The *temperature * of the spaceship would completely depend on how much energy it absorbs, and how much IR energy it emits.
In theory you could have a spaceship that was highly reflective, good IR radiator at Mercury’s orbit and one that was a good collector but bad radiator at Jupiter’s orbit and they could be the same temperature.
This is why the Webb space telescope has a huge multi layer reflector that is very very shiny. Because they need to block the heat of the sun for IR work.
Warning, huge image:
To add to my previous post, The typical “best” radiators the IIS uses right now top out at about 350w of dissipation per square meter.
So Poe Dameron’s X-wing or any other black space ship will have issues. This is why most spacecraft use a white coating that has a high emissivity.
That has only a little to do with it. The ISS is not using a heat pump cycle. (my source is asking an engineer who works on it). So the coolant leaving is just a little warm (heated by the cabin air and cooled equipment). That makes the radiators low temperature, which means their dissipation per square meter is awful because dissipation is proportional to temperature ^4 power. So even making the coolant just a little warmer has an enormous effect on radiated heat. (you can make it warmer quite easily with a heat pump)
My first thought was implanted in me by my mother, dead of winter or dead of summer: always pack a light sweater.
Sure, you could probably design a system that would glow as bright as the sun, that is what “black body radiation” is.
But they most likely made these decisions based on efficiency which tends to degrade rapidly with greater temperature differentials.
But I already did my free to the public math for the day, feel free to google the Carnot cycle it you want to calculate the efficiency curves. But consider the energy costs of air conditioning, which is really just a heat pump.
As long as the size of the sphere is small compared to its distance from the Sun, its size shouldn’t matter.
And in addition to using materials with different albedos in visible and infrared, you can use different materials on different surfaces, and make all of the parts that face the Sun light-colored or shiny, and all of the parts that face away dark. You can also use active cooling (like an air conditioner) to pump heat into the radiators so they’re hotter than the rest of the ship and hence radiate faster, but that’s pretty energy-hungry, so it’s usually easier to just use really big radiators.
Correct. The point is, if you have X watts of heat you must dissipate, and you use a heat pump with a coefficient of performance of 3 to do it, the total amount of heat you have to dissipate goes up by 1.3. If you can double the temperature at the radiators, you get 2^4, or 16 times as much heat dissipated, or 12 times as much once you factor in the heat pump cost.
So hugely worth it if you were, say, making a spacecraft used for interplanetary missions. (since that spacecraft would be frequently performing transfer burns throughout it’s lifespan, and reducing mass means less propellant needed)
But yeah, for one shot spacecraft (such as one used to go to Mars in the foreseeable future), the increased reliability of a simple but heavy radiator loop dominates.
One point is that it doesn’t matter how heavy your radiators are, just how much cross-sectional area they have. If you make your radiator out of very thin mylar foil, supported with a few low-pressure inflated struts of the same material (remember, it doesn’t have to be very strong in zero-g), then you can make a very large radiator with a very low weight.