How hot is the Sun in space? (James Webb related)

[brajesh]

TL;DR Summary

James Webb temperature differential 318 degrees!

I'm reading about the engineering of the James Webb and it surprised me that the temperature difference will be about 318 degrees from one side of the telescope compared to the other (85 C on one side and -233 on the other).

I didn't expect space to be that hot about a million miles further from the sun than the Earth is!
I guess our atmosphere really helps to reduce the heat of the sun!

Is there a formula to calculate how hot the sun is in space, suppose X million miles from the sun?

 

[Orodruin]

It is not about how hot the Sun is (the radiation from the Sun always has the same temperature), but about the solar irradience which follows the inverse square law.

Once you have irradience you need to consider how an object absorbs the sunlight and cools itself by radiating.

 

[russ_watters]

...but about the solar irradience which follows the inverse square law.

...which is about 1000 Watts / sq meter.

 

[collinsmark]

I didn't expect space to be that hot about a million miles further from the sun than the Earth is!

Compared to the distance between the Earth and the Sun, a million miles isn't that far. The Sun-Earth L2 point, which is roughly 930,000 miles from Earth, is only about 1% farther way from the sun than the distance from the Sun to the Earth (about 93,000,000 miles).

For comparison, if you were about 100 yards (almost 100 meters) away from a flagpole/streetlight/something, and that distance represents the distance between the Earth and Sun, you could reach out your arm and touch the L2 point.

I guess our atmosphere really helps to reduce the heat of the sun!

In terms of overall energy, not so much. About 2/3 of the Sun's energy reaching Earth is absorbed by Earth (i.e., only about 1/3 the energy is reflected). As a matter of fact, the atmosphere keeps the Earth warmer than it otherwise would be, on average, due to the greenhouse effect.

What keeps the Earth from getting hotter than it already is, and the same applies to the the JWST, is that empty space (such as what the side of an object not illuminated by the sun "sees") is really, really cold.

 

[sophiecentaur]

Summary:: James Webb temperature differential 318 degrees!

I guess our atmosphere really helps to reduce the heat of the sun!

I assume you mean by "the heat", the radiation arriving at the surface. This is more or less constant as Earth is always pretty much the same distance from the Sun (nearly circular orbit).
What happens to the Earth is just a more complicated version of what is happening in JWST. There are a range of temperatures on Earth and in the atmosphere and the rates of gain and loss of Energy for Earth would be constant if the system were in complete equilibrium. Obvs, daytime and nighttime temperatures are different but there is a lot of thermal lag and the atmosphere contributes to this. Cloudy nights are warmer than cloudless nights etc..
JWST has no atmosphere and it always faces in the same direction relative the Sun. There is still some heat admitted by the heat shield and some small transfer through the craft but a Delta T of 300C is pretty useful.

BTW, the Earth and Moon are nearby 'hot' bodies and the heat shield also helps to reduce the effect of heat radiation from both. (Plus, man made interference, I imagine).

 

[snorkack]

Summary:: James Webb temperature differential 318 degrees!

I'm reading about the engineering of the James Webb and it surprised me that the temperature difference will be about 318 degrees from one side of the telescope compared to the other (85 C on one side and -233 on the other).

I didn't expect space to be that hot about a million miles further from the sun than the Earth is!
I guess our atmosphere really helps to reduce the heat of the sun!

Is there a formula to calculate how hot the sun is in space, suppose X million miles from the sun?

It is not so much atmosphere, but rotation of Earth.
Earth is a sphere. Which means that Earth has 4 times the area of its cross-section.
Yes, the poles are turned away from Sun. But Earth rotates. The circumference of a circle is still π times its diametre, so even equator has night half of day, and most of the rest the Sun is not at zenith.
If you consider a surface that faces Sun all the time, and is perfectly insulated on its back, then all the incoming sunlight must be radiated back as infrared. Which is 4 times more than Earth gets on average, and π times more than equator gets on average. Thus calculate fourth root of π, because the infrared radiation is proportional to T⁴.

 

[sophiecentaur]

all the incoming sunlight must be radiated back as infrared.

That would only apply to a black surface. A very reflective surface would only absorb less than 1% and reflect the rest. On JWST the multiple layers help further. It can be really handy to have your own built in vacuum flask!

 

[hutchphd]

-

That would only apply to a black surface. A very reflective surface would only absorb less than 1% and reflect the rest

As long as the surface is gray (shows no wavelength preference) and one considers only radiative processes, the steady state equilibrium temperature is independent of emissivity. By detailed balance the emissivity and absorbtivity must be the same.
A gray Earth and a black Earth will each end up at 300K.

 

[brajesh]

...which is about 1000 Watts / sq meter.

Shouldn't this get less as you get further away from the sun?

 

[Astronuc]

Shouldn't this get less as you get further away from the sun?

JWST is not going that far from earth, only about 1 million miles, compared to the distance from the sun ~93 million miles (~149.6 million km). The "1000 Watts / sq meter" is at Earth's distance from the sun.

 

[sophiecentaur]

As long as the surface is gray (shows no wavelength preference)

Yes. You are strictly right about that. I actually meant almost totally absorbent. That would mean a very very dark colour, whichever you would call the colour - looking at it - or the "wavelength preferences".

 

[russ_watters]

JWST is not going that far from earth, only about 1 million miles, compared to the distance from the sun ~93 million miles (~149.6 million km). The "1000 Watts / sq meter" is at Earth's distance from the sun.

And if we want to be pedantic, I pulled that number out of the air without checking. Checking now, it is 1,360 for Earth so using the inverse square law that gives 1,125 w/sq m at L2.

 

[brajesh]

JWST is not going that far from earth, only about 1 million miles, compared to the distance from the sun ~93 million miles (~149.6 million km). The "1000 Watts / sq meter" is at Earth's distance from the sun.

ok. So how could I use the distance in a formula to get the watts per meter squared?

What is it at 50 million miles?
At 200 million miles?
etc?

 

[russ_watters]

ok. So how could I use the distance in a formula to get the watts per meter squared?

What is it at 50 million miles?
At 200 million miles?
etc?

The factor is 1 divided by the distance delta, squared (per my previous).

 

[brajesh]

The factor is 1 divided by the distance delta, squared (per my previous).

Thank you!

 

[Astronuc]

What is it at 50 million miles?
At 200 million miles?
etc?

One would take the total radiance of the sun (in kW) and divide by the surface area of the spherical area (4πR²) at distance R from the sun.

Consider that Skylab needed a sun shade while in Earth's orbit.
https://www.space.com/15705-skylab-sun-shade.html

 

[OmCheeto]

And if we want to be pedantic, I pulled that number out of the air without checking. Checking now, it is 1,360 for Earth so using the inverse square law that gives 1,125 w/sq m at L2.

I get 1333 w/sq m at L2. (using 1360 w/m^2 @ 149.6e6 km and L2 = 151.1e6 km)

...

Is there a formula to calculate how hot the sun is in space, suppose X million miles from the sun?

Have you written down the formula yet? I'm curious if it matches mine. Thanks!

 

[hutchphd]

And if we want to be pedantic, I pulled that number out of the air without checking. Checking now, it is 1,360 for Earth so using the inverse square law that gives 1,125 w/sq m at L2.

L2 is 1 million miles farther. $$\delta=\frac 1 {93}$$
$$(1+\delta)^{-2}\approx(1-2\delta)$$ so the difference should be about 2% which is $27W/m^2$

OOPs:minus

 

[snorkack]

How much of Sun´s disc shows at L2? How much of Sun´s luminosity, accounting for limb darkening? Some does, because Earth is denser than Sun, so L2 is in antumbra, not umbra.
Uranus is less dense than Sun. Is Uranus´ L2 a place where the sun does not shine?

 

[JLowe]

How much of Sun´s disc shows at L2? How much of Sun´s luminosity, accounting for limb darkening? Some does, because Earth is denser than Sun, so L2 is in antumbra, not umbra.
Uranus is less dense than Sun. Is Uranus´ L2 a place where the sun does not shine?

Nasa website says that Webb's orbit at L2 will be similar in size to the Moon's orbit around the Earth. I don't know how much sun would be at the center of L2, but I know Webb will not be in the center.

 

[hutchphd]

There are issues for radio communication if the Earth is not angularly separated from the sun for the Webb antennas. I had originally thought they would use the Earth to eclipse the sun but quick consideration of lunar eclipses tells you the Earth ain't big enough.

 

[256bits]

Summary:: James Webb temperature differential 318 degrees!

I'm reading about the engineering of the James Webb and it surprised me that the temperature difference will be about 318 degrees from one side of the telescope compared to the other (85 C on one side and -233 on the other).

I wonder from where they got the 85C on the hot side.
What heat sources are they including?

A gray Earth and a black Earth will each end up at 300K.

Additional source would be radiation ( and reflection ) from the moon and earth, which they did mention - about 30 watts or so.
And heat from the cool side.
The article did not go into that aspect.

 

[hutchphd]

I wonder from where they got the 85C on the hot side.

It essentially the temperature on sunward Earth if it didn't rotate. The radiation emitted would be then cut in half and the Temperature increased by $\sqrt[4] 2$ so $T=(300K) \sqrt[4] 2=357K$ . Not complicated.

 

[sophiecentaur]

It essentially the temperature on sunward Earth if it didn't rotate.

That would apply to the Earth's surface (not the atmosphere) in a very simple model. The effective mean temperature of the Earth has to be such as to produce thermal equilibrium. But there's nowhere actually at that temperature.
A fast spinning metal satellite would end up at 300K (+/-). The conductivity of a body in orbit has to be taken into account and also the colours (reflectivity) of the faces. So, if you take a satellite with a 365day rotation period, paint it white on the sunny side and black on the dark side, the core temperature would be much lower (and vice versa). The 'average' temperature over the surface would still be 300K. Inside, you could arrange for anything in between. JWST's heat shield gets VERY hot but that allows the cold side to be very cold.

 

[russ_watters]

Note, the moon rotates slowly, has no atmosphere and is pretty black. Its sunny-side temp is 120C.

 

[sophiecentaur]

Note, the moon rotates slowly, has no atmosphere and is pretty black. It's sunny-side temp is 120C.

A very simple situation compared with us but it makes the point.
The side in shadow will be pretty darn cold. But a month is quite a short time for thermal matters in a big rock.