Hi jasonwjs,
this is by no means an easy question you are asking! First I will have
to make a few assumptions for the calculation to be possible at all:
- Temperature in deep space usually only reaches 3°C above absolute
zero. However, you are asking about an object in earth orbit. If I
assume that it is actually orbiting the earth, (planetary and) solar
radiation would heat up an object that perfectly absorbs and emits to
280° K (7° C) (or about 5° C if planetary radiation and shielding
effects are taken into account) - which is what I will assume as your
environmental temperature. (
http://www.faqs.org/faqs/astronomy/faq/part4/section-14.html )
- I will assume that your black nickel has an emissivity of 0.06 and
an absorbtivity of 0.93 of the incoming radiation (
http://www.efdlant.navfac.navy.mil/criteria/documents/MH/1003_13A.PDF
)
- I will also assume that one surface is hit by the radiation at a 90°
angle.
To calculate an equilibrium temperature we will have to use the
Stefan-Boltzmann Law:
Radiation Power ( * absporptivity ) = emissivity * Stefan's Constant *
Radiation Area (Temperature of Radiator^4 - Temperature of
surroundings^4)
In your case we assumed emissivity as 0.06, absorptivity 0.93 ,
Stefan's Constant is 5.6703*10^-8 watt/((m^2)(K^4)), Radiation Area is
assumed to be 1 m^2 for this calculation and Temperature of
surroundings is 280°K. To achieve equilibrium we also have to assume
that the emission of energy is equal to the solar radiation: 1350
watt/m^2.
This brings us to the following equation:
1350 W/m^2 * 0,93 = 0.06 * 5.6703*10^-8 W/((m^2)(K^4)) * 1 m^2 * ( T^4
- 6146560000 K^4 )
If we bring T to the left side and all the other parts to the right,
the equation looks like this:
1350 W/m^2 * 0.93 = 0.06 * 5.6703*10^-8 W*m^2/K^4 * T^4 - 0.06 *
6146560000 * 5.6703*10^-8 W*m^2
T^4 = ((0.93 * 1350 + 0.66 * 6146560000 * 5.6703*10^-8)/0.06 *
5.6703*10^-8) K^4
T^4 = (0.93 * 1350/(0.06 * 5.6703*10^-8) + 6146560000) K^4
T =~ 783° K
However, be aware that if this was a homework question or similar,
that you teacher might have assumed that the temperature in space is
2.7° K due to the famous background radiation only, without taking
planetary and solar radiation into account for the calculation of
surrounding temperature (it is very difficult to say what the
surrounding temperature in a certain area of space really is). If you
insert approximately 3° K into the above equation the result would be
T =~ 779°
The resulting temperatures seem to be very high - that is due to the
black nickel coating being a nearly ideal absorber of solar radiation
while being a much poorer emitter.
I have also found an article describing the equilibrium temperature of
ballistic missiles ( http://www.ucsusa.org/security/CM_apA-E.pdf )
where the scientists took the temperature of the surrounding out of
the equation and instead integrated albedo effects into the equation.
However, it is not so easy to use their equation because it requires a
spherical rotating body.
If you assume that your body is orbiting with one of its thin sides
towards earth and in some distance to it, you could assume that the
radiation from the planet is fairly small and can be ignored. In such
a case you would just have to assume the surrounding temperature as
zero and do the calculation with that in mind:
T^4 = 1350 * 0.93/(0.06 * 5.6703*10^-8) K^4
T =~ 779 K
Please request a clarification if you have any questions about the
answer!
Additional Resources
Stefan-Boltzmann Law
http://hyperphysics.phy-astr.gsu.edu/hbase/thermo/stefan.html#c3
Is Space Hot or Cold?
http://spider.ipac.caltech.edu/staff/waw/mad/mad5.html
Search Terms
equilibrium temperature space diffuse
://www.google.com/search?sourceid=navclient&q=equilibrium+temperature+space+diffuse
temperature of space
://www.google.com/search?sourceid=navclient&q=temperature+of+space
absorption radiation "black nickel"
://www.google.com/search?sourceid=navclient&q=absorption+radiation+%22black+nickel%22
emissivity nickel
://www.google.com/search?sourceid=navclient&q=emissivity+nickel |
