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Estimating the Temperature of a Flat Plate in Low Earth OrbitThis problem is a realworld problem that I first encountered in my daily work at NASA. A flat plate is orbiting the Earth at a mean altitude of 300 km. Its orbital velocity is 7.5 x 10^{3} m/sec eastward. (All spacecraft are launched eastward to take advantage of the Earth's rotational motion.) The plate is in sunlight. Sunlight warms the plate, and the plate radiates thermal energy back into space. Ambient atomic and ionic species are also present. These species are moving at characteristic thermal velocities, which are very small compared to the plate's orbital velocity and may, for all intents and purposes, be ignored. The flat plate may be thought of as 'running into' populations of stationary particles at a velocity of 7.5 x 10^{3} m/sec. From the plate's point of view, these particle populations appear to be ramming it 'headon' at 7.5 x 10^{3} m/sec. When the particles hit the plate, they are assumed to loose their kinetic energy to the plate as heat. Again, as in the case for sunlight, the plate also radiates this energy back into space. The balance between incoming energy and outgoing (radiated) energy causes the plate to come to an equilibrium temperature. The problem is to calculate this temperature. We begin with StefanBoltzmann's law for a black body in sunlight. S_{SUN}
=
T^{4 }W/m^{2}
This law enables us to estimate the temperature T of such an object, assuming that we know the power per unit area S_{SUN} falling on the plate. From direct measurement, we know that at 1 A.U, S_{SUN}
= 1,360 W/m^{2}.
We also know that
= 5.67 x 10^{8} W/(m^{2} K^{4}).
Thus, the nominal temperature of an object, in space and in sunlight, is 394 °K. Next, we must deal with the ambient particle fluxes. In general, random directional particle flux to the surface is calculated from =¼(nV_{PARTICLE})
m^{2}sec^{1}
where n is the number density of the particle species per m^{3}, and V_{PARTICLE} is the average velocity of the individual particles in m/s. Here, we take V_{PARTICLE} = 7.5 x 10^{3} m/sec. Since particle flux to the surface is assumed to be ram directional (i.e., we are assuming that the plate 'sees' all the particles coming headon at 7.5 x 10^{3} m/sec), the usual factor of ¼ (which arises when the directions are random) may be omitted. From an established knowledge of the orbital environment at 300 km, we select the following, dominant particle species: Oxygen
ions: O^{+
}Atomic oxygen: O
Molecular oxygen: O_{2} Molecular nitrogen: N_{2}. Then, for each of these selected particle species, we may define a power flux to the surface: S_{PARTICLE}
= (nV_{PARTICLE})(E_{KINETIC/PARTICLE}) W/m^{2}.
where E_{KINETIC/PARTICLE} is the kinetic energy per particle. The values of n and E_{KINETIC/PARTICLE} must be determined. If (as we have already assumed) the particles simply deliver their kinetic energy to the plate upon impact (totally inelastic collision), then they may be treated as energy fluxes additional to that of the sun, and a final temperature may be estimated from: T^{4}
= S_{SUN} + S_{PARTICLES}
Now, we are ready to calculate. We have already calculated the temperature contribution from the sun: T =
394 °K.
Next, we calculate the kinetic energy of each of the particle species: 1. Oxygen ions: O^{+}: m_{O+}
= 16 x (1.67 x 10^{27} kg) = 2.67 x 10^{26} kg
V_{O+} = 7.5 x 10^{3} m/sec E_{K/O+} = ½ m_{O+} (V_{O+})^{2} = 7.51 x 10^{19} j 2. Atomic oxygen: O: m_{O}
= 16 x (1.67 x 10^{27} kg) = 2.67 x 10^{26} kg
V_{O} = 7.5 x 10^{3} m/sec E_{K/O} = ½ m_{O} (V_{O})^{2} = 7.51 x 10^{19} j 3. Molecular oxygen: O_{2}: m_{O2}
= 32 x (1.67 x 10^{27} kg) = 5.34 x 10^{26} kg
V_{O2} = 7.5 x 10^{3} m/sec E_{K/O2 }= ½ m_{O2} (V_{O2})^{2} = 1.50 x 10^{18} j 4. Molecular nitrogen: N_{2}: m_{N2}
= 28 x (1.67 x 10^{27} kg) = 4.68 x 10^{26} kg
V_{N2} = 7.5 x 10^{3} m/sec E_{K/N2} = ½ m_{N2} (V_{N2})^{2} = 1.32 x 10^{18} j The number densities of each of the particle species at the altitude of 300 km are obtained from published references (cited): n_{O+}
= 5 x 10^{11} /m^{3} (Sp. & Planetary Env., Vol. 1,
pg. 231)
n_{O} = 2 x 10^{14} /m^{3} (U.S. Std. Atm., pg. 30) n_{O2} = 10^{11} /m^{3} (U.S. Std. Atm., pg. 30) n_{N2} = 5 x 10^{12} /m^{3} (U.S. Std. Atm., pg. 30) Using these quantities, we obtain the power fluxes S for each of the particle species: S_{O+}
= 2.82 x 10^{3} W/m^{2
}S_{O} = 1.13 W/m^{2
}S_{O2} = 1.12 x 10^{3} W/m^{2
}S_{N2} = 4.96 x 10^{2} W/m^{2}
For convenience, we sum these contributions: S_{PARTICLES}
= S_{O+ }+ S_{O} + S_{O2} + S_{N2} = 1.18
W/m^{2}.
Recall that S_{SUN} = 1,360 W/m^{2}, so that the temperature contribution from the ram particles alone must be very small in comparison. Let us estimate this ram particle temperature contribution. An estimation of the change in temperature T due to S_{PARTICLES} may be obtained from^{10} S_{PARTICLES}
= 4T^{3}
T
With T = 394 °K, and S_{PARTICLES }= 1.18 W/m^{2}, we find T = 8.51 X 10^{2} K. This value is too small to make a significant difference when compared to the contribution of the sun. We therefore ignore it, and take the plate's temperature to be 394 °K.
^{10} To
obtain this expression, let us rewrite our working equations as follows:
Set S = S_{SUN}, and set S + S
= S_{SUN} + S_{PARTICLES}
. Then, set S = T^{4}
and S + S
= (T
+ T)^{4}
= (T^{4}
+ 4T^{3}T
+ 6T^{2}(T)^{2}
+ 4T(T)^{3}
+ (T)^{4})
(T^{4}
+ 4T^{3}T)
for T<<T.
Subtracting the two expressions and substitting S
= S_{PARTICLES}
gives the expression in the text.


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