00:01
All right, so let's say we have a planetary probe that has a diameter of two meters, and the temperature of the surrounding region in space is like absolute zero, and we're told that the probe sensors are getting a reading of negative 40 celsius, so 233 kelvin.
00:18
And we're told that the probe has an emissivity of 0 .9, and it has a power production internally of 100 watts per, cubic meter.
00:30
Let's call this q 100 watts per cubic meter.
00:36
And we wanted to find the power input of the incident radiation from all this.
00:39
And it has an absorption of just one minus the immissivity, so 0 .1 in this case.
00:45
So what we're going to have is like the power output of the probe is going to be, you know, q times the volume of the probe.
00:55
So this is pi times d cubed over six, and this should be 418 .8 watts.
01:03
And so the power input of radiation, called this pn, times the absorption rate, so this is 1 minus epsilon, plus this power output, sorry, should be equal to the net power, which is the emissivity times, according to stuff in bolton law, sigma times t to the fourth, times the surface area of our probe.
01:32
And the surrounding temperature is zero, so when we, you know, we don't have to worry about any power transfer or energy transfer from that...