Timer Notes Evaluate Feedback Print Course Contents Lecture #13: Heat Bonus Ques
ID: 1419960 • Letter: T
Question
Timer Notes Evaluate Feedback Print Course Contents Lecture #13: Heat Bonus Question: Spinning Asteroid A small spinning asteroid is in a circular orbit around a star, much like the earth's motion around our sun. The asteroid has a surface area of 9.50 m2. The total power it absorbs from the star is 4300 W. Assuming the surface is an ideal absorber and radiator calculate the equilibrium temperature of the asteroid (in K) A: 1.83x102 OB: 2.07x102 oc: 2.34x102 OD: 2.65x 102 o E: 2.99x 102 OF: 3.38x 102 OG: 3.82x 102 H: 4.31x102 Submit Answer Tries 0/20 his discussion is closed. Send FeedbackExplanation / Answer
E=A sigma T^4 where E is the energy emitted/sec, A is the surface area of the emitting object, sigma is a constant the Stefan Boltzman's constant = 5.67x10^(-8) W/m^2/K^4 and T is the radiating temperature in degrees Kelvin; perfect radiators absorb all the energy incident upon them and then radiate all that energy to space
the statement that the asteroid is spinning means that we can assume all parts of it are at the same temperature (in other words, there is not a warm side and cold side at very different temps)
so, if the asteroid absorbs 4300W, then assuming it is a perfect emitter/absorber, we know also that the energy emission is 4300 J/s (1W=1J/s), we have
4300W=9.5m^2 x 5..67x10^(-8) W/m^2/K^4 x T^4
(are you sure about the area, that's a really small asteroid, did you omit the exponent?)
anyway, usingthe data given here, we solve for T:
T^4=4300W/(9.5m^2 x 5.67x10^(-8)W/m^2/K^4)
T^4=7.9829x10^9 K^4
T=299K which is a reasonable temperature for the inner solar system
or T=2.99*10^2K
applying Wien's law to this problem would allow us to compute the wavelength at which the asteroid emits the peak amount of energy; this would occur at a wavelength of about 1 micron, which is deep in the infrared portion of the spectrum
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