Radiation in space (Thermodynamics)

[alivedude]

Homework Statement

A space probe carry an instrument that detects radiation in space, the area of the probe is ##5000 \, cm^2## and its collecting data for 10 minutes. The information sent back to Earth is given from this image below, how much radiation energy is it for 10 minutes?

(frequency on the x-axis)

Homework Equations

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Plancks radiation law? Maybe?

The Attempt at a Solution

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I have no idea how to approach this problem, the answer is ##10-50 \, MJ## but I have no hints or anything so I'm lost. Been playing around with the dimensions a bit but its pretty much just guessing. Any hints?

 

[haruspex]

Yes, please quote Planck's radiation law. Does its general shape match the graph? If so what values for the unknowns in the equation would make it match the graph exactly?

 

[rude man]

The y-axis units don't make sense to me. Especially the m^-3 factor.

 

[alivedude]

Yes, please quote Planck's radiation law. Does its general shape match the graph? If so what values for the unknowns in the equation would make it match the graph exactly?

Planck's radiation law:

##u(\nu , T)=\frac{8 \pi h \nu^3}{c^3} \cdot \frac{1}{e^{\frac{h \nu}{kT}}-1}##

I have really no idea to be honest.

 

[alivedude]

The y-axis units don't make sense to me. Especially the m^-3 factor.

exactly, clueless

 

[haruspex]

The y-axis units don't make sense to me. Especially the m^-3 factor.

The Y axis units are power per unit area, as given by Planck's radiation law. See https://en.m.wikipedia.org/wiki/Planck's_law

 

[haruspex]

Planck's radiation law:

##u(\nu , T)=\frac{8 \pi h \nu^3}{c^3} \cdot \frac{1}{e^{\frac{h \nu}{kT}}-1}##

I have really no idea to be honest.

Compare your graph with the family of curves at https://en.m.wikipedia.org/wiki/Planck's_law. Looks promising to me.
What variables in the general law can you tune to try to match your given curve?

 

[alivedude]

Compare your graph with the family of curves at https://en.m.wikipedia.org/wiki/Planck's_law. Looks promising to me.
What variables in the general law can you tune to try to match your given curve?

Im thinking like this:

integrate ##u(\nu) d\nu## and then multiply by the volume given by ##A \cdot c \cdot t ## where A is the arean of the probe, ##c## is the speed of radiation(light) and ##t## is the time (10 mins)

Is this a crazy idea? I think that integrating that #u(\nu)# is beyond the course but maybe i can approx the area under the curve?

 

[haruspex]

Im thinking like this:

integrate ##u(\nu) d\nu## and then multiply by the volume given by ##A \cdot c \cdot t ## where A is the arean of the probe, ##c## is the speed of radiation(light) and ##t## is the time (10 mins)

Is this a crazy idea? I think that integrating that #u(\nu)# is beyond the course but maybe i can approx the area under the curve?

Yes indeed, you need the area under the curve. But the method I was leading you to is to identify the source temperature that gives rise to the observed spectrum. Armed with that, no approximation is necessary. The integral has an analytic solution (which you can look up at the page I linked to).
However, on closer inspection I'm beginning to suspect there is an error in the question. It seems to me that the frequencies on the given graph should multiples of 10¹⁵, not 10¹⁴.

If you feel happy just approximating the area by hand, feel free.

 

[alivedude]

Yes indeed, you need the area under the curve. But the method I was leading you to is to identify the source temperature that gives rise to the observed spectrum. Armed with that, no approximation is necessary. The integral has an analytic solution (which you can look up at the page I linked to).
However, on closer inspection I'm beginning to suspect there is an error in the question. It seems to me that the frequencies on the given graph should multiples of 10¹⁵, not 10¹⁴.

If you feel happy just approximating the area by hand, feel free.

Im going to check this out, but I think that I understand now. Thanks a lot!

 

[rude man]

Im thinking like this:

integrate ##u(\nu) d\nu## and then multiply by the volume given by ##A \cdot c \cdot t ## where A is the arean of the probe, ##c## is the speed of radiation(light) and ##t## is the time (10 mins)

Is this a crazy idea? I think that integrating that #u(\nu)# is beyond the course but maybe i can approx the area under the curve?

The data from the spacecraft has to be a spectral energy density, which would be J - s which, when integrated over the frequencies gives the required units which is J. That's what the problem states: "how much radiation energy is it for 10 minutes?"

The fact that the integration time is over 10 minutes and over an area of 5000 cm² is of no computational consequence.
So the units on the given chart, which was represented as the acquired data, should have been J Hz^-1 or J-s.

Of course, an on-board computer might have taken area and integration time into account, sending out B which would have the units J m^-2 s^-1 s = J m^-2. Again, though, B ⋅ 5000 cm² ⋅ 10 minutes integrated over frequency would give J. This B has the same units as in the wikipedia article: "The SI units of Bν are W·sr⁻¹·m⁻²·Hz⁻¹".
Dimensionally:
W = J s^-1
Hz^-1 = s
sr is dimensionless (steradians)

 

[haruspex]

The data from the spacecraft has to be a spectral energy density, which would be J - s which, when integrated over the frequencies gives the required units which is J. That's what the problem states: "how much radiation energy is it for 10 minutes?"

The fact that the integration time is over 10 minutes and over an area of 5000 cm² is of no computational consequence.
So the units on the given chart, which was represented as the acquired data, should have been J Hz^-1 or J-s.

Of course, an on-board computer might have taken area and integration time into account, sending out B which would have the units J m^-2 s^-1 s = J m^-2. Again, though, B ⋅ 5000 cm² ⋅ 10 minutes integrated over frequency would give J. This B has the same units as in the wikipedia article: "The SI units of Bν are W·sr⁻¹·m⁻²·Hz⁻¹".
Dimensionally:
W = J s^-1
Hz^-1 = s
sr is dimensionless (steradians)

Yes, I realized later my mistake here, by which time I'd no internet.
The graph given has units for ##B_{\lambda}## on the Y axis, not ##B_{\nu}##. Unless that is a mistake, the area under the curve will not give the right answer. You would need to convert the graph to have the X axis in terms of wavelength first.