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Space probe problem

  1. May 9, 2014 #1
    Thanks for any help in advance. I'm currently doing random physics problems in preparation for my exam. I came across this one and thought I'd give it a go. The question at hand is,

    A space probe has a 100W thermal energy source and is in deep space after some years of travel.
    The surface of the probe is black and has an area of approximately 6m^2. The power is internal to the space probe. Determine approximately:

    1. The temperature of the unmodified space probe
    2. The temperature of the space probe if a thin thermal shield (also black) surrounds the space probe. The shield is attached close to the probes surface by a number of thermally insulating supports.

    For the first part i assumed the power source was at the center of the space craft and radiated outwards, hitting the entire spacecraft evenly. I calculated the temperature using Stefan-Boltzman law.

    for part 2. I thought this was to do with the law of conduction but I can't really seem to get anywhere with it and the more time I spend on it the more I'm convinced the way I thought I had solved the first part is wrong...

    Some guidance would be greatly appreciated.

    2. Relevant equations
    P = σAT^4
    dq/dt = Ka dT/dx
  2. jcsd
  3. May 9, 2014 #2
    The solution to the first part seems right. Can't tell for sure since you didn't post it. Showing your works is always better than describing it. The 2nd part is solved similarly but don't forget that the thermal shield emits radiation according to its temperature in both directions - outward to space and inward back to the probe.
  4. May 9, 2014 #3
    Thanks Dauto. I realized that shortly after i posted. Interestingly, if you keep adding shields you end up with a geometric sequence. Some of the numbers are bizarre though.
  5. May 9, 2014 #4

    D H

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    Another hint, if you haven't already solved the problem: the probe has been in deep space for a long time. That means it is in a steady state where the thermal energy generated by the heat source and thermal energy radiated into deep space balance one another out.
  6. May 9, 2014 #5
    Thanks for the response. That is only relevant before you add a shield, correct? I originally modeled it so when the power is emitted from the probe (black body) and hits the shield, the shield absorbs the total power then re-emits it in two directions, hence half of the power is 'reflected' back to the probe. Once the probe absorbs the half that was remitted it then re-emits with 1.5 of the power is started with (100w) the same process happens again and again.. hence a geometric series.
  7. May 9, 2014 #6

    D H

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    You've just made an over unity device! Does that make any sense?

    The probe is still going to reach a steady state temperature, just not the same temperature as the probe without a heat shield.
  8. May 9, 2014 #7
    Well, I assumed the power the probe had to 'remit' once a shield was added would be greater and as such the temperature of just the probe would increase. Otherwise, wouldn't it be pretty useless adding insulation? The only time the temperature would decrease is if the initial thermal source stopped radiating.

    What I mean is you have the original temperature of the probe when there is just a thermal source, when you add a shield it will absorb the power being emitted from the thermal source and remit half of it back to the probe(which is the thermal source), hence increasing the temperature. However, a consequence of the probe being a black body means it will absorb this remitted power (from the shield) and then remit it the initial power it had (as a thermal source) + the power it absorbed from the shield remission.

    This makes sense doesn't it? As you add more shields the temperature of the probe increases.

    I hope that makes sense...

    Thanks for the replies.
  9. May 9, 2014 #8

    D H

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    The probe is still going to reach some finite steady state value. The probe temperature will not keep rising and rising without bound.

    You haven't shown your work yet.
  10. May 10, 2014 #9
    Sorry for the delay. I'll put my working on here shortly.
  11. May 10, 2014 #10
    Here it is,

    So I start by assuming my probe is a black body. For the first part one can use the Stefan-Boltzmann law to re-arrange for T, since,

    Pprobe,i = σAprobeT4probe.

    This is pretty trivial but you find that the temperature of the probe is approximately 131K.

    Now we add a heat shield. Since this is also treated as a blackbody, the shield absorbs the power given off by the probe,

    Pprobe,i = Pshield.

    The shield then re-emits the power in both directions, hence half of it is reflected back at the probe.

    The final power of the probe is then,

    Pprobe,f = Pprobe,i + 1/2 Pprobe,i.

    The probe is in equilibrium with the shield, since the intial power was absorbed by the shield. From this point we therefore only consider the 1/2 Pprobe,i that was 'reflected' back to the probe. This is then re-emitted by the probe, it hits the shield, again the shield remits it in both directions and the 'reflected' power is therefore,

    1/4 Pprobe,i.

    This keeps happening and as such we end up with a geometric series,

    Pprobe,total = Pprobe,i [1 + [itex]\sum[/itex] ([itex]\frac{1}{2}[/itex])n] (sum from 1 to infinity)

    This geometric series converges to 1.

    Therefore, the temperature when adding a shield is given by,

    2Pprobe,i = σAprobeT4probe

    I found an extension to this problem. How many shields would be required to keep the probe at 250K. This is where it starts to get a lot trickier. Since as you add more shields the splitting becomes more complicated. Maybe you could offer some advice on how to do this in a 'nicer' way?

  12. May 10, 2014 #11

    D H

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    There's a much easier way to get that factor of 21/4 increase in temperature due to the heat shield. Everything is in a steady state. That means the heat shield must be at about 131K. This also means the heat shield radiating 100 watts out into space and is radiating 100 W back to the probe. What about the probe? It's receiving 100 watts from from the heat source and 100 watts the shield, for a total of 200 watts. To be at steady state, it must be radiating 200 watts, and thus is at a temperature of 21/4*131K.

    With two heat shields, the outer shield will still be at 131K, the inner shield will be at 21/4*131K, and the probe itself will be at 31/4*131K. Note well: There's an implicit assumption here that probe and the heat shields all have the same surface area.
  13. May 11, 2014 #12
    I see, so you've used the fact that the shield and probe are in thermal equilibrium (which is why you were mentioning this a few days ago). If that is the case you also know that the shield must be radiating the probe with as much power as the probe is radiating the shield with - 'so a body at temperature T surrounded by a cloud of light at temperature T on average will emit as much light into the cloud as it absorbs.' - in essence it's this right? Hence, if the cloud is our shield and the body is our probe, the probe must be absorbing 100W from the shield. Hence the power has increased by 2-fold.

    I keep going over this, but I'm getting terribly confused. I can see how it works for just 2 shields (i think). However, once you add the third shield I'm struggling to see how this method progresses. It's probably how I'm thinking about it.

    With the two shields, I was thinking that the outer shield and the probe are initially in thermal equilibrium (both at 131k). They both radiate in equal directions (this is only relevant for the shields not the probe), hence the shield trapped between the outer shield and the probe is being hit by the radiation from the shield and the probe, hence 2^1/4 * 131K. This results in the probe being the original 100W + the 200W that are coming back from the middle shield. It's tricky to think of it this way (assuming its even correct) once you add a third shield...
    Last edited: May 11, 2014
  14. May 11, 2014 #13
    Also, when you add the 2nd shield, does that mean the middle shield isn't in a steady state with everything else?
  15. May 11, 2014 #14

    D H

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    You're overthinking this. Since you've been struggling with this for a while, and since the answer is so blindingly simple once you see it, I'll show you the solution.

    Think of the probe, the heat shields, and empty space as being radiative layers. I'll denote empty space as the zeroth layer, the outermost heat shield as layer #1, and so on down to the probe itself which is layer #N. The nth layer is at a temperature Tn that makes it radiate with power Pn. The probe radiates with power PN outwardly. The nth heat shield radiates Pn inwardly and Pn outwardly. The space layer, layer #0, doesn't radiate1.

    The steady state condition is very simple: A given layer must transfer 100 watts, net, to the layer just outside it:
    [tex]P_n-P_{n-1} = 100\,\text{W}[/tex]Since the space layer doesn't radiate1, P0=0, which in turn means P1=100 watts, and in general, Pn=n*100 watts. This in turn means the temperature of the nth layer is
    [tex]T_n = \left(\frac{P_n}{\sigma A}\right)^{1/4} = \left(\frac{n*P_1}{\sigma A}\right)^{1/4} = n^{1/4}T_1\,\,\text{where}\,T_1=\left(\frac{100\,\text{W}}{\sigma A}\right)^{1/4}[/tex]
    Let's do a sanity check to see if this simple result, Pn=n*100 watts, makes sense from an energy balance perspective. The outermost heat shield receives 200 watts from heat shield #2, and radiates 100 watts into space and 100 watts back to heat shield #2. The probe itself receives 100 watts from the energy source and (N-1)*100 watts from the innermost heat shield, and radiates N*100 watts toward the innermost heat shield. The intermediate level heat shield at layer n receives (n+1)*100 watts from the layer below and (n-1)*100 watts from the layer above, and radiates n*100 watts both above and below. In all cases, the incoming and outgoing energies are equal. In other words, we have an energy balance.

    Note that the above assumed that the probe and each heat shield have the same area, 6 square meters in this case. The surface area of the nth heat shield has to be a bit more than that of the layer just below. In practice, the assumption of constant surface area across all heat shields can be made to be almost true by making consecutive heat shields separated by a very small distance.

    1 Strictly speaking, that empty space does not radiate is not true. The cosmic microwave background means empty space has an effective temperature of 2.73K. The outermost head shield will receive about 18 microwatts from empty space. That 18 microwatts is very tiny. For this problem it's fair to say that empty space doesn't radiate.
    Last edited: May 11, 2014
  16. May 11, 2014 #15
    Thanks for all the help D H. Do you happen to know of another problem that is similar to this one so I can try all the things you've brought to light?
  17. May 11, 2014 #16

    D H

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    There is no general one-approach-fits-all technique. One approach might work much better than another to solve one problem but the exact opposite will be the case for another problem. You should have a number of tools in your toolkit and you should be ready and willing to switch tools when your first guess tool leads to a mess.

    That said, the conservation laws are incredibly powerful, so long as you don't care about minutia such as the mechanisms by which a system reaches steady state. Sometimes those "minutia" are extremely important; they are what determine if a system breaks or remains intact. Other times, those minutia are just that, little trivial inconsequential details that detract you from finding the solution. That's the case with this problem. How the heat source + probe + heat shield system reaches a steady state is irrelevant. What's important is that steady state solution.
  18. May 11, 2014 #17
    Thanks for your response and again for all the help you've provided, albeit a bit painful once I saw your solution. I do see exactly what you mean by the steady state solution in this problem now and I certainly made it considerably more difficult for myself by trying to keep a track of every interaction building up to the said steady state.
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