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What parameters control the amount of thermal energy a space object must possess for it to be detectable by a sensor also in space (i.e. one that does not have to deal with interference from a planet's atmosphere)?

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It's not the heat that matters, but the brightness. Sedna, discovered in 2003, has an apparent magnitude of 21.3. Wikipedia lists an apparent magnitude of 27 as the faintest object observable from an eight-meter ground-based telescope, and magnitude 36 as the faintest observable with a 40-meter telescope under construction.

These objects are not seen by their thermal radiation, like stars, but because they have nonzero albedo and reflect sunlight. Both incident and reflected sunlight fall off like the square of the distance. When Sedna moves from perihelion (now) to aphelion (in 11,000 years) its distance will increase by about a factor of ten; the sunlight incident on the surface will decrease by a factor of 100 (which is five magnitudes) and the brightness at Earth of the reflected light will decrease by another factor of 100. At aphelion, then, Sedna will have an apparent magnitude of around 37 and might or might not be visible from Earth's surface using a telescope the size of a baseball diamond. That's at a distance of 0.015 light-year.

So we don't have to worry about seeing Oort cloud objects by their reflected light at a light-year. For blackbodies (like stars) there is a straightforward relationship between absolute magnitude, distance, and size: a big, cold object can emit more light and be more luminous than a small, hot object. Absolute magnitudes are based on brightness at a distance of 10 parsecs or 32 light-years, so our barely-detectable magnitude-37 blackbody at one light-year would have an absolute magnitude of 42.5 or so. Such an object would either be a stray planet (of which, apparently, there are many) or a black dwarf. I'm not certain whether a brown dwarf could be that cool and still retain its atmosphere, and a brown dwarf that lost its atmosphere would just be a rocky stray planet. (I may come back later and edit in a calculation, hmmmm.)

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  • $\begingroup$ Some helpful equations: for a spherical blackbody, the luminosity is related to temperature T and radius R by the Stefan-Boltzmann relation, $L = 4\pi \sigma R^2 T^4$ where $\sigma$ is the Stefan-Boltzmann constant. Once you know L, you can use the formulas on this page to get the "absolute magnitude" M and "apparent magnitude" m, where the apparent magnitude determines how difficult it is to actually see the object. $\endgroup$ – Hypnosifl Jan 8 '15 at 17:54
  • $\begingroup$ Also, a formula given on that page is that absolute magnitude M is related to apparent magnitude m by M = m + 5 - 5*log(D), where D is the distance in parsecs...since 1 light year = 0.3066 parsecs, I get the answer that a body with an apparent magnitude of 37 would have an absolute magnitude of M = 44.57, a little different than the 42.5 you gave. $\endgroup$ – Hypnosifl Jan 8 '15 at 18:17
  • $\begingroup$ @Hypnosifl I was using the approximation $0.3066^2 = 0.01$ (or $3^2 = 3.26^2 = 10$) so I could do the math in my head. $\endgroup$ – rob Jan 8 '15 at 19:25
  • $\begingroup$ Yeah, I figured it was probably an approximation, just thought I'd give the method for an exact calculation is user89 wasn't sure. And continuing on that line, another element to the calculation would be figuring out the lumosity L given the absolute magnitude M--the page I referenced said M = +4.77 - 2.5 log(L / Lsun), so (L / Lsun) = 10^[(4.77 - M)/2.5], so with M=44.57 this gives (L / Lsun) = 1.202 * 10^-16. And the luminosity of the Sun is 3.846 * 10^26 Watts, so the luminosity of this object would be 4.623 * 10^10 Watts. $\endgroup$ – Hypnosifl Jan 8 '15 at 22:22
  • $\begingroup$ Then the last step would be to plug that value for L into the Stefan-Boltzmann relation, and if you divide both sides by $4\pi \sigma$ with $\sigma$ = 5.670373 * 10^-8 W/(m^2 K^4), you get that $R^2 T^4$ is equal to 6.488 * 10^16 m^2 K^4. So for example if the temperature of a body was 300 K, then $R^2$ would be (6.488 * 10^16)/(300^4) = 8.01 * 10^6 m^2, so R would be 2830 meters--so blackbody radiation from something that small with a temp. of 300 K and a distance of 1 light year would be visible with a telescope that could see objects with apparent magnitudes up to 37. $\endgroup$ – Hypnosifl Jan 8 '15 at 22:34
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There is no straightforward answer to this question without specifying the sensitivity as a function of wavelength of your proposed instrument.

Assuming a blackbody (and many cool objects, like brown dwarfs, depart significantly from this), the flux received at the Earth is given by $$ f_\nu = \frac{R^2}{D^2} \pi B_{\nu}(T,\nu),$$ where $B_\nu$ is the Planck function (temperature dependent), $D$ is a light year and $R$ is the radius of the (assumed spherical) object.

To see whether this is detected you have to convolve this with the sensitivity function $S(\nu)$ of your detector, which will be a function of telescope aperture, detector technology etc. The signal, $S$ in any wave band $\nu_1$ to $\nu_2$ will be $$ S = \frac{\pi R^2}{D^2}\int_{\nu_1}^{\nu_2} S(\nu) B_{\nu}\ d\nu $$

Hypnosifl in his comments suggests simply scaling the solar absolute magnitude and solar luminosity to calculate the luminosity of an object with a magnitude equal to the threshold for detection with a particular instrument. This approach cannot work as stated. The problem here is that the calculation assumes that the object has a similar spectrum to the Sun, which if it is cooler, is obviously not the case. His calculation tells you that if you have an instrument with a bolometric magnitude threshold of 37 (i.e. integrated over all frequencies*), then one could detect an object with $4\pi\sigma R^2 T^4 = 4.6\times 10^{10}$ Watts. But real detectors do not work like this. They work in restricted wavebands.

But let's take the 300K example. A telescope that detects visible objects to 37th magnitude will be useless. Essentially none of the flux from a 300K object appears in the visible part of the spectrum. One has to apply something called a bolometric correction. The bolometric correction is at a minimum for an object near the Sun's temperature, but then increases rapidly for hotter and cooler objects. It is calculated using an equation of the form I wrote above.

For a blackbody at 300K, and the correction between bolometric magnitude and V-band magnitude I can't even find a calculation that goes cool enough. The coolest things I can easily find are brown dwarf models from Baraffe et al. (1998) where a 500Myr old brown dwarf of $0.02M_{\odot}$ has a luminosity of $3\times 10^{21}$ Watts, a temperature of 900K and an absolute V-band magnitude of 44! The bolometric correction is about 24 magnitudes. i.e. Such an object would have a visual magnitude that is 24 magnitudes larger than its bolometric magnitude!

EDIT: An alternative way to do this is use examples of objects that have been detected and scale them. So, an object the temperature of the Sun can be dealt with using hypnosifl's formulae. For a colder object we use the example of the coldest known brown dwarf (Luhman & Esplin 2014), this is 7.5 light years away, with a temperature of about 250K. This object will have (from reasonably well-understood theory) a radius roughly similar to Jupiter. It lies at the edge of what can be currently studied. It has an apparent magnitude of 14 in the 4.5 micron-band of the near-IR WISE survey, but has a near-IR J-band magnitude of 25 (right on the hairy 2.6 sigma edge with the 6.5m Magellan telescope) and is utterly undetected at optical wavelengths. So, if you put this object at 1 light year, you could also reduce the radius by a factor of 7.5 (to about 10,000 km) and still be able to see it.

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  • $\begingroup$ Thanks for this nice answer! You write: "but real detectors do not work like this. They work in restricted wavebands" -- if I had an array of sensors such that each one covered the others' weaknesses then Hypnosifl's assumption would be okay? (I would have a sum of convolutions, and if I assume $S(\nu)$ to be the same constant over each range of wavelengths, then simplifications would show that I have an equivalent integral over all frequencies) $\endgroup$ – user89 Feb 8 '15 at 18:52
  • $\begingroup$ I also want to make sure I understand the following correctly: "Such an object would have a visual magnitude that is 24 magnitudes larger than its bolometric magnitude!" -- bolometric magnitude is the brightness of an object, as would be detected from a "super-sensor" (so, the signal is captured over all possible wavelengths the object might be radiating over), while the visual magnitude is the brightness of an object over the visible portion of the spectrum. The visible magnitude being 24 magnitudes larger implies that the object would be 24 magnitudes dimmer visually? $\endgroup$ – user89 Feb 8 '15 at 19:01
  • $\begingroup$ @user89 The mooted 37 magnitude threshold is for an optical/IR telescope and will be for the faintest possible detection in the V or possibly B-band where the sky is darkest. Yes in principle if a detector worked at all wavelengths and had a bolometric mag. threshold of 37 then hypnosifl's calc. is correct. I assure you that the detection thresholds at other wavelengths are much poorer than this. Yes you are right, the hypothesised 300K object with that luminosity would be more than 24 magnitudes fainter in the V-band. $\endgroup$ – Rob Jeffries Feb 8 '15 at 20:06
  • $\begingroup$ @user89 Much confusion is generated by the fact that the bolometric magnitude of the Sun is similar to the V-magnitude of the Sun. The zeropoints of the system are set so that this is the case. For objects of different temperatures, they are completely different. $\endgroup$ – Rob Jeffries Feb 8 '15 at 20:07

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