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Suppose we made a solar sail out of a highly reflective material. How big would that solar sail have to be for the Hubble Space Telescope to detect it visually at the average distance of Pluto?

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    $\begingroup$ Good job specifying a instrument. Alas, I suspect the answer depends on what focal plane device they're using. $\endgroup$ – dmckee Feb 2 '15 at 4:32
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    $\begingroup$ If the sail uses a very good reflector and is not reflecting towards the telescope, isn't it essentially "stealthed?" $\endgroup$ – C. Towne Springer Feb 2 '15 at 4:46
  • $\begingroup$ I am assuming an inbound sail that is using the solar wind to decelerate, so the reflective surface is facing the sun. $\endgroup$ – Robert Madsen Feb 2 '15 at 20:54
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The practical detection limit for HST is about a visual magnitude of 30 - that sort of number was reached in the ultra-deep field. Assuming that the solar sail kept reasonably stationary for the 100 hours or so of required exposure then we could do a calculation based on that. There is absolutely no need to resolve the object in order to detect it.

If you have a perfect mirror, then you will only stand a a chance of seeing it when the alignment between Sun Earth and Sail is absolutely perfect (I'll come back to that). In that case you are seeing a reflection of the Sun and so effectively observing from a distance of the sum of Sun-Pluto + Pluto-Earth. I'll leave it to you to find the exact details of this, obviously it depends on what time of year you look and where Pluto is in its orbit, but it is going to be about $d=40+40=80$ astronomical units (au).

Let's assume the mirror is a disc. We view an image of the Sun in that disc. At 80au, the Sun will appear $\theta_0=5.8\times 10^{-5}$ radians across. Assuming the solar disc is uniformly bright and the image is centred in the mirror, the flux received at the Earth will be given by

$$ f = \frac{L_{\odot}}{4\pi d^2} \times Max(1,\left(\frac{\theta}{\theta_0}\right)^2), $$ where $\theta$ is the actual angle subtended by the mirror at the Earth and the flux will not get any bigger if $\theta > \theta_0$.

Assume the reflected light has the same spectrum as the Sun, that the Sun has an apparent magnitude of -26.74, and that the solar constant flux at 1 au is 1360 W/m$^2$. The apparent magnitude $m$ is related to $f$ by $$ f = 1.36 \times 10^3 \times 10^{-(26.74+m)/2.5}$$

Combining the two equation for $f$ $$ \theta = 36.9 d\theta_0 \left(\frac{4\pi }{L_{\odot}} 10^{-(26.74+m)/2.5}\right)^{1/2}$$

If I try $m=30$, $d=80 au$, this gives $\theta = 2.1\times10^{-14}$ radians, far smaller than the solar image. At a distance of $\sim 40\ au$, the mirror is less than a metre across.

Could this be true? I think it is down to the unrealistic assumptions of a perfect mirror and specular reflection. On the other hand, we know that relatively small reflectors do give rise to bright images in the sky. An example would be the Iridium flares that are seen from satellites in low-earth orbit from "door sized" polished antennae. These flashes last a few seconds for a stationary observer on Earth, reaching magnitude -8.

So a more realistic scenario for detecting the sail might be that the alignment is optimal for only a few seconds. In which case using $m=30$ is crazy. I had a play withthe HST ACS exposure time calculator. A 10s exposure can get a SNR of 10 on an object with $m=21$. Putting this value of $m$ in gives a mirror diameter of 8 metres. Still remarkably small, but the Sun is bright, even at Pluto. It would also be interesting to know what the limits are on making mirrors that could give a specular reflection that pure. I suspect that might be the deal breaker, but have no knowledge of this at all.

If you make the reflection Lambertian, the mirror needs to be larger. You are effectively scaling to the area of Pluto, which has quite a high albedo (maybe 0.5). It is visual magnitude roughly 14, with an effective emitting area of $4.4\times10^{12}\ m^2$. Scaling this to $m=30$ (you wouldn't need to observe it at just the right moment) gives a required mirror diameter of about 1 km.

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  • $\begingroup$ Thank you so much for your extensive answer. This result is quite amazing. Of course, I realize that the mirror and alignment are not going to be perfect, so the 2km estimate is still quite satisfying considering that current research into interstellar capable solar sails puts them in the kilometer range. $\endgroup$ – Robert Madsen Feb 2 '15 at 21:00
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A back of the envelope calculation:

The dimmest objects detectable by the HST have absolute magnitude $\sim30$, or power $3\times10^{-20}\text{W}$. Assuming the sails have an albedo of 1, we have an expression for the power of the returned light of a solar sail at Pluto given by: $$ \frac{L_\oplus}{D_\text{SP}}\times \frac{A}{D_\text{EP}}= 3\times10^{-20}\text{W}\, . $$ Putting in values for luminosity of the Sun ($L_\oplus=3.8\times10^{26}\text{W}$) and current values for Earth-Pluto ($D_\text{EP}$) and Sun-Pluto ($D_\text{SP}$) distances, we get a value of $A=625000\text{m}^2$, corresponding to, e.g., a square sail of side length $800\text{m}$.

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    $\begingroup$ Thank you so much for your time. Your calculations are right in line with the answer given above and it provides a great confirmation. $\endgroup$ – Robert Madsen Feb 2 '15 at 21:00
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From Wikipedia:

Current maps [of Pluto] have been produced from images from the Hubble Space Telescope (HST), which offers the highest resolution currently available, and show considerably more detail, resolving variations several hundred kilometres across, including polar regions and large bright spots... The two cameras on the HST used for these maps are no longer in service

An image of Pluto. It looks like the surface spans about 12 to 15 pixels.

It would appear that the solar sail about a tenth of the size of Pluto would cover a couple of pixels in the HST instruments (which are not longer usable, unfortunately).

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    $\begingroup$ Wouldn't the limit of detectability be determined by intensity, not the number of pixels? A telescope with a larger mirror might find a fainter object than the Space Telescope. $\endgroup$ – mmesser314 Feb 2 '15 at 6:06
  • $\begingroup$ @mmesser314 True, but I didn't want to get into that. The albedo of Pluto varies from 0.49 to 0.66. If the albedo of the solar sail is 0.99, then the solar sail can be half the size for the same amount of light from Pluto (when the albedo is 0.5). I don't know how sensitive the instruments are, but reducing the light 50 times (for a 2x2 area) would probably put it very close to being not detectable. $\endgroup$ – LDC3 Feb 2 '15 at 6:20

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