If I wanted to image an Earth-like planet at a resolution where one could read the lips of a human speaking on the surface 1, 10, and 1000 light years away, how large would the telescope reflector need to be (the diameter) in each case?

Another way to ask my question is: Given a parabolic mirror/reflector the size of, say Earth's orbit, what would be the farthest distance one could read someone's lips or read printed text (a secondary concern would be imaging the entire face of the planet at such resolution)?

Assume the telescope is in space (no atmospheric interference) and outside of any asteroid belts or dust clouds, etc. I understand the mirror would have to track the movement of the planet.

My main objective is to get a layman's grasp of the relationship between the size of a telescope's reflector and its ability to see small details very far away. How big of an image can it produce and at what level of detail?

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    $\begingroup$ See the last section of this answer: worldbuilding.stackexchange.com/a/48901/20204 Short answer is "The telescope can't exist" for significant light years away The answer links to this Reddit post: reddit.com/r/explainlikeimfive/comments/2qvna6/… $\endgroup$
    – Ranger
    Feb 8 '17 at 21:56
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    $\begingroup$ Also, is this World Building? This seems to be a pure science/engineering question. $\endgroup$
    – Ranger
    Feb 8 '17 at 21:57
  • $\begingroup$ I read those. They don't seem to address my question. Basically, how far can you see and at what detail given an unlimited mirror size? $\endgroup$ Feb 8 '17 at 21:58
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    $\begingroup$ I'd like to know the limits so I can place my world at the proper range ;-) $\endgroup$ Feb 8 '17 at 22:00
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    $\begingroup$ metabunk.org/… a link that may help you on your journey. Similar discussion but for an actual telescope being used. $\endgroup$
    – ggiaquin16
    Feb 8 '17 at 22:41

When talking about telescopes there are two quantities to take into account; the “light gathering power” and the “resolving power” of the telescope. “Light gathering power” is just how much light can be collected by the telescope. “Resolving power” is a measure of the smallest angle that the telescope can reliably detect.

Because light is a wave it has a way of spreading out (technically: diffracting). The smaller the telescope the more the waveness becomes a problem.

What’s a little weird is that this isn’t just limited to the size of the mirror or lens of a single telescope. By cleverly networking telescopes together you can make them act like a single large telescope.

Visible light has a wavelength of about half a micrometer (one two-millionth of a meter), people are about a meter across (assuming a spherical person), and the Earth is about 13,000,000 meters across. So, using ground-telescopes that are perfectly constructed and networked, we could spot something person-sized from about 1/400th of a light year away, or about double the distance to Pluto. For comparison, the distance to the nearest star is about 4 light years. So, using ground based telescopes we can’t come even remotely close to seeing a person standing around on a planet in another solar system.

The nearest known, reasonable, candidates for being an Earth-like planet (as of April 2013) are about 20 light years away (HD 20794 d, Gliese 581 c, and Gliese 667C c). Spotting dudes and ladies on one of these worlds requires, at minimum, a telescope array that’s at least 100 million km across. That’s an array more than half the size of Earth’s orbit. The good news is that an array like that (under absolutely ideal circumstances) isn’t that difficult to create. Setting aside that the telescopes would each need to be essentially perfect for their size (Hubble-quality), all we’d need to do is set them up in solar orbits about the size of Earth’s orbit. This is a lot easier than sending them to another planet, and about as hard as sending them to crash on the Moon.

So, assuming that we could set all that up, the problem stops being one of resolving power, and becomes one of light-gathering power.

On a sunny day we’re hit by about 10^21 (1,000,000,000,000,000,000,000) photons (give or take) every second. Assuming that a fair fraction of those escape into space, then that number, which seems large, is all that distant aliens have to work with. Over 20 light years that scant 10^21 photons means you would need a telescope array with an area of more than 500 million square kilometers to catch just one photon per second. That’s the size of the surface area of Earth. In the mean time there’s a lot of other light flying around, and single photons are pretty hard to detect so… the signal-to-noise ratio would be small.

Creating an array capable of seeing big stationary things like rivers and mountains on other worlds wouldn’t be too difficult, because you can just use tremendously long exposures to overcome the whole light-gathering issue. This is a pretty standard trick in astronomy; the Hubble Deep Field took a more than a week of total exposure time. There would be some issues with the fact that those distant planets are moving and whatnot, but there are clever ways around that too.

People, and probably aliens too, move around a lot. So if you want to get a picture of one, you need the exposure to take less than, say, a second. Unless you catch E.T. literally napping. I would wildly guesstimate that you’d need at least a few thousand photons per second to overcome the signal noise enough to say for certain that you’re looking at something real.

So, to answer a somewhat more detailed question; to get a picture of an alien that’s person-sized, standing on a world 20 light years away, so that it takes up one pixel in the image, using an exposure time of about one second, would require an array of telescopes with exposed mirrors and lenses with an area totaling more than several thousand times the Earth’s surface area and spread out over a region about the size of Earth’s orbit.

This isn’t technically impossible, but it would be “expensive”, and would require substantially more materials than are likely to be reasonably found in our solar system. It probably isn’t worth it to get a blurry, tiny picture of some alien picking its nose 20 light years away and 20 years ago.

Of course, if you wanted to see farther, you’d need a much larger telescope array.

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    $\begingroup$ Also there are other things to take into account such as atmosphere thickness (i.e. clouds), planets being aligned correctly, dust clouds in space, and anything that could cause a blockage or distortion of the light from planet A to B such as gravity beasts like Black Holes etc. $\endgroup$
    – Sandy
    Feb 8 '17 at 22:39
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    $\begingroup$ Exposure time! I wasn't thinking in those terms. That "sheds some light" on my thought process. Thanks! $\endgroup$ Feb 8 '17 at 22:48
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    $\begingroup$ Also rain, local pollution from cars etc, remember nothing is impossible just very hard, someone once laughed about having metal ships, flying metal tin cans (aircraft), even going into space, if you are using advanced technology rather than just a standard telescope like today then as long as you have a feasible reason why your advanced telescope can do what it claims thats all good :) $\endgroup$
    – Sandy
    Feb 8 '17 at 22:50
  • $\begingroup$ So given an array the size of Earth's orbit, how far away do you think one could read text on a sheet of paper (or on the side of a spaceship)? $\endgroup$ Feb 8 '17 at 22:51
  • $\begingroup$ yes, I'm actually counting on rain and pollution and clouds and planet rotation to obscure most observations. Even current Earth spy satellites have to deal with that! :-) $\endgroup$ Feb 8 '17 at 22:53

TL;DR: Depending on the fidelity of video you want to shoot of the aliens, you can be limited either by diffraction (low fidelity) or photon counting (high fidelity). Overall, the numbers are really large though. For example, the telescope radius for doing lip reading at a distance of 1 lightyear (~ 63000 AU) is roughly ∼92 AU (!). With a 1 AU telescope, you can do lip reading for folks ~700 AU far away.

Key result

To shoot video that resolves a length scale of $d_\alpha$ at $60 \text{ fps}$ with decent color accuracy (8-bit RGB, error rate < 1/10) when the alien star is sun-like and the system is at a distance $\ell$ from the telescope, the telescope radius $r$ is bounded below due to photon counting considerations by:

$$ r_{\text{min}} \simeq \frac{\ell}{2d_\alpha} (7.27\times 10^{-6}\text{ m}).$$

If one relaxes the demands on the video fidelity, the limiting factor becomes the diffraction of red light:

$$ r_{\text{min}} \simeq \frac{\ell}{2d_\alpha} (6.20\times 10^{-7}\text{ m}).$$


If I wanted to image an Earth-like planet at a resolution where one could read the lips of a human speaking on the surface 1, 10, and 1000 light years away, how large would the telescope reflector need to be (the diameter) in each case?

Say we want to do lip reading. The size of a human mouth is about $5\text{ cm} = 0.05\text{ m}$. Suppose we assume that if we have a video where your mouth is about 20 pixels wide, we can reasonably do lip reading (the vertical direction has the same scale so no worries that lips will look too thin). Then $d_\alpha = 0.05/20 = 0.0025\text{ m} =2.5\text{ mm}$. So for the first question, the telescope radius for doing lip reading at distance of $\ell = 1 \text{ lightyear}$ is roughly (using the first formula) $1.4\times 10^{13}\text{ m} \sim 92\text{ AU}$. For comparison, the radius of Neptune's orbit is about $30 \text{ AU}$. So our telescope has to be really big!

Another way to ask my question is: Given a parabolic mirror/reflector the size of, say Earth's orbit, what would be the farthest distance one could read someone's lips or read printed text (a secondary concern would be imaging the entire face of the planet at such resolution)?

One can similarly answer the second question. For that, you plug in the telescope size $r = 1\text{ AU} = 1.5\times 10^{11}\text{ m}$ and the same $d_\alpha = 2.5\text { mm}$ as before. This gives the distance to be $\ell \sim 10^{14}\text{ m} \sim 700\text{ AU}$. You can see that this is about 23 times the radius of Neptune's orbit.

Imaging the full planet at this resolution is not an issue -- all we are doing is replacing a small telescope (like the Hubble) with a huge telescope. This does not decrease our field of view (physical size represented by picture) but it does enhance the resolution (sharpness of picture) and imaging fidelity (correctness of picture).

My main objective is to get a layman's grasp of the relationship between the size of a telescope's reflector and its ability to see small details very far away. How big of an image can it produce and at what level of detail?

You can tweak the level of detail and image size (I assume you mean pixels) by adjusting the parameter $d_\alpha$ by considering the following: imagine a digital photograph of the object you want to actually image. Calculate the appropriate $d_\alpha$ by dividing the physical length of the object by the minimum width of the "imaginary digital photo" (in pixels) at which the object is clearly identifiable. Basically, $d_\alpha$ is the physical size corresponding to 1 pixel width.

Cosmic videography theory (CVT)

Pursuit - Speed, color and pixels

Suppose we want to shoot the video with a minimum time-scale of $\delta t$. For example, if you want to shoot slow-mo at $200 \text{ fps}$ then use $\delta t = 1/200 {\text{ s}}$. We would like to record RGB values on a linear scale with 256 divisions, where each channel is defined by a pair $(\lambda_X,\epsilon_X)$ ($X=R,G,B$).

The ease of resolution will depend on how small of a background you have. This is common sense -- you can see many stars at night and only the sun during the day. For simplicity (and best performance), assume that our sun, telescope and alien are in a straight line (in that order). We put stuff between the sun and the telescope to (effectively) block out all the radiation from the sun. For example, you can use lead blocks to block $\gamma$ rays etc. In principle, it is possible to have zero background if your resolution is good enough and your detectors have no noise of their own (super chilled, superconducting, super everything) so let's stick with that.

You can get rid of the background but you cannot get rid of Rayleigh's criterion. Let the diameter of the alien subject be $d_\alpha$, the radius of our Peeping Tom Telescope Array (PTTA) be $r$ and the distance to the alien subject be $\ell$.

$$ \frac{d_\alpha}{\ell} = \theta \geq \frac{\lambda_R}{2 r} \implies r \geq \frac{\lambda_R \ell}{2 d_\alpha}$$

where we used $\lambda_R$ as it is the largest wavelength out of the three and hence needs the largest PTTA.

As pointed out in @SandyBeach's nice answer, we need to consider both resolution and exposure. We have already discussed resolution, we now proceed to counting photons.

Intro - Photon generation and redirection

Definitions: power = energy per unit time, intensity = power per unit area, intensity spectrum = intensity per unit wavelength, flux = number per unit time, flux density = flux per unit area, flux density spectrum = flux density per unit wavelength.

The intensity spectrum $dI(\lambda)/d\lambda$ of the alien's sun is given by Planck's law; we can use this to get the flux density spectrum $d \sigma_\Phi(\lambda)/d\lambda$ using $I = E(\lambda) \sigma_\Phi = hc \sigma_\Phi/\lambda $ for photons:

$$\frac{d P(\lambda)}{d \lambda} = \frac{2 hc^2}{\lambda^5}\frac{1}{\exp\left[\frac{hc}{\lambda k_B T}\right]-1} \implies \frac{d \sigma_\Phi(\lambda)}{d \lambda} = \frac{2 c}{\lambda^4}\frac{1}{\exp\left[\frac{hc}{\lambda k_B T}\right]-1}$$

The total flux density is $\Sigma_\Phi = \int_0^\infty \frac{d\sigma_\Phi(\lambda)}{d\lambda} d \lambda$. Similarly the fraction of flux density in a window of wavelengths $(\lambda-\epsilon,\lambda+\epsilon)$ will be

$$f(\lambda, \epsilon) = \frac{1}{\Sigma_\Phi} \int_{\lambda-\epsilon}^{\lambda+\epsilon} \frac{d\sigma_\Phi(\lambda')}{d\lambda'} d\lambda'.$$

Suppose the subject of our video is illuminated by a star and has a response function $\alpha(\lambda)\geq 0$, i.e., if the incident flux is $\Phi(\lambda)$ then the outgoing flux is omnidirectional (to first approximation) with magnitude $\alpha(\lambda)\Phi(\lambda)$. Assuming that the subject is not on fire, $\alpha(\lambda)\leq 1$. If the subject is blue in the face, then $\alpha(450\text{ nm})\simeq 1$ and $\alpha(600\text{ nm})\simeq 0$. If the subject is a mango, then $\alpha(450\text{ nm})\simeq 0$ and $\alpha(600\text{ nm})\simeq 1$.

Note: We ignore atmospheric effects (on the alien planet) which may alter the spectrum in a nontrivial manner.

Incident - Poisson statistics

In an ideal scenario, the RGB values recorded would correspond directly to $\alpha(\lambda_X)$. For example, if the scale is from 0 to 255, then we would (ideally) record R=0 if $\alpha(\lambda_R) \leq 1/256$, B=255 if $\alpha(\lambda_B) \geq 255/256$ and so on. However, the minimum of the three $f(\lambda_X,\epsilon_X)$ values is a bottleneck for measurement. Why? Since our fluxes may be small, we must apply Poisson statistics instead of taking a uniform flux. The flux density of the star $f(\lambda_X,\epsilon_X)\Sigma_\Phi$ is related to the incident flux on the alien subject by a wavelength-independent geometric factor, say $Z$, so $\Phi(\lambda_X) = Z f(\lambda_X,\epsilon_X)\Sigma_\Phi$. Then the expected flux in channel $X$ in the time $\delta t$ will be

$$\mu(\lambda_X) = \delta t\cdot Z_T \alpha(\lambda_X) Z f(\lambda_X,\epsilon_X)\Sigma_\Phi$$

where $Z_T = \pi r^2/(4\pi \ell^2) = r^2/4 \ell^2$ is an additional geometric factor and (like before) $r$ is the radius of the PTTA and $\ell$ is distance from the PTTA to the subject.

The probability of getting a count of $k$ will be

$$P_X(k) = \frac{\mu(\lambda_X)^k e^{-\mu(\lambda_X)}}{k!}\quad k = 0, 1, 2,\ldots$$

As long as the subject is not too dark (i.e. small $\alpha$ for all $\lambda_X$), we can approximate the Poisson distribution by a Gaussian of mean $\mu(\lambda_X)$ and standard deviation $\sqrt{\mu(\lambda_X)}$:

$$P'_X(k) = \frac{1}{\sqrt{2\pi \mu(\lambda_X)}} \exp\left[-\frac{(x-\mu(\lambda_X))^2}{2\mu(\lambda_X)}\right]$$

Fit of Gaussian distribution to Poisson distribution

Graph showing fit of Gaussian (line) to Poisson distribution for mean values $64,128$ and $256$.

Reasoning - The Measurement Process

Let us now discuss the actual measurement process. Naively, one would think that $\mu$ should at least be $256$ for $\alpha = 1$ so that the full color range of 0-255 is used. In a time step $\delta t$, count the photons and assign the value to the corresponding color. However, the counting statistics make this complicated. For example, if $\alpha = 0.5 \leftrightarrow \mu=128$, it is quite possible that the actual counts observed in successive time steps will look like {120, 133, 131, 125, 134} (because the Gaussian standard deviation is $8\sqrt{2}$) and therefore the $\alpha$ value will be incorrect at all the five time steps. ☹

Special Investigation - A quest for better color reproduction

Suppose we have a tolerance $p$ for errors ($0<p<1$), i.e., at most $Np$ out of $N$ measurements are incorrect. We demand that this condition holds separately for each bin from B=0 to B=255 and similarly for G and R. Without loss of generality, let the alien's sun have $f(\lambda_B,\epsilon_B)$ as the least value out of the three. For simplicity, let $m=\mu(\lambda_B)$ when $a(\lambda_B)=1$, so $m$ is the maximum number of "blue photons" received in an interval $\delta t$ (as $a(\lambda)\leq 1$).

As shown above, it is clear that having the value $m = 256$ will not satisfy $p<1$ as most results will be erroneous. Therefore, $m$ has to be much larger than $256$. In that case, the B value as a function of count $n_B$ is set using

$$ B(n_B) = \begin{cases} \mathrm{floor}(256 n_B/m) & 0\leq n_B < m \\ 255 & n_B \geq m \end{cases} $$

(We will account for the error due to the truncation in our simulation.)

Using a larger average value will reduce our relative error:

Peak narrowing.

This graph shows the reducing rate of error as $m$ is increased ($m=2^9, 2^{10},2^{11}$). Each individual peak shows the probability density of some actual B value being mistaken for another B value; the larger the spread, the higher the chances of error. The relative error for the Gaussian scales as $1/\sqrt{m}$ (the standard deviation scales slower than the mean) which is reflected in the thinner peaks for larger $m$.

One way to numerically find a suitable value of $m$ is to use Monte Carlo simulation. The basic idea is described in pseudocode below:

poisson_to_gaussian_cutoff = 15

function right_wrong(m, n_runs):
    right = array(elem = 0, length = 256)
    wrong = array(elem = 0, length = 256)
    A = random_number_array(uniform_dist(0,1), length = n_runs)
    B = []
    for a in A :
        if a >= poissonToGaussianCutoff / m :
            B.append(256/m * random_number(gaussian_dist(mean = m*a, stdev = sqrt(m*a))))
        else :
            B.append(256/m * random_number(poisson_dist(mean = m*a)))
    for (b,a') in zip(B, floor(256 * A)):
        # array numbering begins at zero
        if (0 <= b <= 255 and b == a'):
            right[b] += 1
        else if (0 <= b <= 255 and b != a'):
            wrong[b] += 1
        else if (b < 0 and a' == 0):
            right[0] += 1
        else if (b < 0 and a' != 0):
            wrong[0] += 1
        else if (b > 255 and a' == 255):
            right[255] += 1
        else if (b > 255 and a' != 255):
            wrong[255] += 1
    return (right, wrong)

M = [256, 512, 1024, ...] # candidate m values to test
N_runs = 2^20             # seems to work well
binwise_p_list = []
for m in M:
    mright, mwrong = right_wrong(m, N_runs)
    binwise_p_list.append(mwrong / mright) # element-wise division

We plot $p$ values for each bin for different choices of $m$.

Color accuracy - counting rate

For a value of $p=0.1$, $m=2^{22}$ is okay. For super-duper high quality at $p=0.01$, $m=2^{28}$ is reasonable (although it is kind of hard to see in the graph).

Action - Numbers

Let's stick with $p=0.1$ so $m=2^{22}$. For a smooth viewing experience, we demand $\delta t\leq 1/60\text{ s}$ ($\geq 60\text{ fps}$). The value for $f(\lambda_B,\epsilon_B)$ can be found using numerical integration. I could not find a good way to fix the values of the wavelength windows so I've set them as: $\lambda_B = 450 \text{ nm}$, $\lambda_G = 530 \text{ nm}$, $\lambda_R = 620\text{ nm}$ and $\epsilon_B = \epsilon_G = \epsilon_R = 40\text{ nm}$ (roughly based off the table in Ref 1.). For a sun-like star ($T=5772 \text{ K}$), the values of $f$ and $\Sigma_\Phi$ are

$$ f_B = 0.0491401,\,f_G = 0.0593743,\,f_R = 0.0634818\text{ and } \Sigma_\Phi = 9.281\times 10^{25} \text{ m}^{-2}\text{s}^{-1}.$$

Since the value of $f_B$ is the lowest, we will use it. (We also used $f_B$ in the preceding discussion.)

For a subject chilling at about $1.008 \text{ AU}$ and the alien sun having radius equal to $R_\odot$, we have the geometric factor

$$Z = \pi d_\alpha^2/4 \left[\frac{R_\odot^2}{1.008 \text{ AU}}\right]^2 = 1.67\times 10^{-5}d_\alpha^2$$

where $d_\alpha$ is (like before) the diameter of the alien subject.

We assume that the typical value of $\alpha(\lambda_X)\sim 1$; we found out the $m$ value(s) using this. (Otherwise, we could rescale $m$ somehow to account for the maximum achievable $\alpha(\lambda_X)\ll 1$...)

We put everything together, recalling that $m$ is the lower limit for $\mu(\lambda_X)$:

$$\mu(\lambda_X) \leq \delta t\cdot Z_T \alpha(\lambda_X) Z f(\lambda_X,\epsilon_X)\Sigma_\Phi $$ $$m \leq \delta t \frac{r^2}{4\ell^2}\cdot 1.67\times 10^{-5} d_\alpha^2 \cdot 0.0491\cdot 9.281\times 10^{25}$$ $$ r \geq \frac{\ell}{2 d_\alpha}\sqrt{\frac{16\cdot 2^{22}}{(1/60)\cdot 1.67\times10^{-5}\cdot 0.0491\cdot 9.281\times 10^{25}}}$$ $$ r \geq \frac{\ell}{2d_\alpha} (7.27\times 10^{-6}\text{ m}) $$

This expression is an order of magnitude higher than the resolving-power based expression which has $\lambda_R \sim 6\times 10^{-7} \text{ m}$ instead of $7.27\times 10^{-6}\text{ m}$.

Let us do some quick sanity checks for the numbers:

  1. The total flux density incident on the subject if it were Earth-sized ($d_\alpha = 12600 \text{ km}$) would be

    $$\frac{\Sigma_\Phi \times Z}{\pi d_\alpha^2/4} \sim 10^{26} \cdot 1.6 \times 10^{-5} \cdot (4/\pi) \sim 2\times 10^{21} \text{ m}^{-2}\text{s}^{-1}$$ which agrees with the standard number up to order of magnitude (also present in @SandyBeach's answer).

  2. Consider putting a $1\text{ cm}\times 1\text{ cm}$ plate of aluminium coated with Vantablack, one of the darkest materials known, in the sun (here on Earth). The number of photons reflected by it in a time of $1/60 \text{ s}$ will be $10^{21} \cdot 0.01\cdot 0.01 \cdot 1/60 \cdot 0.00035 \sim 6 \times 10^{11}$. This number is still very big relative to the figure we obtained for $p=0.01$, $m = 2^{28} \sim 2.7\times 10^8$.

  3. The values of $f_X \sim 0.06$ are not too shabby given that the solar spectrum peaks at . I think that the value for red is higher than green because one has to multiply the intensity spectrum by $\lambda/hc$ to get the flux density spectrum, thereby red-shifting the peak of the distribution.

Horror - Low accuracy, low speed attempted compromise

If we set $p = 0.6$ (for R=0,...,128, not for all R=0,...,255), we can get away with $m = 2^{17}$.

Compromise plot

We reduce the framerate to a crawling $1 \text{ fps}$. If we try to scale up $\Sigma_\Phi$, the planet will move further away and $Z$ will decrease in order to avoid cooking the alien. With these new parameters, we have $$ r \geq \frac{\ell}{2d_\alpha} \frac{(7.27\times 10^{-6}\text{ m})}{\sqrt{2^5 60}} = \frac{\ell}{2d_\alpha} (1.66\times 10^{-7}\text{ m}).$$

So if you try to degrade the "stream" quality by too much then you are limited by diffraction, not photon counting.

Suspicion - Something feels off

  1. The wavelength calculations are wrong because the light will be blue-shifted/red-shifted.

    Well, the relative speed of the alien subject to its own star is much less than $c$. You are right about the light being Doppler shifted. That just changes what kind of detector is used and does not affect our calculations.

Dirge - References

  1. https://en.wikipedia.org/wiki/Visible_spectrum#Spectral_colors
  • 1
    $\begingroup$ This went into more detail than my puny human brain can understand. I got some numbers at the end (of the first section) that were staggeringly large and had to smile and nod. $\endgroup$ May 5 '17 at 3:18
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    $\begingroup$ Your answers are incredible. +1 from me and again, good to have you on board! $\endgroup$
    – Secespitus
    May 5 '17 at 9:55
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    $\begingroup$ Results indeed seems off, and I'd like to see this figured out, but you gave really solid foundation for this. Also, could you please highlight the actual sizes, as a summary on the beginning or the end, for people who can't understand all the math or simply don't have time or patience to do it? That way your answer will be less intimidating with no harm to accuracy. $\endgroup$
    – Mołot
    May 5 '17 at 9:56
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    $\begingroup$ @Mołot, turns out I made a calculation mistake in the "Answers". Sorry! If someone could just double-check the calculation I've given under "Answers", that'd be awesome. I've gotten it wrong twice already :( $\endgroup$ May 5 '17 at 11:55
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    $\begingroup$ @JerryTheC, I think I kind of follow the explanation, but not fully... Why don't you submit an answer with the calculation done using your method? I'd love to see how the numbers compare. I would also like to see how you model the color accuracy bit, amongst other things. $\endgroup$ May 6 '17 at 19:49

According to this, you'd need a telescope, the size of "the Hubble space telescope" to read the numbers on a licence plate or recognise someones face on the distance of 10km.

So I guess, you need a telescope, the size of a planet, or even bigger to achieve what you want. Even if you have your telescope in space, the other person is almost certainly on a planet with an atmosphere (if you want to keep track of the planet and read lips), it is almost impossible (see the link above) to see anything. Maybe you have the biggest chance trying to do this at dawn. You also want to consider weather on your chosen target planet.

Maybe an altered version of the deathstar will do the trick.

By the way such telescopes usually don't come with autofocus and people tend to move a lot.

You also have to consider, that light bends (around large objects, like stars and planets). Having a lot of moving parts in space and a distance of 10 light years, will probably make it pretty hard to keep track of your target, because a slight wobble close to the origin light source, will have a giant influence on the other end.

As a sidenote, you might also want to consider how you treat close light sources, that will "over expose" your image.


To see the same for twice the distance, you neet to double the zoom. According to photo.stackexchange.com/questions/13717/…, you can calculate zoom from focal-length, from a given lens (in our case we take the HST specifications). The HST hat a focal lenght of 57.6m and can see with the mentioned resolution 10km. For one lightyear, that would give you a focal length of 5.472 * 10^13m. The earth has a diameter of 12'742 km. The (mid-) Distance from Sun and Saturn is according to this 1.427 * 10^12m. So the focal length is about thirty eight times the distance of the solar system. To keep the f-stop of f/22 of Hubble, that would make a lens diameter of 2.487 * 10^12m. Which is still 1.7 Times the distance from Sun to Saturn.

These Calculations, however are for lenses, Hubble doesn’t have a lens. Like all large telescopes, Hubble uses a curved mirror to focus starlight. This mirror is located deep inside the telescope.

You might want to protect this precious piece from asteroids.

You can do the Math for 10 and 100 lightyears.

  • $\begingroup$ A wild guess? No real knowledge or calculation? $\endgroup$
    – JDługosz
    May 4 '17 at 5:11
  • $\begingroup$ I'm referencing the mentioned source, with the calcilation. $\endgroup$
    – Frezzley
    May 4 '17 at 5:42
  • 1
    $\begingroup$ That gives the resolution of HST. How do you get “size of a planet or bigger” for the answer? $\endgroup$
    – JDługosz
    May 4 '17 at 6:47
  • $\begingroup$ That's much better! Your “update” is really what should be the answer, beyond the first paragraph. Do you see how what you wrote now is quantitatively different from the original? The original (after the first paragraph) goes off on irrelevant tangents and only has one sentence of Answer. I'd suggest deleting those old paragraphs. $\endgroup$
    – JDługosz
    May 4 '17 at 11:54

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