The possibility though remote, is intriguing as we may be able in the future to actually "see" our own planet's history. Though sounding science fiction, if we are able to detect bodies in space that are able of reflecting light emitted from our planet earth, using amplification systems and filters,this may - if possible, give us a tool of utmost importance.

  • $\begingroup$ Related: physics.stackexchange.com/q/11940/2451 $\endgroup$
    – Qmechanic
    Commented Jan 21, 2013 at 0:51
  • $\begingroup$ Yes. It is possible to do so. $\endgroup$
    – Apoorv
    Commented Jan 21, 2013 at 1:03
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    $\begingroup$ We'd want to calculate the number of photons that are incident & then reflected from the earth (from the sun, the brightest source in our vicinity), then assume a $\frac{1}{r^2}$ law to see how many of those reflected photons hit the distant planet, then determine what fraction of those photons are reflected from the distant planet, give it another $\frac{1}{r^2}$ law on the way back, and see how many photons get back to us. I'm guessing it's <<1 per year, but haven't done the math. At that rate, it's going to be hard to separate that expected photon each year from noise. $\endgroup$
    – Will Cross
    Commented Jan 21, 2013 at 3:55
  • $\begingroup$ There is a huge difference between detecting total illumination and resolving images. Astronomers are detecting light echos of events, but there is a lot of integration (averaging) going on. You can get a (time convoluted) light curve, but resolving detail seems out of bounds. $\endgroup$ Commented Jul 26, 2014 at 4:00

4 Answers 4


Not in any practical sense. First of all, the intensity will be very low: inverse-r-squared going out to the reflector, and inverse-r-squared coming back. As @WillCross pointed out, it will be at very low levels. Secondly, the resolution will be very poor, as there's nothing to focus it and the subtended angle is very very small (in both directions).

Consider that we already do have such a reflector, about 1.5 light seconds away: the dark side of the Moon, illuminated by Earthshine. It somewhat lights up the lunar surface, and variations in the total illumination can be detected that correlate to cloud cover (increased reflection) on Earth. Even being so close, there's no way to see any sort of useful images as you envision. Maybe you could detect a massive nuclear war by reflections from the Moon or nearby planets, but at light year distances, it would be hard to sort out from the noise.

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    $\begingroup$ Just a quibble: the dark side of the Moon usually refers to the side that never faces Earth, even though this surface is regularly illuminated by the Sun. The reflecting surface you're talking about is the colloquial "light side of the Moon", just at a moment when it is not illuminated by the Sun. Still, +1. $\endgroup$
    – Kyle Oman
    Commented Jul 25, 2014 at 21:49
  • 1
    $\begingroup$ @Kyle, although it's common to use the "Dark Side of the Moon" to describe the side of the moon that never faces the Earth, it's certainly not physically accurate to do so. As a result, even though the way the phrase in the answer rubs against colloquial usage of the term, it's more accurate physically. The "Dark Side of the Moon" is quite literally, the side that is not receiving sunlight. $\endgroup$
    – Sean
    Commented Dec 12, 2014 at 14:23

The average albedo of the Earth is about 0.3 (i.e. it reflects 30 percent of the light incident upon it).

The amount of incident radiation from the Sun at any moment is the solar constant ($F \sim 1.3 \times 10^3$ Wm$^{-2}$) integrated over a hemisphere. Thus the total reflected light from the Earth is $L=5\times 10^{16}$ W.

Let's assume this has the same spectrum as sunlight and let's assume that this light gets reflected from something which is positioned optimally - i.e. it sees the full illuminated hemisphere. In that case, roughly speaking, the incident flux on any reflecting body will be $L/2\pi d^2$ (because it is scattered roughly into a hemisphere of the sky).

Now we have to explore two divergent scenarios.

  1. There just happens to be a large object a 1000 light years away that is highly reflective.

Let's be generous and say it is a perfect reflector, but we can't assume specular reflection. Instead let's assume the reflected light is also scattered isotropically into a $2\pi$ solid angle. Thus the radiation we get back will be $$ f = \frac{L}{2\pi d^2} \frac{\pi r^2}{2\pi d^2} = \frac{L r^2}{4\pi d^4},$$ where $r$ is the radius of the thing doing the reflecting.

To turn a flux into an astronomical magnitude we note that the Sun has a visual magnitude of $-26.74$. The apparent magnitude of the reflected light will be given by $$ m = 2.5\log_{10} \left(\frac{F}{f}\right) -26.74 = 2.5 \log_{10} \left(\frac{4F \pi d^4}{L r^2}\right) -26.74 $$

So let's put in some numbers. Assume $r=R_{\odot}$ (i.e. a reflector as big as the Sun) and let $d$ be 1000 light years. From this I calculate $m=85$.

To put this in context, the Hubble space telescope ultra deep field has a magnitude limit of around $m=30$ (http://arxiv.org/abs/1305.1931 ) and each 5 magnitudes on top of that corresponds to a factor of 100 decrease in brightness. So $m=86$ is 22 orders of magnitude fainter than detectable by HST. What's worse, the reflector also scatters all the light from the rest of the universe, so picking out the signal from the earth will be utterly futile.

  1. A big, flat mirror 1000 light years away.

How did it get there? Let's leave that aside. In this case we would just be looking at an image of the Earth as if it were 2000 light years away (assuming everything gets reflected). The flux received back at Earth in this case: $$ f = \frac{L}{2\pi [2d]^2} $$ with $d=1000$ light years, which will result in an apparent magnitude at the earth of $m=37$.

OK, this is more promising, but still 7 magnitudes below detection with the HST and perhaps 5 magnitudes fainter than might be detected with the James Webb Space Telescope if and when it does an ultra-deep field. It is unclear whether the sky will be actually full of optical sources at this level of faintness and so even higher spatial resolution than HST/JWST might be required to pick it out even if we had the sensitivity.

  1. Just send a telescope to 1000 light years, observe the Earth, analyse the data and send the signal back to Earth.

Of course this doesn't help you see into the past because we would have to send the telescope there. But it could help those in the future see into their past.

Assuming this is technically feasible, the Earth will have a maximum brightness corresponding to $m \sim 35$ so something a lot better than JWST would be required and that ignores the problem of the brightness contrast with the Sun, which would be separated by only 0.03 arcseconds from the Earth at that distance.

Note also that these calculations are merely to detect the light. To extract anything meaningful would mean collecting a spectrum at the very least!


There are several reflectors that were placed on the moon during the Apollo mission and the number of photons detected (after being reflected) is very small compared to the number of photons the laser emits. See the section "Details"


Rather late, but as LDC3 resurrected this question from 2013....

This has already happened. Radio waves reflected from what is presumed to be an asteroid swarm about 25 LY away have returned to earth and been identified as early TV broadcasts. Quite appropriately, the first identified signals were episodes of Dr. Who.

However, radio and visible light are rather different things. If you are expecting to watch Julius Caesar walk up the steps of the Portico, it won't happen. We can barely do that from low orbit.

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    $\begingroup$ Do you have a citation to back up the claim that we have detected our own radio signals? I'm reasonably sure we couldn't detect our own radio signals even 1 ly away with even the most powerful radio telescopes. $\endgroup$
    – user10851
    Commented Jul 26, 2014 at 2:09

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