Say I put a satellite in orbit with several LEDs on the outside: red, yellow, blue, white and green. Could I see them from the ground with consumer-grade binoculars or telescopes?

If I can see them, could I differentiate between the colors? (e.g. would the color affect how well I could see them?)

Say I have 20/20 vision with no color blindness and I travel to an ideal dark observation point on the earth.


2 Answers 2


You cannot see them.

Let's put some number. You did not specify the type of LED you want to use, so I decide to go for 1W led (such as this one https://www.adafruit.com/product/518). The specification of this LED say that it emit 90 lumens over a cone of 140 degree, that means over $\simeq 11$ steradians.

Low earth orbits range from 180km to 2000km. So let's say our satelitte is 200km over our heads.

Then the illuminance $L$ of our LED at sea level is then of $$L=\frac{90 \text{ lm}}{11 \text{ sr} (200.10^3 m)^2} = 2.10^{-10} \text{ lm.m}^{-2} $$ And this assume full transparency of the atmosphere. (I found transparency value of about 82% otherwise)

Now we have to compared this to eye sensibility. According to Wikipedia (https://en.wikipedia.org/wiki/Illuminance), the sensibility of naked eye under starlight is $5.10^{-5}\text{lm.m}^{-2}$. So with naked eye, I would be impossible to see the LED.

Even, with consumer-grade binoculars or telescopes, we are still out of reach reaching. Standard binocular allows only for a improvement of a factor 40 (see below for the estimation of this factor) in visible luminosity. Even high grade telescope of 75cm of aperture only give you an increase of $10^4$, that bring you closer but you are still below.

For the question on the color, you have two factor, first human eye does not have the same sensibility for every color (mostly green is the most visible color) and LED does not have the same efficiency to emit light for every color.

Estimation of gain factor for binocular and telescope.

I estimate the gain from the difference in apparent magnitude for star from naked eye and binocular. The magnitude $m$ is defined as $$ m= m_{ref} -2.5\log_{10} \frac{L}{L_{ref}}$$ Accordint to this website (http://www.stargazing.net/david/constel/howmanystars.html), human eye can seen star up to magnitude 6.5, whereas binocular allow you to see start up to magnitude 10.5, from the formula on the magnitude you get a factor $\simeq 40$ in illuminance. Top grade telescope allow you to reach magnitude 16, hence the factor.

Edit: A bit more digging give me different value for the sensibility of the naked eye. We have an approximate formula to convert illuminance to apparent magnitude used in astronomy. $$ m =-14.18-2.5\log_{10}(L). $$ An accepted value for the limit in magnitude for naked eye is magnitude 6.5, using the formula give you a limit of $5.10^{-9} \text{ lm.m}^{-2}$, you are still out of reach but with more powerful LED this seems possible to reach with binocular.


It really depends on the resolution of your binoculars...


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