I'm the Attitude Control System team lead on a cube satellite project, and we're trying to use cheap lux sensors to determine the body sun vector (vector adding the lux from each sensor). I want to filter out signals from the Earth, and to do so would require a priori knowledge of the lux values encountered by the Sun and Earth. I have done some research and the Sun should be around 140,000 lux at maximum, while the earth would be around 42,000 at maximum due to the .3 albedo.

Can anyone confirm that these values are valid?

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    $\begingroup$ I'm no expert but is lux the right measurement for this? I thought lux was more related to human eye vision while radiance is typically measured in watts per square meter or in apparent magnitude. You can probably find values for watts per square meter and then convert them to lux as w/m^2 are used often for solar power calculations $\endgroup$
    – Dragongeek
    Commented Nov 17, 2019 at 1:01

3 Answers 3


The lux value of the Sun is ok. The albedo of Earth is not constant, it depends on cloud cover and the location above oceans or land.

Dragongeek is right, lux is defined by using the standard sensitivity of the human eye. Radiance as watts per square meter does not depend on the human eye's sensitivity for different wavelength.

The radiance above the atmosphere is 1.3608 ± 0.0005 kW/m². The distance of Earth to the Sun varies between 0.98327 AU and 1.017 AU. If we define 100 % radiance at 1 AU, there is a varation between 103.4 and 96.7 %. There is no precise well defined value for the albedo of Earth.

Conversion between W/m² and lux is not easy, the definition 1 W/m² = 683 lux is valid only for light of the frequency 540·10^12 Hz or the wavelength 555 nm.

For sunlight above the atmosphere it is 1 W/m² = 98 lux. So 1.3608 kW/m² is equivalent to 133,358 lux.


Typical A, B problem answer, but:

I think it would be better not to depend on a fixed hard coded value of a property of the earth. What you are interested in is the detected value. You could calculate and update this value as you go.

For example, do a rough pass to calculate the relative position of the sun (vector add/integrate all light detected), then detect all light not from a sensor near this orientation. That would give you a measure of all non-sun light sources. You could keep a averaged version of this to try and detect and provide a backup value for when the sun and earth aligned well enough that the above method fails.

The advantage of this in addition to not needing a hard-coded value is it can keep track of changes over time (both outside world and changes to the state of the sensors).


Just to add to Uwe's answer. The radiance depends upon distance to the Sun which varies from perihelion to Aphelion. Perihelion is January 3rd-ish and, I think its 3% higher than the mean distance. However its small beer compared to the presence or otherwise of Earth Albedo.

Sun sensors of the type you mention are usually made with solar cells.

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    $\begingroup$ I think this should be an edit to @Uwe's answer. $\endgroup$
    – Star Man
    Commented Nov 17, 2019 at 19:51
  • $\begingroup$ ...or if not, then a comment, but not a stand-alone Stack Exchange answer. Also, it shouldn't be of the form "I think it's x...", it should be fact-based. It will vary as $1/(1 \pm \epsilon)^2$. Earth's orbital eccentricity is 0.01671 so it ranges from +3.43% to -3.26 relative to 1 AU. $\endgroup$
    – uhoh
    Commented Nov 17, 2019 at 22:51
  • $\begingroup$ @StarMan it could be, but doesn't have to be. I've introduced two extra bits of information, I didn't have to put the first sentence "Just to add" there at all, though I chose to as I'm giving credit to Uwe for an already good answer. $\endgroup$
    – Puffin
    Commented Nov 17, 2019 at 23:14
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    $\begingroup$ @uhoh lots of people write "I think"; its a way of taking responsibility for uncertainty in the answer. It doesn't mean there isn't a fact there. $\endgroup$
    – Puffin
    Commented Nov 17, 2019 at 23:16

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