From an object's apparent magnitude is it possible to calculate the ojects illuminence on Earth? How?

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Approximately, yes. The logarithmic visual magnitude scale has been tied to illuminance measured in lux for zero magnitude, so it's just a matter of raising 2.512 to the power of the magnitude, which will give you how much brighter or dimmer than zero magnitude your star is, and then multiplying the lux value. I don't have the conversion figure to hand, but I should be able to dig it out.

Grant Hutchison

At visual magnitude zero, a star gives us 3.92e-9 erg/(square centimeter second Angstrom) at 5500 A, or 3.81e-23 watt/(square meter Hertz). These values change somewhat for stars with extreme colors, since the standard V filter band has appreciable width so the flux-weighted mean wavelength changes. This calibration traces to comparison of bright stars (especially Vega) with laboratory blackbodies.

An easy rule is that a V=0 star gives us (above the atmosphere) about 1000 photons per square centimeter per second per Angstrom of bandwidth. (As I work it out, the nmber is closer to 1084).

These absolute calibrations are still uncertain at the level of a percent or slightly less. The magnitude scale is internally better defined than this, which is one reason it continues in use for many purposes - at this point, some measurements would see an increase in error bars if converted to energy flux.

Light diminishes with the square of the distance, you can work out the luminosity and apparent Magnitude using these rules, therefore our Sun shines much lower magnitude from Alpha Centauri , mag is a log base 2.5118...from the centauri system the Sol shines at about +0.4 magnitude

There is also a (similarly approximate, and for the same reasons) conversion from apparent visual magnitude to units of illuminance, lumen/square metre (=lux).
I have the conversion figures at home, if no-one comes up with them in the next few hours.

Grant Hutchison

Here we go:

Illuminance from a star of apparent visual magnitude zero = 2.54e-6 lux

Worked example:

Apparent visual magnitude of the Sun: -26.78

Illuminance = 2.512^26.78 x 2.54e-6 lux = 131000 lux


Grant Hutchison

Strictly speaking, shouldn't you divide the lux value by 2.511886...^(magnitude)?

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Oops. Thanks.
I was sure I'd typed "to the power of the magnitude difference" in my first post, which would have neatly left the onus on everyone else to keep track of the minus sign.

Grant Hutchison

Most of you already know this, but for the sake of some of our younger members, every five magnitudes is exactly a factor of a hundred in total luminosity. So this 2.51... number is the fifth root of 100.

If for some reason you need it to even more places, simply raise 100 to the 0.2 power on a calculator.

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