I'm working on a planet and I want to know what is the maximum amount of sunlight my planet gets in lux (for Earth It's 120000 lux).

My planet orbits an M4V class red dwarf that has a mass of 0.22 M☉, radius of 0.16 R☉ and a temperature of 3000 K. The planet's semi-major axis is 2.3 AU.

Please also include the formula or program you used to calculate the answer. If you need any additional info I'll add it.

  • 2
    $\begingroup$ Your requirement of lux makes this complicated. Calculating radiative flux is simple math, but luminous flux depends on the interplay of the stars spectrum and human eye. $\endgroup$
    – b.Lorenz
    Aug 19, 2018 at 16:56
  • $\begingroup$ Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic. $\endgroup$
    – Ash
    Aug 19, 2018 at 17:01
  • $\begingroup$ Also, because you have a planet orbiting a red dwarf (smaller than sun) further than the distance that mars is from, it is likely (read: certain) that the planet is too cold for liquid water to form on. So, if you want life, move it a lot closer (my estimate is about 10-20 times closer) $\endgroup$ Aug 19, 2018 at 21:17
  • $\begingroup$ Based on this question and its answers I think it should be moved to Physics SE. $\endgroup$
    – kettlecrab
    Aug 19, 2018 at 22:58

1 Answer 1


Calculating flux

I think a slightly more helpful quantity to calculate is the flux received by the planet - the power per unit area from the star. The mean flux on Earth is the solar constant, $F_e=1.36\times10^3\text{ W m}^{-2}$. If a planet orbits a star of luminosity $L_*$ at a distance $r_p$, the flux received is $$F_p=\frac{L_*}{4\pi r_p^2}$$ The ratio of the planet's flux $F_p$ to the solar constant $F_e$ is $$\frac{F_p}{F_e}=\frac{L_*}{L_{\odot}}\left(\frac{r_p}{1\text{ AU}}\right)^{-2}$$ For an M4V dwarf, I'd expect $L_*\approx0.006L_{\odot}$. Plugging this in, we get $$F_p=1.1\times10^{-3}F_e=1.50\text{ W m}^{-2}$$ for an M4V star.

Why should we use flux instead of lux?

  • It's easier to calculate. And when I say easier, I mean much easier. Lux is a lot more complicated.
  • It's great for calculating things like the effective temperature of a planet.
  • It's much more commonly used in this sort of scenario.

I used stellar models by Eric Mamajek to find the star's luminosity - he gives a $\log L_*/L_{\odot}=-2.2$, so $L_*\approx0.006L_{\odot}$. It's worth noting that those models list an M4V star as having $T_{eff}\approx3200\text{ K}$ and $R\approx0.258R_{\odot}$ - so this is based on those figures. Using your numbers (which are for a smaller, cooler star) and the Stefan-Boltzmann law, I get $L_*\approx0.00186L_{\odot}$, leading to a flux of $F_p\approx0.48\text{ W m}^{-2}$. I believe your values are slightly off for the given spectral type.

I've written a Python program to calculate the flux received on a planet based on a given spectral type, using those models. It should be a quick shortcut for you in the future.

Astronomical lux

It's claimed that you can convert between a star's apparent magnitude in the V-band ($m_V$) and its illuminance ($I_V$) - the value I think you're looking for. Wikipedia's formula is $$I_V=10^{(-14.18-m_V)/2.5}$$ which matches values from the National Park Service. The absolute magnitude of an M4V star is - according to the same stellar models - $M_V\approx12.80$. Apparent magnitude can be calculated from absolute magnitude by $$m_V=M_V+5\log\left(\frac{r_p}{10\text{ parsecs}}\right)$$ Putting this together for our case yields $m_V\approx-16.96$, and, finally, we get $I_V\approx12.7\text{ lux}$.

  • $\begingroup$ Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway. $\endgroup$
    – Ash
    Aug 19, 2018 at 17:09
  • $\begingroup$ @Ash I tend to use a specific grid of stellar models - I assume various ones differ by a factor of maybe two or three. An M5V dwarf is listed as being at $\approx0.003L_*$; the value is sensitive to the exact choice of spectral type. $\endgroup$
    – HDE 226868
    Aug 19, 2018 at 17:11
  • $\begingroup$ Yeah based on that data set it looks like the given spectral type and the specifications are incompatible. $\endgroup$
    – Ash
    Aug 19, 2018 at 17:13
  • $\begingroup$ @Ash Ah, fair point. I've added in additional numbers based on the OP's specifications. $\endgroup$
    – HDE 226868
    Aug 19, 2018 at 17:18
  • $\begingroup$ You are right. Lux is only useful to determine how bright the sky would seem to human eyes. Any other derivative effect - solar cell power, plant life, weather... depends on radiative flux. $\endgroup$
    – b.Lorenz
    Aug 19, 2018 at 18:05

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