# How do I calculate the amount of sunlight a planet gets?

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.

• 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. Aug 19, 2018 at 16:56
• Yeah what b.Lorenz said, I can give you the stellar luminosity very easily but the next bit is a lot more problematic.
– Ash
Aug 19, 2018 at 17:01
• 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) Aug 19, 2018 at 21:17
• Based on this question and its answers I think it should be moved to Physics SE. Aug 19, 2018 at 22:58

# 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 $$I_V\approx12.7\text{ lux}$$.

• Calculated luminosity for the OP's star is actually .002 solar luminosity, according to this calculator anyway.
– Ash
Aug 19, 2018 at 17:09
• @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. Aug 19, 2018 at 17:11
• Yeah based on that data set it looks like the given spectral type and the specifications are incompatible.
– Ash
Aug 19, 2018 at 17:13
• @Ash Ah, fair point. I've added in additional numbers based on the OP's specifications. Aug 19, 2018 at 17:18
• 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. Aug 19, 2018 at 18:05