# 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.

• 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. – b.Lorenz Aug 19 '18 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 '18 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) – JSCoder says Reinstate Monica Aug 19 '18 at 21:17
• Based on this question and its answers I think it should be moved to Physics SE. – person27 Aug 19 '18 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 '18 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. – HDE 226868 Aug 19 '18 at 17:11
• Yeah based on that data set it looks like the given spectral type and the specifications are incompatible. – Ash Aug 19 '18 at 17:13
• @Ash Ah, fair point. I've added in additional numbers based on the OP's specifications. – HDE 226868 Aug 19 '18 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. – b.Lorenz Aug 19 '18 at 18:05