# How close to a brown dwarf would a planet need to be to receive as much light as Earth?

How close to a brown dwarf would an orbiting planet need to be in order to receive as much sunlight as Earth receives from the Sun?

• There's a wide range in brown dwarf surface temperature. upload.wikimedia.org/wikipedia/commons/2/23/… Without specifics on the surface temperature of the brown dwarf, this isn't answerable. You also might want to specify if you mean visible light or heat. Brown dwarfs emit a lot of light in the infrared spectrum. Commented Dec 10, 2017 at 1:09

Brown dwarfs are born moderately hot and luminous and then they contract and cool. Thanks to electron degeneracy in their cores, they never become hot enough to ignite hydrogen (though there is a brief deuterium burning phase) and as a result their fate is to cool and fade.

The plot below (from Burrows et al. 1997) shows how the luminosity behaves as a function of time since birth (age) for objects of various mass. The brown dwarfs are the green and red tracks (some would call the red tracks giant planets). The curves are labelled in Jupiter masses. The highest mass brown dwarf is 73 Jupiter masses, the next one down is 70$M_J$ and then the green curves count down in steps of 5$M_J$ to 15$M_J$, then the first red curve is at 13$M_J$ and the red curves count down in steps of 1$M_J$.

The axes are logarithmic (to base 10). Thus when the y-axis says -2, it means the brown dwarf is one hundredth the luminosity of the Sun, -3 = one thousandth etc. The x-axis is logarithmic in units of billions of years. Most brown dwarfs observed in the Galaxy will be in the range 1-10 billion years (i.e. between 0 and 1 on this x-axis).

The flux received by a planet in orbit around a star/brown dwarf will be proportional to the its luminosity divided by orbital radius squared. From this we can write that the orbital radius where a similar flux to that of the Earth around the Sun will be received is $$R = \left( \frac{L_{BD}}{L_{\odot}} \right)^{1/2}\ {\rm au}$$ $$\log\left(\frac{R}{1\ {\rm au}}\right) = 0.5 \log \left(\frac{L}{L_{\odot}}\right)$$

This you can get you answer immediately from the graph below. Choose the mass of your brown dwarf and its age. Find the log luminosity on the y-axis and halve this value. That is the the log of the orbital radius in astronomical units where a planet would receive the same amount of flux as the Earth from the Sun.

e.g. Consider a 50$M_J$ brown dwarf that is a billion years old. Find the appropriate green curve and see that the y-axis value for x=0 (1 billion years) is -4.2. The log of the orbital radius (in au) at which a planet would receive the same amount of flux from the brown dwarf as does the Earth from the Sun would be -2.1. In linear units, this is $10^{-2.1} = 0.0079$ au, or 1.19 million km.

A further issue to consider is that the brown dwarf surface will be far cooler than the Sun and also changes (cools) with time. This means that the radiation received from the planet will push further and further into the infrared as the brown dwarfs gets older and hardly any of the received radiation will be in the visible part of the spectrum.

Brown dwarves don't have fusion in their core, thus they don't have their own light. They may have a little heat production from other processes (for example, contraction, or radioactive decay in their core). These aren't enough strong to heat even the dwarf significantly.

Many brown dwarve were found orbiting a star closely, they may have surface temperature even in the order of some thousands K. But they get their temperature from their stars.

For example, the Jupiter's temperature is only 40K more warm as it would be reasoned by the Sun.

In the case of red dwarves, the situation is better. The distance depends on the star. You can very easily calculate it: check its absolute luminosity, so you can see its total power output, compared to the Sun.