# What is the habitable zone around my star?

I have built a star that is loosely based on a real-world star. It has the following properties:

• Spectral class G
• Mass: 1.03 M$_\odot$
• Radius: 1.02 r$_\odot$
• Luminosity: 1.05 L$_\odot$
• Surface temperature 5,792 K

Now, I'm trying based on Calculating the Habitable Zone to calculate the inner and outer boundaries of the habitable zone around the star, but I simply can't seem to wrap my head around the calculations.

How do I calculate the habitable zone based on the above, and what are the values for inner and outer orbital radius around this star? Or do I need to decide on some additional parameter, and if so which?

• Your star actually has the same mass as Kepler-452! it's just a little less luminous than Kepler-452 :) – Stephanie May 17 '17 at 3:18

You have the two equations you need on the linked page under the heading "Stage two": $$r_i=\sqrt{\frac{L_{\text{star}}}{1.1}},\quad r_o=\sqrt{\frac{L_{\text{star}}}{0.53}}$$ where $$r_i$$ and $$r_o$$ are the inner and outer radii of the habitable zone, in astronomical units and $$L_{\text{star}}$$ is the star's luminosity, in solar luminosities. In your case, then, with $$L_{\text{star}}=1.05L_{\odot}$$, we have $$r_1=\sqrt{\frac{1.05}{1.1}}=0.978\text{ AU},\quad r_o=\sqrt{\frac{1.05}{.053}}=1.408\text{ AU}$$ LSerni's answer has already done these calculations. But where do those parameters, $$1.1$$ and $$0.53$$, come from? The page states that they are "constant value(s) representing stellar flux" at those radii. I believe that they are values of a parameter denoted $$S_{eff}$$, the effective solar flux (see an updated version in Kopparapu et al. (2013)), the value required to make the solar constant, the flux density on a certain surface, lead to a specific stable surface temperature.

$$S_{eff}$$, according to this new paper (I cannot find the text of the cited book, so I'm looking at a newer version of the analysis), depends on the effective temperature of the star ($$T_{eff}$$) and four constants ($$a$$, $$b$$, $$c$$, and $$d$$). These four constants determine what kind of a planet will develop. The specific equation is $$S_{eff}=S_{eff,\odot}+aT_*+bT_*^2+cT_*^3+dT_*^4$$ where $$T_*=T_{eff}-5780\text{ K}$$. $$5780\text{ K}$$ is, of course, the effective temperature of the Sun - and you can check that $$S_{eff}=S_{eff,\odot}$$ when $$T_{eff}=5780\text{ K}$$. The paper then derives Equation (3): $$r=\sqrt{\frac{L_{\text{star}}/L_{\odot}}{S_{eff}}}\text{ AU}$$ The values $$S_{eff}=1.1$$ and $$S_{eff}=0.53$$ correspond, I believe, to two sets of choices for $$a$$, $$b$$, $$c$$, and $$d$$ and some $$T_{eff}$$. The paper gives examples in Table 3. I do not know the values chosen to yield the inner and outer values of $$S_{eff}$$. I do, though, invite you to play around with the constants. Keep in mind that they should all be relatively small, and that they are good for "nice" values of $$T_{eff}$$ - specifically, for $$2600\text{ K}\leq T_{eff}\leq7200\text{ K}$$.

So, habitable zone calculations are a pain. For a start, most habitable zone calculations make some key assumptions:

• The orbiting planet(s) is/are Earth-like, or at least similar to a more welcoming Venus or Mars.
• The orbits remain fully inside the habitable zone.
• The planets don't have any freaky axial tilts.
• The luminosity of the star remains constant (many calculations do look at how the habitable zone changes over a stars life, but variable stars could have much shorter oscillations in luminosity). Essentially, we want to orbit well-behaved stars.
• We want liquid water on the surface.

These assumptions do not cover all of the possible scenarios in which life could arise. For instance, they ignore the possibility of life on moons orbiting gas giants, where tidal forces could provide heat (hello, Europa and Enceladus!). They also imply that life must be carbon-based, using water as a solvent. Essentially, the term "circumstellar habitable zone" should really be "circumstellar this-seems-about-right-for-Earth-and-humans-to-live-don't-you-think zone".

The boundaries are also highly dependent on climate models, as we saw earlier - the section on Wikipedia detailing various predicted Solar System habitable zones should convince you of this. Choices of the four constants for $$S_{eff}$$, for instance, have dramatic effects on a planet, changing it from a Venusian hell to a cold Martian twin. Models from first-principles need to take into account, for instance, the greenhouse effect (radiative forcing, anyone?).

So, here's how to determine the habitable zone, in a nutshell:

1. Choose the properties of your star at a given time - essentially, luminosity.
2. Choose the physical properties of the sort of planet you want, early in its life. These include atmospheric composition, mass and radius (maybe), albedo, etc.
3. Create models of the evolution of the planet depending on the incident stellar flux.
4. Determine the range of fluxes in which such worlds can be habitable.
5. Calculate the radii at which the stellar flux will take these values.

All of this, for the best models, is extraordinarily complicated. I don't know how to do most of it. However, we can look at one type of case which is really simple: the idealized greenhouse model. A simple derivation can be found here.

Let $$T_s$$ be the surface temperature and $$T_a$$ be the atmospheric temperature (assuming that both are roughly uniform across the planet). The planet itself has albedo $$A$$, and its atmosphere has an absorption constant $$f$$, which depends on its composition. The incident flux is $$F_s$$. The energy balance equation for the planet is $$\frac{F_s(1-A)}{4}=(1-f)\sigma T_s^4+f\sigma T_a^4\tag{Planet}$$ The equation for the atmosphere is $$f\sigma T_s^4=2f\sigma T_a^4\tag{Atmosphere}$$ Putting these two together yields $$\frac{F_s(1-A)}{4}=(1-f)\sigma T_s^4+\frac{f}{2}\sigma T_s^4=\left(1-\frac{f}{2}\right)\sigma T_s^4$$ Rearranging, we get $$T_s=\left[\frac{F_s(1-A)}{4\sigma\left(1-\frac{f}{2}\right)}\right]^{\frac{1}{4}}\quad\text{or}\quad F_s=\frac{4\sigma T_s^4\left(1-\frac{f}{2}\right)}{1-A}$$ The latter is probably more helpful to us if we want to find the boundaries of the habitable zone, although because it is so idealized, it still doesn't take into account more complicated effects like radiative forcing. I should also note that setting $$f=0$$ makes $$T_s$$ simple the planet's effective temperature. However, it is almost never true that $$f=0$$; on Earth, $$f\simeq0.77$$.

I suspect I may have gone into a little more detail than you needed. As I said at the beginning, you really only need those two equations to figure out the rough boundaries of the habitable zone. Still, I hope that the rest of this answer was a little - dare I make this pun - illuminating.

• I think that's the longest "Short answer" I have every seen. – Secespitus Apr 30 '17 at 23:16
• I'm curious what the "Long" answer looks like! – Andon Apr 30 '17 at 23:36
• @Andon I've added it in. It actually turned out to be pretty short. – HDE 226868 May 1 '17 at 14:17

You have already the data you need - $L_\odot$ . That is the star's absolute luminosity. At that point you can simply apply the square root formula.

The habitable zone goes from $\sqrt{\frac{L_\odot}{1.1}}$ to $\sqrt{\frac{L_\odot}{0.53}}$.

In your case that is $\sqrt{\frac{1.05}{1.1}} = 0.977$ AU to $\sqrt{\frac{1.05}{0.53}} = 1.41$ AU.

There is also this calculator where you can plug your data and it will give 0.974 AU for the minimum habitable zone (runaway greenhouse) limit, and 1.717 AU for the maximum habitable zone limit. Optimistic habitability goes from 0.769 ("Recent Venus limit") to 1.809 ("Early Mars limit").

First, some basics. Luminosity and Power are the same thing. They are both energy over time.

The intensity of light is given by $$\text{Intensity}=\frac{\text{Luminosity}}{\text{Area}}$$ The area in question is the area of a sphere (the power is being spread out over bigger and bigger spheres as the distance from the star increases). $$A_{\text{sphere}}=4 \pi r^2$$ So $$\text{Intensity}=\frac{\text{Luminosity}}{4 \pi * \text{radius}^2}$$ or $$I=\frac{L}{4 \pi r^2}$$ If we are comparing two situations where we want the intensity to be the same, the we have, $$I_1=I_2$$ $$\frac{L_1}{4 \pi r_1^2}=\frac{L_2}{4 \pi r_2^2}$$ or, more simply $$\frac{L_1}{r_1^2}=\frac{L_2}{r_2^2}$$ Solving for $r_2$, we get $$r_2=r_1 \sqrt{\frac{L_2}{L_1}}$$ Now we plug in the orbital radius of the Earth and the Luminosity of the Sun (in those base units) and we get. $$r=r_{\text{Earth}} \sqrt{ \frac{L}{L_{\text{Sun}}}}$$ $$r=r_{\text{Earth}} \sqrt{ \frac{1.05 L_{\text{Sun}}}{L_{\text{Sun}}}}$$ $$r =\sqrt{1.05} r_{\text{Earth}}\approx 1.02 r_{\text{Earth}}$$ Similarly, the habitable zone pretty close to the one in our solar system, just 2% bigger.

This star is only slightly more massive/luminous than The Sun so expect the habitable zone to be extremely similar to The Sun's, for The Sun it is considered to stretch from 0.7 AU to 1.5 AU, although if a planet has thin atmosphere the inner zone could be 0.5 AU for surface water, and if a planet has a very strong greenhouse effect it could have liquid water past 2.00 AU. If you want a planet like Earth with a similar atmospheric pressure I would suggest moving it out a little further than 1.00 AU, between 1.08 and 1.15 AU sound about right to compensate for the slightly higher luminosity.

• This answer may very well be correct, but I'm looking for the math; sorry. I'll see if I can clarify that in the question. – user Apr 30 '17 at 21:44