What is the minimum size of a planet that could harbor human life? You get to decide what the atmosphere is, but I'm reasonably sure that O2, CO2, and H2O are necessary components of the atmosphere. Assume the planet's surface is flat. I'm not sure what the soil and rocks the planet are made of, even though I know that plays a role in the planet's ability to harbor an atmosphere. (And the atmosphere must be hospitable to plants as well, or else everyone is going to die.)

  • $\begingroup$ You're interested in volume, then, not mass? $\endgroup$ – HDE 226868 Jan 5 '16 at 0:32
  • $\begingroup$ Yes, more specifically, surface area $\endgroup$ – Xandar The Zenon Jan 5 '16 at 0:56

The retention of gases such as oxygen, carbon dioxide, and water depends on the mass and radius of the planet, via its surface gravity. It also depends on the effective temperature, something we can calculate easily. I'd like to quote from an earlier answer of mine:

The planetary equilibrium temperature: $$T = \left(\frac{L_{\odot}(1-a)}{16 \sigma \pi D^2} \right)^{\frac{1}{4}}$$ We can approximation that $L_{\odot} \approx L_{\text{Sun}}=3.846 \times 10^{26}$. As another approximation, $a=0.3$. We also know that $\sigma=5.670 \times 10^{-8}$. Plugging this all in, $$T = \left(\frac{3.846 \times 10^{26}(1-0.3)}{16 \times 5.670 \times 10^{-8}\pi D^2} \right)^{\frac{1}{4}}=9.85856 \times 10^7 \times D^{-\frac{1}{2}}$$ At $D_V$, $D_E$, and $D_M$, this comes out to $$T_V=299.986 \text{ K}$$ $$T_E=254.547 \text{ K}$$ $$T_M=207.515 \text{ K}$$

Next, we can use a chart to figure out what gases will be in the atmosphere. First, however, we must calculate escape velocity.

Seager et al. (2007) proposed a mass-radius relationship for terrestrial planets. $$M\approx\frac{4}{3}\pi R^3\left[1+ \left(1-\frac{3}{5}n \right)\left(\frac{2}{3}\pi R^2\right)^n\right]$$ Here, $n$ is a parameter that depends on the composition of the planet. If we assume a planet composed mainly of silicates (like Earth), then $n\approx0.537$.

Escape velocity is given by $$v_{\text{esc}}=\sqrt{\frac{2GM}{R}}$$ Plugging in our expression, we have $$v_{\text{esc}}=\sqrt{2G\frac{4}{3}\pi R^2\left[1+ \left(1-\frac{3}{5}n \right)\left(\frac{8}{3}\pi R^2\right)^n\right]}$$

Let's go to the chart I mentioned earlier:

Image courtesy of Wikipedia user Cmglee under the Creative Commons Attribution-Share Alike 3.0 Unported license.

At each of the three distances (now temperatures), we have the following requirements to keep all of the gases at the orbital radii of Venus, Earth, and Mars: $$\text{Venus:}\approx4\text{ km/s}$$ $$\text{Earth:}\approx3.5\text{ km/s}$$ $$\text{Mars:}\approx3.25\text{ km/s}$$ Using Wolfram Alpha, I get $$\text{Venus:}\approx177\text{ kilometers}$$ $$\text{Earth:}\approx162\text{ kilometers}$$ $$\text{Mars:}\approx154\text{ kilometers}$$ These numbers are substantially low. They result in part from complicating factors that drastically affect the temperature of the planet, including non-uniform albedo and potential greenhouse effects.

Non-quantitative summary

The main feature of a habitable planet is its composition: the composition of its atmosphere and the planet itself. Certain compounds must be present for human-like life to survive. These compounds can only exist at certain temperatures at certain surface gravities, or else they will escape the planet.

I used a chart and a mass-radius relation, as well as some assumed constants, to get various values for the radius necessary to keep some basic compounds (oxygen and carbon dioxide), assuming a silicate composition. The values differ based on how far away the planet is from its parent star. Note that the planet should remain in the circumstellar habitable zone. The three distances given here are near the inner edge, middle, and outer edge of the Sun's habitable zone.

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    $\begingroup$ Um, I'm reasonably sure that is a great answer, but could you write a summary? I kind of just read science science math math math science, three numbers equal to scientific variables. Some of us aren't quite as smart as others. $\endgroup$ – Xandar The Zenon Jan 5 '16 at 2:15
  • $\begingroup$ @XandarTheZenon Sure. Being concise is not my strong suit. I'm busy at the moment, but I'll condense it later. $\endgroup$ – HDE 226868 Jan 5 '16 at 2:18
  • $\begingroup$ Thanks HDE, also, I know this is a little off topic, but why do people do the @insertname when addressing people? $\endgroup$ – Xandar The Zenon Jan 5 '16 at 2:26
  • $\begingroup$ @XandarTheZenon It "pings" them, sending a message to their inbox. People are always automatically "pinged" by comments on their own post, but not on other people's posts. $\endgroup$ – HDE 226868 Jan 5 '16 at 2:33
  • $\begingroup$ I see, this makes sense. $\endgroup$ – Xandar The Zenon Jan 5 '16 at 2:39

This depends mainly on two factors: size of the planet, and density. Assuming that the planet is composed largely of osmium, the densest stable element (with an atomic mass of 190), and not nickel (atomic mass 58.7), the planet's core would be significantly less than the 2400 km ball of molten nickel and iron under our feet.

A good estimate for the core size might be around 2.56 times smaller than the Earth's core, or 2/5 of the size of the Earth, 960 km in diameter. The crust, however, would need to be about 1300 km thick (to put this into scale, the crust is usually about 45 km thick on Earth) and not very dense at all, so that humans could survive at a relatively comfortable gravitational level of about 2 gravitational forces.

This brings the total diameter of the new Earth to around 3500 km, sans atmosphere.

Disclaimer: I am not a physicist, nor am I Randall Munroe, author of xkcd. However, utilizing equations and Wikipedia, I have come up with this answer for you.

  • 1
    $\begingroup$ Can you show the steps you used to get to these figures? Thanks! $\endgroup$ – HDE 226868 Jan 5 '16 at 1:57

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