What would happen if Earth's oxygen reached the Sun?

Let's calculate the mass of this huge oxygen atmosphere, shall we?

I can use a modified version of the barometric formula to calculate the density at any altitude $h$ above a reference point - in this case, at Earth's surface: $$\rho=\rho_0 \exp \left[\frac{-g_0M(h-h_0)}{RT} \right]$$ where $_0$ denotes a quantity at $h=0$. To find the average value, we have to use a formula: $$f(x)_{\text{avg}}=\frac{1}{b-a} \int_a^b f(x)dx$$ In our case, $a=h_0$ and $b=150,000,000,000$ (both in meters). $\rho$ is a function of $h$, so we have $$\rho_{\text{avg}}=\frac{1}{150,000,000,000-h_0}\int_{h_0}^{150,000,000,000}\rho_0 \exp \left[\frac{-g_0M(h-h_0)}{RT} \right]dh$$ $$\rho_{\text{avg}}=\frac{1}{150,000,000,000-h_0}\int_{h_0}^{150,000,000,000}\rho_0 \exp \left[\frac{-g_0Mh+gMh_0}{RT} \right]dh$$ $$\rho_{\text{avg}}=\frac{1}{150,000,000,000-h_0}\int_{h_0}^{150,000,000,000}\rho_0 \exp \left[\frac{g_0Mh_0}{RT}-\frac{g_0Mh}{RT} \right]dh$$ Using $u$-substitution: $$u= \to du=-\frac{g_0M}{RT}dh$$ $$\rho_{\text{avg}}=-\frac{g_0M}{RT}\frac{1}{150,000,000,000}\int_{h_0}^{150,000,000,000}\rho_0 \exp u du$$ $$\rho_{\text{avg}}=-\frac{g_0M}{RT}\frac{1}{150,000,000,000} \rho_0 \left[\exp u \right]_{h_0}^{150,000,000,000}$$ $$\rho_{\text{avg}}=-\frac{g_0M}{RT}\frac{1}{150,000,000,000} \rho_0 \left[\exp \left[ \frac{g_0Mh_0}{RT}-\frac{g_0Mh}{RT} \right] \right]_{h_0}^{150,000,000,000}$$ $h_0=0$, so $$\rho_{\text{avg}}=-\frac{g_0M}{RT}\frac{1}{150,000,000,000} \rho_0 \left[\exp \left[ -\frac{g_0Mh}{RT} \right] \right]_0^{150,000,000,000}$$ Substituting in some variables, I get $$\rho_{\text{avg}}=-\frac{9.81 \times 0.0289644 }{8.31432 T}\frac{1}{150,000,000,000} \rho_0 \left[\exp \left[ -\frac{9.81 \times 0.0289644 h}{8.31432 T} \right] \right]_0^{150,000,000,000}$$ The tricky bit is temperature, which is not a function of altitude. Even worse, I've estimated that the lapse rate is zero! Okay, well, it's an approximation, so I'll say that $T$ is Earth's average air temperature - 288 Kelvin, according to Wikipedia. Plugging that in, I get $$\rho_{\text{avg}}=-\frac{9.81 \times 0.0289644 }{8.31432 \times 288}\frac{1}{150,000,000,000} \rho_0 \left[\exp \left[ -\frac{9.81 \times 0.0289644 h}{8.31432 \times 288} \right] \right]_0^{150,000,000,000}$$ $$\rho_{\text{avg}}=-7.91 \times 10^{-16} \rho_0 \left[\exp \left[ -\frac{h}{1.19 \times 10^4} \right] \right]_0^{150,000,000,000}$$ At $h=0$, $p_0=1.225$, so $$\rho_{\text{avg}}=-9.69 \times 10^{-16} \left[\exp \left[ -\frac{h}{1.19 \times 10^4} \right] \right]_0^{150,000,000,000}$$ $$\rho_{\text{avg}} \approx 9.69 \times 10^{-16}$$ Pretty low, right? But let's find the mass. given that $r=150,000,000,000$, we get $$M=V \rho_{\text{avg}}$$ $$M=1.37 \times 10^{19} \text{ kilograms}$$ That's only about double the current mass of our atmosphere.

Unless I made an error somewhere - and I very well might have - the gravitational effects on the Solar System will be about nil. Nothing will happen to the Earth, the Sun, the Moon, or anything else. My result conflicts with that of 2012rcampion, though that assumed constant density. See the power of $e^{-x}$ (pun intended)?

The giant extended atmosphere, though, will most likely leave Earth. There's a reason Earth isn't a gas giant - it isn't massive enough. Oxygen molecules at such great distances from Earth can easily fall victim to atmospheric escape, and will soon leave. This envelope would be depleted even if Earth was all by itself.

But it isn't all by itself. There are other nearby bodies, and they would accrete some gas - potentially. Venus and Mercury could move through the sphere, as could Mars. They might each take up some of the oxygen. Most, though, would go to the Sun, or simply float around.

The oxygen wouldn't burn, because a combustion reaction couldn't happen! A typical combustion reaction is $$\text{Hydrocarbon}+x\text{O}_2 \to y\text{CO}_2+z\text{H}_2\text{O}$$ The problem is, there's no fuel! Pure oxygen atmospheres are prone to fire - you only have to look as far as Apollo 1 - but you still need fuel. You'll be hard-pressed to find that. Even near the Sun, where there's a lot of heat, you still won't find much fuel, because while the Sun is rich in hydrogen and helium, there's not a lot of carbon - at least, not relative to the other elements.

Summary

First off, would the atmosphere grow, or would it just become much more dense?

Whoops; I kind of assumed that the extra oxygen was instantaneously placed there. Well, as the others said, it would be extremely difficult to create. So it's rather unfeasible without a bit of handwaving.

Given the circumstances, if the atmosphere grows, could it eventually reach the Sun?

No, because, as I said earlier, it would easily escape from Earth.

What will happen once it gets close enough? Would the Sun burn up all the oxygen leading to Earth, then burn everything on Earth and Earth itself?

Perhaps some will burn, but not a lot. You simply don't have the fuel needed.

posted over 9 years ago

HDE 226868‭

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