# Increasing Earth's orbital radius to stop global warming

Followup from Dropping ice in the ocean to stop global warming as promised.

In the Futurama episode "Crimes of the Hot", All earth robots fire their engines at the same time at the Galapagos islands. Earth's orbital radius is increased by enough to make Earth's orbital period 372 days instead of 365.256 days (rounded because if you're going to move Earth, might as well get rid of leap days as well), as a method to cool the earth and stop global warming.

How workable is this particular solution? Wondering both about the practicality and effectiveness.

• Don't forget that if you went to move the Earth back a little, you would also have to coordinate the movement of the Moon as well. Feb 23, 2017 at 14:33
• Read Brandon Sanderson's Mistborn series. (Spoiler alert: This did not work out very well for the Lord Ruler.) Feb 23, 2017 at 18:44

Effectiveness 10/10. Move further from the sun, you get less incoming solar radiation, so you cool the planet.

Practicality 0/10. Planets are heavy.

More specifically, the energy needed to increase the earth's orbit would be absolutely astronomical (all puns intended). The solution used in Futurama as you describe it is also utterly impractical. The atmosphere would stop the engines from doing anything except generating more heat into the atmosphere (the exhaust has to actually reach escape velocity and leave the planet for it to change the orbit) and the amount of mass you would need to throw out is extreme.

You would also need to fire your engines once per day or have some sort of very fast moving gantry for them since otherwise the spin of the planet would cause it to cancel itself out.

Also if you wanted a circular orbit instead of an elliptical one you would need to do two course corrections. One to switch into an elliptical orbit to rise away from the sun then another to circularize that orbit once you were at the desired distance.

• If you just want to reduce solar radiation, placing orbital solar mirrors/shades are much easier than moving the planet's orbit. Just bounce part of the radiation away. Of course, this only solves the temperature problem and not the ocean acidification issues. Feb 23, 2017 at 11:46
• Technically planets are not "heavy." Planets contain a lot of mass (compared to the objects typically found on the planet's surface). However, a planet's weight is relative to the acceleration of gravity on its mass. So, technically, the "weight" of the planet would change as you move it (though the mass would stay the same). Just me nitpicking. Feb 23, 2017 at 14:30
• @CharlesCaldwell Actually if you really want to nit pick the planet is experiencing the gravitational pull of the star...so it is heavy. Feb 23, 2017 at 16:04
• @TimB Hence why I said the weight would change as you move the planet since the gravitational force on the planet would change with the distance. It wouldn't change by much but it would change. Feb 23, 2017 at 16:08
• Indeed. It would also change by the weight of the reaction mass ejected to change the orbit. It would still be heavy though. I wasn't going for precision, I was going for snappy punchline :p Feb 23, 2017 at 16:10

# There are many reasons this is a bad idea

Distance of Earth at 372 day rotation

First we have to find the radius of our new 372 day orbit. By Kepler's third law, orbital period is $$T = 2\pi\sqrt{\frac{a^3}{\mu}}$$ where $\mu$ is the standard gravitational parameter of the sun ($1.327\times10^{20} \text{ m}^3/\text{s}^2$), $a$ is the semi-major axis of the Earth's orbit ($1.496\times10^{11} \text{ m}$, at least for now), and $T$ is in seconds. If we plug in the numbers above we get $T = 31560349$ in seconds, or 365.28 days. Pretty good!

Lets set $T = 32140800$ for 372 days and solve for $a$. We get $1.514\times10^{11}$ meters, a marginal increase.

Temperature drop due to this distance

Now lets find out how hot this makes our planet. The effective temperature of a planet at a certain distance from the sun is given by $$T = \left(\frac{L(1-A)}{16\pi\sigma a^2}\right)^{1/4}$$ where $L$ is the luminosity of the sun ($3.828\times10^{26} \text{W}$), $a$ is the distance to the planet, $A$ is the albedo of the planet (0.3), and $\sigma$ is the Stefan-Boltzman constant, $5.67\times10^{-8} \text{ J /(K}^{4}\text{m}^2\text{s}$).

If we plug in the characteristics of Earth, we get 254.59 K. Not exactly super-accurate, mostly because there already is a good greenhouse effect keeping some of that heat in. However, if we calculate the temperature at our now longer orbital distance, we get 253.07 K. This at least gives us an estimate of the temperature delta we are looking at: if we can move the Earth out to this longer orbit then we might drop global temperatures by about 1.5 K. Since warming vs. pre-industrial is already around the 1 K range, this isn't that effective in the first place.

Energy to move the Earth's orbit

But wait! We still have to have spent the energy to move the Earth in the first place! Specific orbital energy can be calculated by $$E = \frac{\mu}{2a}$$ with symbols as above and ignoring the sign. For our current orbit we then have a specific energy of $4.435\times10^{8}\text{ m}^2\text{/s}^2$. For the farther orbit we have $4.382\times10^{8}$ for a difference of $5.273\times10^{6}\text{ m}^2\text{/s}^2$. Last we multiply this number by the mass to the Earth ($5.972\times10^{24} \text{ kg}$) to find that we need $3.15\times10^{31} \text{ J}$ to move the Earth to this new orbit.

Using the world's greatest internet page we find that this is equal to the sun's daily energy output, or about a million years worth of Solar energy on the surface of the Earth. This is going to be tough.

Even if we did move the Earth...

Now, assuming that we move the Earth, there will be some waste heat in this process transferred to the Earth. Lets make the very unreasonably low assumption that only 1% of the energy expended to move the Earth is transferred to the Earth as waste heat. That means of the $3.15\times10^{31} \text{ J}$ needed to move the Earth, $3.15\times10^{29} \text{ J}$ will be delivered to Earth's atmosphere and hydrosphere. This much energy is enough to raise the oceans (mass = $1.4\times10^{21}\text{ kg}$; specific heat $3850 \text{ J/kg}\cdot\text{K}$) by about 6 million degrees Kelvin.

Put another way, the waste heat from moving the earth over the course of 11 years is enough to raise the temperature of the oceans by 1 degree K every minute for the entire 11 years. That is a heavy price to pay for 1.5 K of cooling.

# Conclusion

This is not a thing that will work.

• Something went wrong with one of your formulas. Feb 24, 2017 at 9:07
• I think you have a typo explaining σ: "σ is the emissivity of the planet (0.612), and σ is the" (upvoted anyway, because this is a good answer and that's a small typo) Mar 1, 2017 at 9:38

It's a complicate loop the one you are trying to change...

Higher distance from the Sun means less power reaching the surface, and that seems good. But less power reaching the surface means also less efficient photosynthesis (which is the only vast scale known process that can capture CO2 from the atmosphere).

Searching for internet I found this explanation regarding your question:

For millions of years, levels of carbon dioxide have risen and fell between the ice ages. Orbital patterns initiate warm-up periods that cause changes in ocean circulation. These changes cause the carbon dioxide-rich water from the ocean depths to rise to the surface, where carbon dioxide is released as gas returning to the atmosphere. The increase in atmospheric carbon dioxide then creates a further warm-up, and finally, the orbital pattern changes again and the amount of solar heat reaching the earth decreases. (Donna Hesterman, University of Florida)

So yes, it is possible change the orbital pattern of the earth this can "to brake" the climatic change, but this isn't a definitive solution since in the current state of the earth, we can't be known concretely that other consequences would have this fact.

That's why I do not see a feasible solution, if not a simple hypothetical case.