How Cold Will Earth's Equator Be at a Martian Distance?

Question

At 93 million miles from the sun, Earth is in a perfect place for liquid water to exist, which is crucial for the evolution of life. Currently, it averages in at 14 degrees Celsius (57 degrees Fahrenheit).

Its most famous neighbor, Mars, couldn't be any more different. Orbiting the sun from a distance of almost 130 million miles, it is colder and, just as importantly, dimmer. Indeed, Martian sunlight is only 44% as bright as Terran sunlight.

In an alternate universe, Earth orbits the sun from the same distance as Mars and has the same length of day as Mars--25 hours compared to our 24, but extra luminosity is crucial to the question--but other than that, it is the exact same--same mass, same diameter, same axial tilt, same atmospheric thickness (the last part is also crucial to the question.) In that respect, how cold will that extra distance and extra hour of the rotation make Earth's equator, the warmest latitude (due to the sun's light being 100% direct)?

Answer 1

We can use the concept of effective temperature to make an estimate.

Effective temperature is the theoretical temperature a body will have without an atmosphere. Given we're just moving Earth we can simplify the formula to this :

$$\frac {T_1} { T_2} = \sqrt { \frac {R_2}{R_1} }$$

The temperatures should be in Kelvin, and the distances can be in any unit you like.

For Earth we have for the effective temperature $T_1=252^\circ K$ and $R_1=1 AU$. Mars orbits between $1.4 AU$ and $1.7 AU$.

So the "new" Earth ("Mearth" ?) effective temperature would be between $193^\circ K$ and $212^\circ K$.

To allow for the effective of Earth's atmosphere we'll just take an easy estimate based on the percentage difference between Earth effective mean temperature and actual mean temperature, which is about $287^\circ K$. So we get a boost of about $14$ % from our atmosphere (this is why we should treat it with respect, kiddies !).

Apply this to "Mearth" we get a mean temperature between about $218^\circ K$ and $242^\circ K$ or $-55^\circ C$ and $-31^\circ C$ depending on where in the Martian orbit it is.

And that's the hot part !

For reference Mars has an average temperature of about $-63^\circ C$ and "Mearth" would be be about $-43^\circ C$

Answer 2

There are a whole lot of second-order effects that will make this answer wrong (e.g., cooling down the planet as a whole does not result in uniform cooling everywhere--weather patterns change, and heat distribution changes with them).

However, as a first approximation, assuming greenhouse heating is unaffected, with the only practical difference being the reduction in solar constant (the extra hour in a day does not appreciably change thermal emission characteristics), it should be between 40 and 50 degrees Celsius cooler.

Average afternoon temperatures at low altitudes along Earth's equator are around 31C--so at Mars's distance, they'd be reduced, as a first order approximation, to between -10 and -20C. That means a lot of freezing, which will change the Earth's albedo and make things even colder.

If you want the Earth to remain habitable to surface life recognizably similar to what we have now, you will need to massively increase greenhouse heating. The easiest way to do that is to massively increase the amount of CO2 in the atmosphere, which will of course make the Earth uninhabitable to us, without major genetic engineering.

Answer 3

As noted by @a4android in the comments, a hypothetical Earth in a Mars-like orbit probably would have developed much differently (especially with respect to biological activity). I don’t think that’s what you’re curious about, so in this answer I’ll consider what would happen if modern Earth was transplanted into Mars’ orbit.

The sun emits radiation radially outward, and we can imagine a spherical “shell” of radiation emitted at any particular moment inflating with the sun at its center. At a distance $R$, the radiation emitted from the sun is distributed over a sphere with surface area $4\pi R^2$. Since the same amount of radiation is spread over a greater and greater area as it moves away from the sun, intensity decreases proportionally to $1/R^2$. Therefore, since moving Earth into Mars’ orbit increases its distance from the sun by a factor of about $1.4$, the intensity of sunlight striking its surface is multiplied by a factor of $1/1.4^2\approx 0.51$.

If $T$ represents temperature in Kelvin, then light intensity is proportional to $T^4$, meaning that multiplying intensity by $0.51$ results in $T$ being multiplied by $(0.51)^{1/4}\approx 0.85$. Thus, the immediate result of the distance increase will be an approximate $15\%$ decrease in average surface temperature.

However, it will actually be a lot more severe than that, because of two self-reinforcing climatological cooling loops:

• Snow (often) has a high albedo. Regions that receive more snow as a result of cooling will also reflect more of the sun’s radiation energy. This means that the overall proportion of solar energy absorbed decreases, causing greater overall cooling.
• Water is a greenhouse gas. At lower temperatures, more water will be liquid or solid. This means less water in the atmosphere to help trap heat, causing even greater planetary cooling.

This will probably end up being much more harsh than you were hoping. If so, you might consider either increasing the sun’s luminosity (if distance from the sun is crucial in your alternate reality) or putting Earth a bit closer (if a slight but manageable decrease in global temperature is what you’re after).