What would happen to Earth's temperatures without an atmosphere?

[Darkmisc]

Hi everyone

I've read that if Earth had no atmosphere it'd be completely frozen over with temperatures around 255K. Why is this the case when the moon has daytime temperatures of 373K (and night time temperatures of 100K)? Why wouldn't the oceans thaw during the day? Are they just too big to do that?

Thanks

 

[Cutter Ketch]

Hi everyone

I've read that if Earth had no atmosphere it'd be completely frozen over with temperatures around 255K. Why is this the case when the moon has daytime temperatures of 373K (and night time temperatures of 100K)? Why wouldn't the oceans thaw during the day? Are they just too big to do that?

Thanks

It’s hard to comment without seeing the context of what you were reading. I think your observation regarding the moon is spot on. The side of the Earth facing the sun would get hot and the side away would get cold. The Earth rotates once per day rather than every 28 days so the excursions from the mean would be less. However the daily swings would be huge. You will freeze to death at night. You would roast in the day time.

There would be no oceans. The would evaporate.

 

[anorlunda]

I've read that

On PF we always try to give the links. Where did you read that?

 

[russ_watters]

I've read that if Earth had no atmosphere it'd be completely frozen over with temperatures around 255K. Why is this the case when the moon has daytime temperatures of 373K (and night time temperatures of 100K)?

What's the average of 373 and 100?

Note that the lunar day is 28x as long as the Earth day, which helps enable the temperature extremes.

Why wouldn't the oceans thaw during the day? Are they just too big to do that?

Way, way, way too big. It takes all day just to thaw a big steak on my counter!

 

[Darkmisc]

https://www.lpl.arizona.edu/~showman/greenhouse.html

255K was obtained by using the following formula:

T = (F/&sigma)^1/4.

Plugging in F=240 Watts/meter^2 and &sigma=5.67 x 10^-8 Watts/meter^2 Kelvin^4, we find that T=255 K, which corresponds to a temperature of -18oC or 0oF. (Sorry, the indices and superscript didn't copy and paste properly). 240 Watts/meter^2 is the amount of sunlight absorbed. 5.67 x 10^-8 Watts/meter^2 Kelvin^4 is the Stephan-Boltzmann constant.

Thanks. I didn't know a lunar day was 28 days.

The average of 373 and 100 is close to 255K, but it looks like the formula only applies when the Earth is receiving sunlight. If that's true, why do we get 255K rather than something like 373K, given the moon and Earth are roughly the same distance from the sun?

 

[russ_watters]

The average of 373 and 100 is close to 255K, but it looks like the formula only applies when the Earth is receiving sunlight. If that's true, why do we get 255K rather than something like 373K, given the moon and Earth are roughly the same distance from the sun?

240 w/m^2 is the average over the entire globe, including the part that isn't receiving sunlight.

 

[Darkmisc]

Cheers. Thanks