The moon's days are one month long. At night, the surface rocks have time to cool down, losing all the heat they gained during the day. They get so cold they draw heat out of the moon's interior, and this flow of heat from the depths of the moon is what stops the moon's surface from dropping below −190°C. During the day, the rocks heat up until they start to radiate almost as much heat as is arriving from the Sun. But some heat flows into the depths of the moon to make up for the heat drawn out during the previous night.
The moon has no atmosphere, so the only thing stopping it from cooling down to absolute zero at night is the heat capacity and thermal conductivity of its surface rocks. On Earth, we have the heat capacity of the atmosphere to keep the surface warm during the night. Above each square meter of the Earth's surface is 10,000 kg of air (see Atmospheric Weight). The heat capacity of dry air is is roughly 1 kJ/K/kg. So the air above each square meter of the Earth has heat capacity 10 MJ. The Earth and its atmosphere radiate something like 250 W/m2 into space. Suppose this 250 W/m2 were extracted uniformly from the entire column of air above each square meter of the surface. It would take 10 MJ ÷ 250 W = 10 hrs to cool the air by 1°C.
We see that the heat capacity of the atmosphere is significant in reducing the temperature changes caused by the setting of the sun or the formation of thick clouds. Indeed, the temperature changes we observe from day to night on Earth are more than ten times greater than our simple heat capacity calculation suggests. Perhaps if we consider the individual 3-km layers of the atmosphere, we will come up with a different answer.
As day turns to night, the Sun no longer warms the Earth, but the Earth and its atmosphere continue to radiate heat into space. The following table gives the escaping power in Watts per square meter from each 3-km layer of the Earth's atmosphere up to the tropopause, as calculated by our TEP2 program under conditions we describe here.
--------------------------------------------------The surface of the Earth radiates 390 W/m2. Of this, 80 W/m2 escapes directly into space. The rest of it is absorbed by the atmosphere. But the atmosphere radiates heat towards the surface also. The 0-km layer (the first 3 km of the atmosphere) radiates 260 W/m2 upwards and downwards. We may have thought only of the upward component in the past, but the downward component exists also, and is equal to the upward component.
Name Temp BB Layer Escaping
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Surface 290.0 401.1 388.8 80.0
0-km 280.0 348.5 262.6 46.1
3-km 270.0 301.3 184.8 39.7
6-km 250.0 221.5 111.5 44.6
9-km 230.0 158.7 50.7 26.4
12-km 220.0 132.8 22.2 10.1
15-km 220.0 132.8 14.4 5.5
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Total: 252.5
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Let us ignore the heat passing up through the atmosphere by radiation and convection, and assume that the Earth's surface starts to lose 80 W/m2 when the sun sets. Suppose the surface itself is water 1 m deep. The heat capacity of water is roughly 4 kJ/K/kg and its density is 1,000 kg/m3. Beneath each square meter is fluid with heat capacity 4 MJ/K. This water will cool by only 1 °C in 10 hrs.
The 0-km layer radiates 46 W/m2 directly into space. The mass of air in this first 3-km layer is a little less than 3,000 kg, with heat capacity roughly 3 MJ/K. At a loss of 46 W/m2, this layer will cool down by only 0.5°C in ten hours.
The changes in temperature we observe at the surface of the Earth from day to night are much greater than we would expect from a consideration of the heat capacity of the atmosphere and total escaping power alone. What is going on when the sun sets, that makes the Earth's surface cool down so fast? We will hope to answer this question in our next post.
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