# Tag Info

7

How can the air in the Earth's shadow scatter blue light if it doesn't get any direct illumination, and any indirect illumination getting back there should be VERY depleted in blue (i.e. mostly red light)? You're completely right. If you simulate light scattering only, you'll get exactly this result: a sandy-colored sky, with a bit redder belt of Venus, and ...

3

First, let's acknowledge this fact: $$c_p=R+c_v \tag{1}$$, where $R$ is the specific gas constant. This means that $x=R/c_p$. Rearranging the equation, we can see that $$P+I=\int{c_p\left(\frac{P}{P_{00}}\right)^{R/c_p}\Theta dM}\tag{2}$$ Notice that $\left(\frac{P}{P_{00}}\right)^{R/c_p}$ is the Exner function. By extension,$$P+I=\int{c_p T dM}\tag{3}$$ If ...

3

In theory, the methane emissions would warm the earth about 25 times more than the carbon dioxide emissions. That's because of something called the global warming potential. The global warming potential of a greenhouse gas is how much it warms the planet compared to carbon dioxide's base value of 1. As you might expect, methane's GWP is 25. Sources: https://...

2

Any relationship or correlation? That is quite a broad standard. Invoking Tobler's first law of geography, there is a relationship. I would question if there is such a thing universally, but there are some instances that we can see an overlap. For example, many deserts receive plenty of sunlight, but observe little rain. One of the first things that comes to ...

1

The back radiation looks to be calculated by summing all radiation components that the atmosphere (including clouds) absorbs and then subtracting the radiation it emits upwards. What remains is the radiation it emits downwards (i.e. back radiation). In the figure it is indeed:  (67\ \mathrm{Wm^{-2}} + 24\ \mathrm{Wm^{-2}} + 78\ \mathrm{Wm^{-2}} + 350\ \...

1

I highly doubt there's anything at all which would fulfil the criteria you're looking for due to the huge spectral range involved and an overlap between the two emission spectra. You'd need something which doesn't absorb anything between 0.2 µm and 70 µm. Alternatively you'd need something with two absorption bands of equal spread and magnitude sitting on ...

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