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Let's suppose the Earth is a black body. Usually, when determining its energy balance, the following approximation is taken into consideration:

$$ \pi R_{Earth} ^2 I_{Sun} = \sigma T_{Earth} ^4 4\pi R_{Earth} ^2$$

However, I am failing to understand why we consider that the effective area of the Earth is a disk and not $ 2 \pi R_{Earth}^2$.

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    $\begingroup$ Because the sun’s radiation isn’t perpendicular to the Earth’s surface over the half of the Earth that is in sunlight. When you take this into account, the flux is reduced by another factor of $1/2$. Try integrating the solar flux over the half-Earth. $\endgroup$
    – G. Smith
    Commented Jun 9, 2019 at 23:41
  • $\begingroup$ I will try, appreciate it $\endgroup$
    – miniplanck
    Commented Jun 9, 2019 at 23:50

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When we talk of black body it is called so when it absorbs all the radiation falling on it. So if we take half the surface area of the earth, this means that the surface reciwving the radiation from the sun must be taken into account, and the other side does not need to be considered as the other side facing planets, does not recieve any compararabke amount of radiation,as planets don't have their own light. As the black body radiates out the heat it absorbs, so, we take the surface facing the sun only to know the radiation power. The temperature taken must also be for the side of the earth facing sun due to uneven heating of earth

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