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This problem is from the book by Arthur Beiser, Concepts of Modern Physics, example 1.7.

Solar energy reaches the earth at the rate of about $1.4$ kW per square meter of surface perpendicular to the direction of the sun. By how much does the mass of the sun decrease per second owing to this energy loss? The mean radius of the earth’s orbit is $1.5 \times 10^{11}$ m.

The solution provided by the book uses the following

$$P=\frac{P}{A}\times A = \frac{P}{A}\times 4\pi r^2=(1.4\ \text{kW})\times 4\pi (R_{\text{earth}})^2$$

Once we get the power, one can simply convert that amount of energy loss per time into mass loss per time, through the identity $m=E/c^2$.

But since Earth only gets solar radiation on its half of the surface, the area has to be $A=2\pi r^2$, isn't it?

Why one should use a full area of the Earth?

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Note that value you are given in the question for $R_{earth}$ is the radius of the Earth's orbit, not the radius of the Earth itself. The expression $\frac P A \times 4 \pi r^2$ is calculating the power radiated by the sun across the area of a sphere as large as the Earth's whole orbit, not just the area of the Earth.

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  • $\begingroup$ Thank you very much. I just totally misunderstood. $\endgroup$
    – MrTanorus
    May 2 '20 at 11:46

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