I recently tried to derive an expression for the total power received by the earth from the sun, using integration. However, I am stuck at an integration step. Would appreciate if anyone could help with this.
Consider the Earth located at a distance of Ro from the sun. Consider the sun as a point source of power, emitting power of $P_0$.
Intensity at every point on the surface of the earth is not the same, as each point on the earth is located at a different distance from the sun. Only concentric rings are located the same distance from the sun, and thus will capture the same intensity.
Intensity expression:
Let the radius of a concentric ring be $R$. Let the depth of the concentric ring from the surface of earth be $x$.
Intensity, $$I = \frac{P_0}{4\pi[(R_0 + x)^2 + R^2)]} \tag{1}$$
Finding an expression for $x$ in terms of $R$:
$$R_e^2 = R^2 + (R_e – x)^2,$$ where $R_e$ is the radius of the Earth
$$x = R_e – \sqrt {R_e^2 – R^2} \tag{2}$$
Subst (2) into (1): $$I = \frac{P_0}{4\pi[(R_0 + R_e – \sqrt {R_e^2 – R^2})^2 + R^2]} \tag{3}$$
Area expression:
Next, I find an expression for area of an infinitesimally thin concentric ring, $dA$. I can slice the concentric ring once, to make it into a rectangle.
Thus, area of concentric ring, $$dA = 2\pi R\ dR \tag{4}$$
Power collected integration expression:
$$\begin{align}
&\text{Power collected on earth} \\
&= \int I\ dA \\
&= \int_0^{R_e} \frac{P_0}{4\pi[(R_0 + R_e – \sqrt {R_e^2 – R^2})^2 + R^2]} 2\pi R\ dR \\
&= \frac{P_0}{2} \int_0^{R_e} \frac{R}{ 2R^2 – 2(R_0+R_e)\sqrt{R_e^2 – R^2} + R_e^2 + (R_0 + R_e)^2}\ dR
\end{align}$$
I am now stuck, don't know how to integrate this further. The variable is $R$. Appreciate any insights or ways to proceed. Please help.
Best Answer
I know you are trying to solve the problem by integrating. But you can avoid the bothersome integration altogether by using the concept of solid angle.
(image from Wikipedia - Solid angle)
Viewed from the sun, the earth spans a solid angle $\Omega$. And the whole sphere around the sun has the solid angle $4\pi$. Thus the power $P$ received by the earth is the fraction of the total power $P_0$ emitted into the solid angle $\Omega$: $$P=\frac{P_0\Omega}{4\pi} \tag{1}$$
So the remaining work is to calculate $\Omega$.
Viewed from the sun, the earth appears with an half angle $\theta$.
From the right triangle we see $$\sin\theta=\frac{R_e}{R_0} \tag{2}$$
The solid angle of a circular cone (with half angle $\theta$) is known to be $\Omega=2\pi(1-\cos\theta)$ (see Wikipedia - Solid angle - Solid angles for common objects). So with using (2) we get $$\Omega=2\pi(1-\cos\theta) =2\pi\left(1-\sqrt{1-\sin^2\theta}\right) =2\pi\left(1-\sqrt{1-\frac{R_e^2}{R_0^2}}\right) \tag{3}$$
Using (1) and (3) we finally get the power $$P=\frac{P_0\Omega}{4\pi} =\frac{P_0}{2}\left(1-\sqrt{1-\frac{R_e^2}{R_0^2}}\right).$$
For $R_e \ll R_0$ this can be approximated by $$P\approx \frac{P_0R_e^2}{4R_0^2}.$$