# How much of the total energy that leaves the sun makes it to earth? Why?

Aug 19, 2016

We intercept $\frac{\pi \cdot {d}^{2}}{4} = {\text{113,097,335 km}}^{2}$

#### Explanation:

We're $\text{150,000,000 km}$ apart on average. Earth is only $\text{12,000 km}$ in diameter. The sun emits in all directions.

We subtend a tiny surface when viewed from the sun, with a small fraction in the line of its energy.

At $\text{150,000,000 km}$, the surface is the area of that sphere, which is equal to

${\text{area" = 4 * pi * "150,000,000 km"^2 = 2.82743 * 10^17 "km}}^{2}$

We intercept

(pi*d^2)/4 = "113,097,335 km"^2 = +- 0.00000004%

Aug 19, 2016

To workout the problem we need to understand the concept of solid angle.

Solid angle $\Omega$ subtended by sphere's segment area $a$ at the centre of sphere of radius $R$ is equal to the ratio of $a$ to the square of the sphere's radius $R$.
$\Omega = \frac{a}{R} ^ 2$
Total solid angle at the centre of sphere is $4 \pi .$

Considering Sun to be situated at the centre of sphere whose radius is equal to the average distance between sun and earth, which is $1.496 \times {10}^{8} k m$.

Solid angle subtended by the area of earth exposed to sun is
${\Omega}_{e} = \frac{\pi {r}_{e}^{2}}{R} ^ 2$ ......(1)
where ${r}_{e}$ is average radius of earth and is $6.371 \times {10}^{3} k m$.
Sun radiates energy in all directions. Therefore fraction of energy reaching earth $\Delta {E}_{e}$ is
$\Delta {E}_{e} = {\Omega}_{e} / \left(4 \pi\right)$
Using (1)
$\Delta {E}_{e} = \frac{\frac{\pi {r}_{e}^{2}}{R} ^ 2}{4 \pi}$
$\implies \Delta {E}_{e} = \frac{{r}_{e}^{2}}{4 {R}^{2}}$
Inserting given values we obtain
$\Delta {E}_{e} = {\left(6.371 \times {10}^{3}\right)}^{2} / \left(4 \times {\left(1.496 \times {10}^{8}\right)}^{2}\right)$
$\Delta {E}_{e} = 4.534 \times {10}^{-} 10$

This is a minuscule fraction of total energy radiated by sun. The reason is a very small solid angle ${\Omega}_{e}$.
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(the problem could also have been worked out by calculating ratio of area of earth's surface receiving energy from sun to total area of the sphere of radius $R$. However to understand "why?" this approach has been adopted.)