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

We're `"150,000,000 km"` apart on average. Earth is only `"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 `"150,000,000 km"`, the surface is the area of that sphere, which is equal to

`"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%`

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=a/R^2`
Total solid angle at the centre of sphere is `4pi.`

Considering Sun to be situated at the centre of sphere whose radius is equal to the average distance between sun and earth, which is `1.496xx10^8km`.

Solid angle subtended by the area of earth exposed to sun is
`Omega_e=(pir_e^2)/R^2` ......(1)
where `r_e` is average radius of earth and is `6.371xx10^3 km`.
Sun radiates energy in all directions. Therefore fraction of energy reaching earth `DeltaE_e` is
`DeltaE_e=Omega_e/(4pi)`
Using (1)
`DeltaE_e=((pir_e^2)/R^2)/(4pi)`
`=>DeltaE_e=(r_e^2)/(4R^2)`
Inserting given values we obtain
`DeltaE_e=(6.371xx10^3)^2/(4xx(1.496xx10^8)^2)`
`DeltaE_e=4.534xx10^-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.)