Question

Calculate the fraction of the total emitted sunlight that reaches Earth. (Hint: Imagine a sphere around the Sun the size of Earth’s orbit (area = 4πa2), then calculate the fraction of that area taken up by the disk of Earth (area = πr2planet). Earth's reflectivity is 29%)

Answer 1

Answer:

1.305 * 10⁻¹⁰

Explanation:

The earth disk is circular, hence the area of the earth disk is given as:

$Area\ of\ earth\ disk=\pi r_{planet}^2\\\\but\ r_{planet}=6371\ km=6371000\ m\\\\Area\ of\ earth\ disk=\pi (6371000)^2\\\\a=1.5*10^8\ km=1.5*10^{11}\ m\\\\Area\ of\ earth\ orbit=4\pi a^2=4\pi (1.5*10^{11})^2.\\\\Therefore:\\\\\frac{Area\ of\ earth\ disk}{Area\ of\ earth\ orbit}=\frac{\pi (6371000)^2}{4\pi (1.5*10^{11})^2} =4.5*10^{-10}$

This ratio represents the fraction that would be blocked if the planet did not reflect any light, hence:

fraction of the total emitted sunlight reflected by earth = 29% of 4.5 * 10⁻¹⁰ = 0.29 * 4.5 * 10⁻¹⁰ = 1.305 * 10⁻¹⁰

Answer 2

The fraction of the total emitted sunlight that reaches Earth is  1.81 e-16

For one, we know that the earth disk is circular. The area of the earth disk is given as:

πr², with r being the radius of the earth, 6371km

Area of the earth's disk = 3.142 * (6371000²)

Area of the earth's disk = 3.142 * 40589641000000

Area of the earth's disk = 127,532,652,022,000

Area of the earth's disk = 127.5 e+12 m

The area that the earth orbit is given as 4πa², where a = area of the earth

Area of the earth's orbit = 4 * 3.142 * (127.5 e+12)²

If the area of the earths disk is ratio to area of the earths orbit, then we have

$\frac{Area of the earth's disk}{Area of the earth's orbit}$ = $\frac{127.5 e+12}{ 4 * 3.142 * 127.5 e+12}$

$\frac{Area of the earth's disk}{Area of the earth's orbit}$ = π6371000²/4π(127.5 e+12)²

$\frac{Area of the earth's disk}{Area of the earth's orbit}$ = 624.2 e-18

If you then factor the earth's reflectivity into it, we have

0.29 * 624.2 e-18 = 181 e-18 or 1.81 e-16

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