Question

Earth travels around the Sun in an elliptical orbit. Because of its elliptical orbit, Earth is not the same distance from the Sun at all times. When Earth is closer to the Sun, the amount of solar power that arrives at Earth's surface is greater. Total solar irradiance, also referred to as solar intensity, is defined as the total solar power received by a unit surface area and is measured in units of watts per square meter. The Sun's power output is 3.86 x 1026 W. The distance from Earth to the Sun is 0.9813 au when Earth is the closest, at perihelion, and 1.0160 au when Earth is the farthest, at aphelion. Calculate the average total solar irradiance Itop at the top of Earth's atmosphere. Express your answer to 4 significant figures. Itop = W/m²

Answer 1

The intensity of the radiation from the Sun reduces as the distance from the Sun increases.

• The solar irradiance at the top of Earth's atmosphere is approximately 1,376.22 W/m².

Reasons:

The solar flux density is given by the formula;

$\displaystyle S_d = \frac{L}{ \left(4 \cdot \pi \cdot d^2 \right)}$

L = 3.86 × 10²⁶ W

The average distance from the Earth to the Sun is found as follows;

$d_{ave } = \displaystyle \frac{0.9813 \ au + 1.0160 \ au}{2} = 0.99865 \ au$

1 au = 1.496 × 10⁸ km

0.99865 au = 0.99865 × 1.496 × 10⁸ km = 149,398,040 km

$\displaystyle S_d = \frac{3.86 \times 10^{26}}{ \left(4 \times \pi \times (1.49398040 \times 10^{11})^2 \right)} \approx 1,376.22$

The solar flux density reaching Earth, $S_d$ = The average total solar irradiance

$I_{top}$ at the top of Earth's atmosphere ≈ 1,376.22 W/m².

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