Question

Our Sun shines bright with a luminosity of 3.828x10^26 Watt. Her energy is responsible for many processes and the habitable temperatures on the Earth that make our life possible.
(a) Calculate the amount of energy arriving on the Earth in a single day
(b) To how many liters of heating oil ( energy density: 37.3x10^6 J/litre) is this equivalent?
(c) The Earth reflects 30% of this energy: Determine the temperature on Earth’s surface.
(d) What other factors should be considered to get an even more precise temperature estimate?
NOTE: The Earth’s radius is 6730km, the Sun’s radius is 696x10^3 km, 1AU is 1.495x10^8

Answer 1

Answer:

(a) 1317.44 W/m²

(b) 1.74×10¹⁵ litres of heating oil

(c) -20.63°C

(d) Energy storage in the Earth and the air

Explanation:

The parameters given are;

Luminosity of the Sun = 3.828 × 10²⁶ Watt

Distance of the Earth from the Sun, d = 152.06 × 10⁶ km

The radius of the Sun = 696 × 10³ km

The radius of the Earth, $r_E$ = 6730 km

The surface area of the Sun = 12000 × Surface area of the Earth

The surface area of the Sun = 6.09 × 10¹² km²

Cross sectional area of the Earth = 1.27 × 10⁴ m² = 0.0127 km²

By the inverse square law, we have;

$R = \dfrac{Luminosity \, of \, the \, Sun}{4 \pi d^2}$

Where:

R = Solar radiation reaching the Earth

Therefore;

$R = \dfrac{3.828 \times 10^{26}}{4 \times \pi \times (1.5206 \times 10^{11})^2} = 1317.44 \ W/m^2$

Hence, the energy, E, reaching the Earth in a day is given as follows;

E = R × 4×π×$r_E$²×60×60×24 = 1317.44 × 4 × π × 6730000² × 60×60×24

E = 6.479×10²² Joules

(b) The number of litres of heating oil is therefore;

6.479×10²² J ÷ (37.3×10⁶ J/litre) = 1.74×10¹⁵ litres of heating oil

(c) 30% of the Energy is reflected, therefore;

0.7 × 6.479×10²² Joules = 4.54×10²² Joules reaches the Earths surface

From Stefan-Boltzmann law, we have;

$T = \left (\dfrac{\left (1 - \alpha \right ) \times R }{4\times \sigma } \right )^{\dfrac{1}{4}}$

Where:

α = 0.3

σ = Stefan-Boltzmann constant = 5.6704×10⁻⁸ W/(m²·K⁴)

Therefore;

$T = \left (\dfrac{\left (1 - 0.3 \right ) \times 1317.44 }{4\times 5.6704 \times 10^{-8} } \right )^{\dfrac{1}{4}} = 252.52 \ K = -20.63 ^{\circ} C$

(d) The heat storage in the Earth and the air