Question

Our Sun shines bright with a luminosity of 3.828 x 1026 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 litres of heating oil (energy density: 37.3 x 106
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 6370 km; the Sun’s radius is 696 x 103 km; 1 AU is 1.495 x 108 km​

Answer 1

Our Sun shines bright with a luminosity of 3.828 x 1026 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 litres of heating oil (energy density: 37.3 x 106

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 6370 km; the Sun’s radius is 696 x 103 km; 1 AU is 1.495 x 108 km​

Explanation:

luminosity of 3.828 x 1026 Watt.

Watt = J/s

S= L/ 4πd²

S = Amount of solar radiation

L= solar luminosity

d = distance between the Sun and the Earth

S=  3.828 x 1026 / 4* 3.14 * (1.495*10¹¹)²

S= 1363.64 j/s.m²

Total energy incident on the Earth will be given by

E = S *  Earth's surface area *  total number of seconds in a day

E = S * 4πr² * 86400

r = radius of earth

When we substitute the values we get:

E = 1363.64 * 4* 3.14* (6.37*10⁶)²  * 86400

E = 6 * 10²² J

∴ Amount of energy coming to earth everyday is 6 * 10²² J

Answer 2

Given that,

Luminosity of sun = 3.828\times10^{26}\ Watt[/tex]

We need to calculate the solar radiation

Using the inverse square law

$S=\dfrac{L}{4\pi d^2}$

Put the value into the formula

$S=\dfrac{3.828\times10^{26}}{4\pi\times(1.495\times10^{11})^2}$

$S=1362.94\ J/s.m^2$

(a). We need to calculate the amount of energy arriving on the Earth in a single day

Using formula of energy

$E=S\times 4\pi r^2\times86400$

Put the value into the formula

$E=1362.94\times4\pi\times(6.37\times10^6)^2\times86400$

$E=6\times10^{22}\ J$

(b). Energy density of heating oil $\rho=37.3\times10^{6}\ J/litre$

We need to calculate the heating oil

Using formula of heating oil

$heating\ oil=\dfrac{E}{\rho}$

Put the value into the formula

$heating\ oil=\dfrac{6\times10^{22}}{37.3\times10^{6}}$

$heating\ oil=1.60\times10^{15}\ Litre$

(c). The Earth reflects 30% of this energy.

$a=0.3$

We need to calculate the temperature on Earth’s surface

Using formula of temperature

$T=\sqrt[4]{\dfrac{S(1-a)}{4\sigma}}$

Put the value into the formula

$T=\sqrt[4]{\dfrac{1362.94(1-0.3)}{4\times5.67\times10^{-8}}}$

$T=254.70\ K$

(d). The temperature is 254.70 K is very low it is not actual temperature of the earth surface.

We also have to take into account the greenhouse effect,

the emissivity of the atmosphere, the absorptivity of the atmosphere and the surface of the Earth.

By taking into account all of these factors,we can get an accurate temperature estimate.

Hence, (a). The amount of energy arriving on the Earth in a single day is $6\times10^{22}\ J$

(b). The heating oil is $1.60\times10^{15}\ Litre$

(c). The temperature on Earth’s surface is 254.70 K