Solar radiation

Storyboard

The origin of the weather is the sun. Its energy reaches the earth by heating in a different way atmosphere and surface creating gradients that are balanced by conduction, convection and winds.

Therefore, the power of the sun must be studied, how it reaches the earth and how it is distributed over the earth's surface.

>Model

ID:(534, 0)



Mechanisms

Iframe

>Top



Code
Concept
Area on earth that captures radiation
Intensity in orbit relative to the sun
Intensity of the sun in orbit
Intensity on the surface of the sun
Planet earth
Power captured by the ground
Radius of the orbit of the earth and the sun
The planets
The sun

Mechanisms

ID:(15660, 0)



The sun

Description

>Top


The source of energy that defines the Earth's climate is the sun.



The key parameters of the sun are:

Parameter Variable Value
Radius $R$ 696342 km
Surface Area $S$ 6.09E+12 km2
Mass $M$ 1.98855E+30 kg
Density $\rho$ 1.408 g/cm2
Surface Temperature $T_s$ 5778 K
Power $P$ 3.846E+26 W
Intensity $I$ 6.24E+7 W/m2

ID:(3078, 0)



Planet earth

Description

>Top


The planet Earth, shown in the following image:



has the following characteristics:

Parameter Symbol Value
Distance from the sun $r$ 1.496E+8 km
Radius $R$ 6371.0 km
Mass $M$ 5.972E+24 kg
Orbit period $T_o$ 365 days
Rotation period $T_r$ 24 hours
Eccentricity $\epsilon$ 0.017
Axis inclination $\phi$ 23.44°

ID:(9990, 0)



The planets

Description

>Top


Below are the images of the different planets, in order: Mercury, Venus, Earth, Mars, Jupiter, Saturn, Uranus, Neptune, and Pluto:



The different planets have a variety of radii, masses, orbital and rotational periods, axial tilts, and distances from the sun, summarized as follows:

Planet Radius* Mass* Distance from Sun* Orbital Period* Rotation Period* Eccentricity Axial Tilt
Mercury 0.382 0.06 0.39 0.24 58.64 0.206 0.04°
Venus 0.949 0.82 0.72 0.62 -243.02 0.007 177.36°
Earth 1.000 1.00 1.00 1.00 1.00 0.017 23.44°
Mars 0.532 0.11 1.52 1.88 1.03 0.093 25.19°
Jupiter 11.209 317.8 5.2 11.86 0.41 0.048 3.13°
Saturn 9.449 95.2 9.54 29.46 0.43 0.054 26.73°
Uranus 4.007 14.6 19.22 84.01 -0.72 0.047 97.77°
Neptune 3.883 17.2 30.06 164.8 0.67 0.0009 28.32°
Pluto 0.186 0.0022 39.482 247.94 1.005 0.2488 17.16°

* data in proportion to Earth's value

ID:(9991, 0)



Intensity on the surface of the sun

Concept

>Top


The radiation intensity on the sun's surface ($I_s$) is defined as the sun power ($P_s$) per unit of the surface of the sun ($S_s$), where the power is represented by:

$ I_s =\displaystyle\frac{ P_s }{ S_s }$



If we model the sun as a sphere with the sun Radio ($R_s$), its surface area is:

$ S_s = 4 \pi R_s ^2$



Therefore, the radiation intensity on the sun's surface ($I_s$) is calculated as:

$ I_s = \displaystyle\frac{ P_s }{4 \pi R_s ^2}$

ID:(15655, 0)



Intensity of the sun in orbit

Concept

>Top


The intensity at orbit distance ($I_r$) is defined as the sun power ($P_s$) per unit of the sphere surface in orbit ($S_r$):

$ I_r =\displaystyle\frac{ P_s }{ S_r }$



If we consider an imaginary sphere with a radius equal to the distance between the sun and the Earth, sphere surface in orbit ($S_r$), we can calculate its cross-sectional area:

$ S_r = 4 \pi r ^2$



This allows us to obtain the intensity at orbit distance ($I_r$):

$ I_r = \displaystyle\frac{ P_s }{4 \pi r ^2}$

ID:(15657, 0)



Radius of the orbit of the earth and the sun

Description

>Top


The Sun's radiation propagates through its surface, which has an area of $4\pi R_s^2$ with a sun Radio ($R_s$) as the Sun's radius, and it is distributed at the distance of Earth's orbit, which has a surface area equal to $4\pi r^2$ with a distance earth sun ($r$) as the distance between the Earth and the Sun:

None

ID:(3082, 0)



Intensity in orbit relative to the sun

Concept

>Top


If we replace the sun power ($P_s$) of the sun, calculated as the radiation intensity on the sun's surface ($I_s$) on the surface of a sphere with a radius of sun Radio ($R_s$):

$ I_s = \displaystyle\frac{ P_s }{4 \pi R_s ^2}$

,

into the equation for the intensity at orbit distance ($I_r$) of sunlight at the distance earth sun ($r$):

$ I_r = \displaystyle\frac{ P_s }{4 \pi r ^2}$

,

we can obtain the relationship between intensities:

$ I_r =\displaystyle\frac{ R_s ^2}{ r ^2} I_s $

ID:(15658, 0)



Power captured by the ground

Concept

>Top


Given that the intensity at orbit distance ($I_r$) reaching the Earth is equal to the power captured by the earth ($P_d$) captured by the section presenting the planet ($S_d$) according to:

$ I_r =\displaystyle\frac{ P_d }{ S_d }$



and that the section presenting the planet ($S_d$) of the disk of the planet radius ($R_p$) is equal to:

$ S_d = \pi R_p ^2$

,

we have:

$ I_r =\displaystyle\frac{ P_d }{ \pi R_p ^2}$

.

ID:(15659, 0)



Area on earth that captures radiation

Description

>Top


The average earth intensity ($I_p$) over the entire surface of the planet radius ($R_p$) is equal to the intensity at orbit distance ($I_r$) captured by a disk of the planet radius ($R_p$), therefore:

$4\pi R_p^2 I_s = \pi R_p^2 I_p$





Therefore, it follows that:

$ I_r =\displaystyle\frac{1}{4} I_p $

ID:(3084, 0)



Model

Top

>Top



Calculations

Variables

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$I_p$
I_p
Average earth intensity
W/m^2
$I_r$
I_r
Intensity at orbit distance
W/m^2
$S_p$
S_p
Planet surface
m^2
$P_d$
P_d
Power captured by the earth
W
$I_s$
I_s
Radiation intensity on the sun's surface
W/m^2
$S_d$
S_d
Section presenting the planet
m^2
$S_r$
S_r
Sphere surface in orbit
m^2
$P_s$
P_s
Sun power
W
$S_s$
S_s
Surface of the sun
m^2

Parameters

Symbol
Text
Variable
Value
Units
Calculate
MKS Value
MKS Units
$r$
r
Distance earth sun
m
$\pi$
pi
Pi
rad
$R_p$
R_p
Planet radius
m
$R_s$
R_s
Sun Radio
m


First, select the equation: to , then, select the variable: to

Calculations

Symbol
Equation
Solved
Translated

Calculations

Symbol
Equation
Solved
Translated

Variable Given Calculate Target : Equation To be used


Equation

#
Equation

$ I_s =\displaystyle\frac{ P_s }{ S_s }$

I = P / S


$ I_r =\displaystyle\frac{ P_s }{ S_r }$

I = P / S


$ I_r =\displaystyle\frac{ P_d }{ S_d }$

I = P / S


$ I_p =\displaystyle\frac{ P_d }{ S_p }$

I = P / S


$ I_r =\displaystyle\frac{ P_d }{ \pi R_p ^2}$

I = P /( pi * r ^2)


$ I_s = \displaystyle\frac{ P_s }{4 \pi R_s ^2}$

I = P /(4* pi * r ^2)


$ I_r = \displaystyle\frac{ P_s }{4 \pi r ^2}$

I = P /(4* pi * r ^2)


$ I_p = \displaystyle\frac{ P_d }{4 \pi R_p ^2}$

I = P /(4* pi * r ^2)


$ I_s =\displaystyle\frac{ R_s ^2}{ r ^2} I_r $

I_1 = ( r_2 ^2/ r_1 ^2)* I_2


$ I_r =\displaystyle\frac{1}{4} I_p $

I_r = I_p /4


$ S_s = 4 \pi R_s ^2$

S = 4* pi * r ^2


$ S_r = 4 \pi r ^2$

S = 4* pi * r ^2


$ S_p = 4 \pi R_p ^2$

S = 4* pi * r ^2


$ S_d = \pi R_p ^2$

S = pi * r ^2

ID:(15671, 0)



Intensity and power (1)

Equation

>Top, >Model


The intensity ($I$) is defined as the amount of the power ($P$) irradiated per unit of the surface of a sphere ($S$). Therefore, the following relationship is established:

$ I_s =\displaystyle\frac{ P_s }{ S_s }$

$ I =\displaystyle\frac{ P }{ S }$

$I$
$I_s$
Radiation intensity on the sun's surface
$W/m^2$
6493
$P$
$P_s$
Sun power
$W$
6494
$S$
$S_s$
Surface of the sun
$m^2$
6499

ID:(9988, 1)



Intensity and power (2)

Equation

>Top, >Model


The intensity ($I$) is defined as the amount of the power ($P$) irradiated per unit of the surface of a sphere ($S$). Therefore, the following relationship is established:

$ I_r =\displaystyle\frac{ P_s }{ S_r }$

$ I =\displaystyle\frac{ P }{ S }$

$I$
$I_r$
Intensity at orbit distance
$W/m^2$
6495
$P$
$P_s$
Sun power
$W$
6494
$S$
$S_r$
Sphere surface in orbit
$m^2$
10360

ID:(9988, 2)



Intensity and power (3)

Equation

>Top, >Model


The intensity ($I$) is defined as the amount of the power ($P$) irradiated per unit of the surface of a sphere ($S$). Therefore, the following relationship is established:

$ I_r =\displaystyle\frac{ P_d }{ S_d }$

$ I =\displaystyle\frac{ P }{ S }$

$I$
$I_r$
Intensity at orbit distance
$W/m^2$
6495
$P$
$P_d$
Power captured by the earth
$W$
6500
$S$
$S_d$
Section presenting the planet
$m^2$
6700

ID:(9988, 3)



Intensity and power (4)

Equation

>Top, >Model


The intensity ($I$) is defined as the amount of the power ($P$) irradiated per unit of the surface of a sphere ($S$). Therefore, the following relationship is established:

$ I_p =\displaystyle\frac{ P_d }{ S_p }$

$ I =\displaystyle\frac{ P }{ S }$

$I$
$I_p$
Average earth intensity
$W/m^2$
6502
$P$
$P_d$
Power captured by the earth
$W$
6500
$S$
$S_p$
Planet surface
$m^2$
10359

ID:(9988, 4)



Surface of a sphere (1)

Equation

>Top, >Model


The surface of a sphere ($S$) of a radius of a sphere ($r$) can be calculated using the following formula:

$ S_s = 4 \pi R_s ^2$

$ S = 4 \pi r ^2$

$\pi$
Pi
3.1415927
$rad$
5057
$r$
$R_s$
Sun Radio
$m$
6492
$S$
$S_s$
Surface of the sun
$m^2$
6499

ID:(4665, 1)



Surface of a sphere (2)

Equation

>Top, >Model


The surface of a sphere ($S$) of a radius of a sphere ($r$) can be calculated using the following formula:

$ S_r = 4 \pi r ^2$

$ S = 4 \pi r ^2$

$\pi$
Pi
3.1415927
$rad$
5057
$r$
$r$
Distance earth sun
$m$
6490
$S$
$S_r$
Sphere surface in orbit
$m^2$
10360

ID:(4665, 2)



Surface of a sphere (3)

Equation

>Top, >Model


The surface of a sphere ($S$) of a radius of a sphere ($r$) can be calculated using the following formula:

$ S_p = 4 \pi R_p ^2$

$ S = 4 \pi r ^2$

$\pi$
Pi
3.1415927
$rad$
5057
$r$
$R_p$
Planet radius
$m$
6501
$S$
$S_p$
Planet surface
$m^2$
10359

ID:(4665, 3)



Intensity depending on power (1)

Equation

>Top, >Model


The intensity ($I$) is calculated as the power ($P$) divided by the surface area of a sphere with a radio ($r$):

$ I_s = \displaystyle\frac{ P_s }{4 \pi R_s ^2}$

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

$I$
$I_s$
Radiation intensity on the sun's surface
$W/m^2$
6493
$\pi$
Pi
3.1415927
$rad$
5057
$P$
$P_s$
Sun power
$W$
6494
$r$
$R_s$
Sun Radio
$m$
6492

The intensity ($I$) is defined as the power ($P$) per unit of the surface of a sphere ($S$):

$ I =\displaystyle\frac{ P }{ S }$



If we consider an imaginary sphere with distance earth sun ($r$), we can calculate its surface:

$ S = 4 \pi r ^2$



This allows us to obtain the intensity ($I$):

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

ID:(4662, 1)



Intensity depending on power (2)

Equation

>Top, >Model


The intensity ($I$) is calculated as the power ($P$) divided by the surface area of a sphere with a radio ($r$):

$ I_r = \displaystyle\frac{ P_s }{4 \pi r ^2}$

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

$I$
$I_r$
Intensity at orbit distance
$W/m^2$
6495
$\pi$
Pi
3.1415927
$rad$
5057
$P$
$P_s$
Sun power
$W$
6494
$r$
$r$
Distance earth sun
$m$
6490

The intensity ($I$) is defined as the power ($P$) per unit of the surface of a sphere ($S$):

$ I =\displaystyle\frac{ P }{ S }$



If we consider an imaginary sphere with distance earth sun ($r$), we can calculate its surface:

$ S = 4 \pi r ^2$



This allows us to obtain the intensity ($I$):

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

ID:(4662, 2)



Intensity depending on power (3)

Equation

>Top, >Model


The intensity ($I$) is calculated as the power ($P$) divided by the surface area of a sphere with a radio ($r$):

$ I_p = \displaystyle\frac{ P_d }{4 \pi R_p ^2}$

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

$I$
$I_p$
Average earth intensity
$W/m^2$
6502
$\pi$
Pi
3.1415927
$rad$
5057
$P$
$P_d$
Power captured by the earth
$W$
6500
$r$
$R_p$
Planet radius
$m$
6501

The intensity ($I$) is defined as the power ($P$) per unit of the surface of a sphere ($S$):

$ I =\displaystyle\frac{ P }{ S }$



If we consider an imaginary sphere with distance earth sun ($r$), we can calculate its surface:

$ S = 4 \pi r ^2$



This allows us to obtain the intensity ($I$):

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

ID:(4662, 3)



Surface of a disk

Equation

>Top, >Model


The surface of a disk ($S$) of ($$) is calculated as follows:

$ S_d = \pi R_p ^2$

$ S = \pi r ^2$

$r$
$R_p$
Planet radius
$m$
6501
$\pi$
Pi
3.1415927
$rad$
5057
$S$
$S_d$
Section presenting the planet
$m^2$
6700

ID:(3804, 0)



Power captured

Equation

>Top, >Model


The intensity ($I$) is calculated by dividing the power ($P$) by the area of the disk with a radius of the radio ($r$), which means:

$ I_r =\displaystyle\frac{ P_d }{ \pi R_p ^2}$

$ I =\displaystyle\frac{ P }{ \pi r ^2}$

$I$
$I_r$
Intensity at orbit distance
$W/m^2$
6495
$\pi$
Pi
3.1415927
$rad$
5057
$P$
$P_d$
Power captured by the earth
$W$
6500
$r$
$R_p$
Planet radius
$m$
6501

Given that the intensity ($I$) is the power ($P$) captured by the surface of a sphere ($S$) according to:

$ I =\displaystyle\frac{ P }{ S }$



and that the surface of a disk ($S$) is the area of the disk of the disc radius ($r$), which is equal to:

$ S_d = \pi R_p ^2$

,

we have:

$ I =\displaystyle\frac{ P }{ \pi r ^2}$

.

ID:(4666, 0)



Intensity depending on solar intensity

Equation

>Top, >Model


The ratio between the intensity at orbit distance ($I_r$) and the radiation intensity on the sun's surface ($I_s$) is equal to the ratio of the surface area of a sphere with a radius of the sun Radio ($R_s$) to the surface area of a sphere with a radius of the distance earth sun ($r$), therefore it is:

$ I_r =\displaystyle\frac{ R_s ^2}{ r ^2} I_s $

$ I_1 =\displaystyle\frac{ r_2 ^2}{ r_1 ^2} I_2 $

$I_2$
$I_s$
Radiation intensity on the sun's surface
$W/m^2$
6493
$I_1$
$I_r$
Intensity at orbit distance
$W/m^2$
6495
$r_1$
$r$
Distance earth sun
$m$
6490
$r_2$
$R_s$
Sun Radio
$m$
6492

If we replace the sun power ($P_s$) of the sun, calculated as the radiation intensity on the sun's surface ($I_s$) on the surface of a sphere with a radius of sun Radio ($R_s$):

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

,

into the equation for the intensity at orbit distance ($I_r$) of sunlight at the distance earth sun ($r$):

$ I = \displaystyle\frac{ P }{4 \pi r ^2}$

,

we can obtain the relationship between intensities:

$ I_1 =\displaystyle\frac{ r_2 ^2}{ r_1 ^2} I_2 $

ID:(4663, 0)



Average intensity emitted by the earth

Equation

>Top, >Model


The average earth intensity ($I_p$) is equal to one-fourth of the intensity at orbit distance ($I_r$) because the surface area of the emitting sphere is four times larger than that of the capturing disk. Therefore:

$ I_r =\displaystyle\frac{1}{4} I_p $

$I_s$
Average earth intensity
$W/m^2$
6502
$I_p$
Intensity at orbit distance
$W/m^2$
6495

ID:(4667, 0)