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)

Description

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

The key parameters of the sun are:

Parameter | Variable | Value

|:-------------|:------------|:--------:

Radius | $R$ | $696,342 km$

Surface Area | $S$ | $6.09 \times 10^{12} km^2$

Mass | $M$ | $1.98855 \times 10^{30} kg$

Density | $\rho$ | $1.408 g/cm^2$

Surface Temperature | $T_s$ | $5778 K$

Power | $P$ | $3.846 \times 10^{26} W$

Intensity | $I$ | $6.24 \times 10^7 W/m^2$

ID:(3078, 0)

Description

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The Sun emits radiation with a spectrum similar to what is known as the blackbody spectrum. Its maximum irradiation occurs at a wavelength close to the color yellow:

None

ID:(3083, 0)

Equation

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The intensity $I$ is defined as the power $P$ per unit surface area $S$ that is radiated. Therefore, we have the following relationship:

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

Conditions

Variables

$P$

Power

$W$

$S$

Superficie

$m^2$

Relation

ID:(9988, 0)

Object

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The sun is a source of radiation that is emitted symmetrically around its center.

ID:(9990, 0)

Equation

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The intensity is defined as power per unit area, where power is represented by

If we model the Sun as a sphere, its surface area is

and thus the intensity is calculated as

$ P =4 \pi R ^2 I_R $

Conditions

Variables

$\pi$

Pi

3.1415927

$rad$

$P$

Power of the Sun

$W$

$I_R$

Radiation Intensity at the Surface of the Sun

$W/m^2$

$R$

Sun Radio

$m$

Relation

.

ID:(4661, 0)

Equation

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When the particles of a medium oscillate, they act like small antennas emitting radiation. The power of this emitted radiation is related to the temperature of the medium and depends on the oscillation itself. This relationship between the emitted power and the absolute temperature is described by the Stefan-Boltzmann law, which states the following for a surface of area $S$:

$ P = \sigma \epsilon S T_s ^4$

Conditions

Variables

$\epsilon$

Emissivity

$-$

$P$

Power of the Sun

$W$

$\sigma$

Stefan Boltzmann constant

1.38e-23

$J/m^2K^4s$

$S$

Surface of the Sun

$m^2$

$T_s$

Surface Temperature of the Sun

$K$

Relation

where $\sigma$ is the Stefan-Boltzmann constant (with a value of $5.67 \times 10^{-8} W/m^2K^4$), $\epsilon$ is the emissivity, and $T$ is the absolute temperature of the medium.

Emissivity is a constant that indicates how perfectly a surface emits radiation and is closely related to its roughness. A smooth surface has an emissivity close to unity, which means it efficiently emits radiation. On the other hand, a surface with lower emissivity, which can have values between 0.2 and 0.4, has difficulty emitting radiation or may even reabsorb it, reducing the total flux of emitted radiation.

ID:(4664, 0)

Equation

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The surface area of a sphere with radius $r$ can be calculated using the following formula:

$ S = 4 \pi R ^2$

Conditions

Variables

$\pi$

Pi

3.1415927

$rad$

$R$

Sun Radio

$m$

$S$

Surface of the Sun

$m^2$

Relation

ID:(4665, 0)

Equation

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The intensity is defined as the power per unit area

If we consider an imaginary sphere with a radius equal to the distance between the sun and the earth, we can calculate its cross-sectional area

This allows us to obtain the resulting intensity:

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

Conditions

Variables

$r$

Distance Earth Sun

$m$

$I_p$

Intensity at a Distance from Earth in the Solar Power Function

$W/m^2$

$\pi$

Pi

3.1415927

$rad$

$P$

Power of the Sun

$W$

Relation

ID:(4662, 0)

Object

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The planet is at a distance from the sun and absorbs and re-emits the radiation it receives from it.

ID:(9991, 0)

Description

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The radiation from the Sun flows through its surface, which has an area of $4\pi R^2$ with a radius of $R$, and is distributed at the height of the Earth's orbit, which has a surface area of $4\pi r^2$:

None

ID:(3082, 0)

Equation

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If we replace the solar power, calculated as the intensity at its surface

in the equation for the intensity of sunlight at the sun-earth distance

we can obtain the relationship between intensities:

$ I_p =\displaystyle\frac{ R ^2}{ r ^2} I_R $

Conditions

Variables

$r$

Distance Earth Sun

$m$

$I_p$

Intensity at a Distance from Earth in the Solar Power Function

$W/m^2$

$I_R$

Radiation Intensity at the Surface of the Sun

$W/m^2$

$R$

Sun Radio

$m$

Relation

ID:(4663, 0)

Description

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The Earth presents a disk with an area of $\pi r_e^2$ exposed to solar radiation:

ID:(3084, 0)

Equation

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Si I_s es la intensidad media recibida sobre toda la superficie 4\pi R^2 de la esfera de la tierra y I_e es la intensidad solar capturada por el disco terrestre de sección \pi R^2 que presenta la tierra al haz solar, se tiene que por conservación de energía que:\\n\\n

$4\pi R^2 I_s=\pi R^2 I_e$

Por la intensidad media sobre la tierra con es

ID:(4667, 0)

Equation

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Since the Earth can only capture a fraction of the intensity from the Sun-Earth distance,

represented by the surface disk with a fraction of

,

the power captured by the Earth is calculated as

$ P_e = \pi R_e ^2 I_p $

Conditions

Variables

$R_e$

Earth Radio

$m$

$I_p$

Intensity at a Distance from Earth in the Solar Power Function

$W/m^2$

$\pi$

Pi

3.1415927

$rad$

$P_e$

Power Captured by the Earth

$W$

Relation

.

ID:(4666, 0)