The Inverse-Square and Stefan-Boltzmann's Laws Recall that the Stefan-Boltzmann law estimates the total amount of emission, per unit area, that is leaving the surface of a blackbody. This is termed the total emittance. The Stefan-Boltzman law is described by the formula Eq (1):                                                              E* = σSB • T 4                 Eq (1) Where E* is the total emittance in W/m2, σSB is the Stefan-Boltzmann constant with a value of 5.67 x 10-8 W/m2•K4 and T is the blackbody temperature in K.   The equation for the inverse-square law is shown below Eq (2).                                                      E*2 = E*1 • (R1 / R2)2                   Eq (2)  Where E*2 is the irradiance at the distance of interest, E*1 is the emission from an emitter (for example, the Sun) or at a reference location (for example, at the orbital distance of a planet) and is equivalent to E* from Eq (1), R1 is the radius of the emitter, and R2 is the distance to the location of interest in meters. Here are some values for the Sun. Note that the following only applies to the Sun, not other objects. E*1 = 62,930,000 W/m2 R1 = 695,700,000 m The nearby blue-white star Fomalhaut has an estimated radius of 1.842 times (184.2%) that of the Sun and a surface temperature of 8590 K. How much irradiance would a planet located around Fomalhaut receive, assuming this world has the same mean orbital distance as the Earth, 149,597,870,700 m? This is a two-step question–start with Eq 1 first and then move on to Eq 2.

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The Inverse-Square and Stefan-Boltzmann's Laws

Recall that the Stefan-Boltzmann law estimates the total amount of emission, per unit area, that is leaving the surface of a blackbody. This is termed the total emittance. The Stefan-Boltzman law is described by the formula Eq (1):

                                                             E* = σSB • 4                 Eq (1)

Where E* is the total emittance in W/m2, σSB is the Stefan-Boltzmann constant with a value of 5.67 x 10-8 W/m2•K4 and T is the blackbody temperature in K.

 

The equation for the inverse-square law is shown below Eq (2).

                                                     E*2 = E*1 • (R1 / R2)2                   Eq (2) 

Where E*2 is the irradiance at the distance of interest, E*1 is the emission from an emitter (for example, the Sun) or at a reference location (for example, at the orbital distance of a planet) and is equivalent to E* from Eq (1)R1 is the radius of the emitter, and R2 is the distance to the location of interest in meters.

Here are some values for the Sun. Note that the following only applies to the Sun, not other objects.

E*1 = 62,930,000 W/m2
R1 = 695,700,000 m

The nearby blue-white star Fomalhaut has an estimated radius of 1.842 times (184.2%) that of the Sun and a surface temperature of 8590 K. How much irradiance would a planet located around Fomalhaut receive, assuming this world has the same mean orbital distance as the Earth, 149,597,870,700 m? This is a two-step question–start with Eq 1 first and then move on to Eq 2.

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Stefan-Boltzmann Law

The Stefan-Boltzmann law estimates the total amount of emission, per unit area, leaving the surface of a blackbody. This is termed the total emittance. The Stefan-Boltzman law is described by the formula Eq (1):

                                                             E* = σSBT 4                 Eq (1)

Where E* is the total emittance in W/m2, σSB is the Stefan-Boltzmann constant with a value of 5.67 x 10-8 W/m2•K4 and T is the blackbody temperature in K.

The nearby sunlike star Tau Ceti has a surface temperature of approximately 5344 K. What is the total emittance from Tau Ceti? 

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