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96 lines
3.1 KiB
Plaintext
96 lines
3.1 KiB
Plaintext
(1) Submitted separately
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(2)
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The Stefan-Boltzmann Law gives the surface brightness B of an object as a function of its temperature T, with dimension energy over distance-squared over time.
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B = σ T⁴, with σ the Stefan-Boltzmann constant.
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σ ≈ 5.6704 × 10⁻⁵ erg cm⁻² s⁻¹ K⁻⁴.
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B = 5.6704 × 10⁻⁵ × T⁴ erg cm⁻² s⁻¹, with T in kelvin.
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For a spherical body, this surface brightness can be related to the total luminosity L by,
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L = 4 π r² B, with r the radius of the sphere.
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The radius is therefore,
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________
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r = / __L__
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√ 4 π B .
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Assuming the solar luminosity L* = 3.828×10³³ erg/s and the solar surface temperature T* = 5800 K, the solar radius can be determined.
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B* = σ × 5800⁴ K⁴ = 6.417 × 10¹⁰ erg cm⁻² s⁻¹.
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L*/B* = 3.828×10³³ erg s⁻¹ / 6.417 × 10¹⁰ erg cm⁻² s⁻¹
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= 5.965 × 10²² cm².
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_____________
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r* = √ L*/B*/(4 π)
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_____________________________
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= √ 5.965 × 10²² cm² / (12.5664)
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_________________
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= √ 4.747 × 10²¹ cm²
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= 6.890 × 10¹⁰ cm.
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(3)
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Given the luminosity of a body, one can determined the energy flux Φ at a shell of arbitrary distance D, because this flux reduces with distance by an inverse square law. This is seen in the expression in (2), as well.
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Φ = L / (4 π D²), compare with the expression in (2) where Φ = B and D = r at the surface of the emitter.
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The distance from the Earth to the Sun D* is one AU, or in cm,
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D* = 14959790000000 cm.
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Φ* = L* / (12.5664 × 2.2380 ×10²⁶ cm²)
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= 3.828 × 10³³ erg s⁻¹ / (12.5664 × 2.2380 × 10²⁶ cm²)
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= 1.361 × 10⁶ erg cm⁻² s⁻¹.
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The energy density ρ is related to the energy flux Φ by
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ρ = Φ / v, and for electromagnetic energy,
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ρ = Φ / c.
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The energy density at Earth's orbit ρ* is therefore,
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ρ* = Φ* / c
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= 1.361 × 10⁶ erg cm⁻² s⁻¹ / (29979245800 cm s⁻¹)
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= 4.540 × 10⁻⁵ erg cm⁻³.
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This relationship is also associated to the Stefan-Boltzman Law, hinted at with the comparison to the expression in (2).
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ρ = σ T⁴ / c = Φ / c, or
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₄________
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T = √ ρ c / σ
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The temperature T` related to the energy density at Earth's orbital radius is therefore,
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₄_________
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T` = √ ρ* c / σ
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⎛ 4.540 × 10⁻⁵ erg cm⁻³ × 29979245800 cm s⁻¹ ⎞1/4
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= ⎜ ------------------------------------------ ⎟
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⎝ 5.6704 × 10⁻⁵ erg cm⁻² s⁻¹ K⁻⁴ ⎠
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= 3.936 × 10² K.
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The average temperature at the Earth's surface is approximate 289.15 K.
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393.6 K - 289.15 K = 104.45 K.
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So, the temperature associated with the electromagnetic energy density at the Earth's orbital radius is sigificantly (>100 K) greater than the temperature at the Earth's surface. Is this due to reflection and other scattering?
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(4)
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Wien's Law allows one to calculate the peak of emission of a blackbody for a given temperature. Using the assumed surface temperature of the Sun T*, this peak wavelength λ* can be found.
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λ[peak] T = 0.28979 cm K; this is Wien's Law in cm and K.
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λ* = 0.28979 cm K / T*
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= 0.28979 cm K / (5800 K)
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= 4.996 × 10⁻⁵ cm.
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