as-500/hw3/text

(1) Submitted separately

(2)

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.

B = σ T⁴, with σ the Stefan-Boltzmann constant.
σ ≈ 5.6704 × 10⁻⁵ erg cm⁻² s⁻¹ K⁻⁴.
B = 5.6704 × 10⁻⁵ × T⁴ erg cm⁻² s⁻¹, with T in kelvin.

For a spherical body, this surface brightness can be related to the total luminosity L by,

L = 4 π r² B, with r the radius of the sphere.

The radius is therefore, 
     ________
r = / __L__
   √  4 π B    .

Assuming the solar luminosity L* = 3.828×10³³ erg/s and the solar surface temperature T* = 5800 K, the solar radius can be determined.

B* = σ × 5800⁴ K⁴ = 6.417 × 10¹⁰ erg cm⁻² s⁻¹.
L*/B* = 3.828×10³³ erg s⁻¹ / 6.417 × 10¹⁰ erg cm⁻² s⁻¹
      = 5.965 × 10²² cm².
      _____________
r* = √ L*/B*/(4 π)
      _____________________________
   = √ 5.965 × 10²² cm² / (12.5664)
      _________________
   = √ 4.747 × 10²¹ cm²

   = 6.890 × 10¹⁰ cm.


(3) 

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.

Φ = L / (4 π D²), compare with the expression in (2) where Φ = B and D = r at the surface of the emitter.

The distance from the Earth to the Sun D* is one AU, or in cm,

D* = 14959790000000 cm.

Φ* = L* / (12.5664 × 2.2380 ×10²⁶ cm²)
   = 3.828 × 10³³ erg s⁻¹ / (12.5664 × 2.2380 × 10²⁶ cm²)
   = 1.361 × 10⁶ erg cm⁻² s⁻¹.

The energy density ρ is related to the energy flux Φ by

ρ = Φ / v, and for electromagnetic energy,
ρ = Φ / c.

The energy density at Earth's orbit ρ* is therefore,

ρ* = Φ* / c 
   = 1.361 × 10⁶ erg cm⁻² s⁻¹ / (29979245800 cm s⁻¹)
   = 4.540 × 10⁻⁵ erg cm⁻³.

This relationship is also associated to the Stefan-Boltzman Law, hinted at with the comparison to the expression in (2).

ρ = σ T⁴ / c = Φ / c, or
    ₄________
T = √ ρ c / σ

The temperature T` related to the energy density at Earth's orbital radius is therefore,
     ₄_________
T` = √ ρ* c / σ

     ⎛ 4.540 × 10⁻⁵ erg cm⁻³ × 29979245800 cm s⁻¹ ⎞1/4
   = ⎜ ------------------------------------------ ⎟
     ⎝       5.6704 × 10⁻⁵ erg cm⁻² s⁻¹ K⁻⁴       ⎠

   = 3.936 × 10² K.

The average temperature at the Earth's surface is approximate 289.15 K.

393.6 K - 289.15 K = 104.45 K.

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?

(4)

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.

λ[peak] T = 0.28979 cm K; this is Wien's Law in cm and K.

λ* = 0.28979 cm K / T*
   = 0.28979 cm K / (5800 K)
   = 4.996 × 10⁻⁵ cm.