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            Radiative Flux

Problem

Data

Useful Equations

 

Solution (A) The Solar Constant

``

Solution (B) Calculating the Flux Density of an Object Observed Through a Filter

To find the Johnson B-band flux of a star, in this case, the Sun, use the following formula. The value of the fν zero-point for the B filter is given in Table 2 of Section 2.3 in the text. Convert this to SI units: 4063*10^(-26) W m-2 Hz-1. The absolute magnitude of the Sun in the B filter is -26.10. Insert the relevant values and solve for fB of the Sun. You will need to use the text found on canismajor.ca to complete this problem.

``

"M[B]=-2.5*(log)[10](f[B]"(star)"/f[B](m[B]=0))":

 

solve(-26.10 = -2.5*log[10](f[B]/(4063*10^(-26))), f[B])

0.1119043122e-11

(5.1)

 

The Sun's B-filter flux is approximately 1.12 * 10-12 W m-2 Hz-1 (= 1.12 * 10-12 N)

 

To calculate the total energy received by the Earth from the Sun in this band, multiply this figure times the central frequency of the B-band (λ = 0.438 * 10-6 m, from Table 2) and then by half the area of the Earth (the hemisphere facing the Sun) (radius = 6.371 * 106 m.)

 

nu = c/lambda

``

nu = 2.9979*10^8*Unit('m')/(Unit('s')*(.438*10^(-6)*Unit('m')))

nu = 0.6844520548e15/Units:-Unit('s')

(5.2)

or about 6.8 * 1014 Hz. So the luminosity at Earth is

 

evalf(2*(6.8*10^14/Unit('s')*1.12)*10^(-12)*Unit('W')*Unit('s')*Pi*(6.371*10^6*Unit('m'))^2/Unit('m')^2)

0.1942325509e18*Units:-Unit('W')

(5.3)

The total luminosity of the Sun in the B band is taken at 1 AU (= 1.496 * 1011 m).

 

L[B, sun] = evalf(2*(6.8*10^14/Unit('s')*1.12)*10^(-12)*Unit('W')*Unit('s')*Pi*(1.496*10^11*Unit('m'))^2/Unit('m')^2)

L[B, sun] = 0.1070951962e27*Units:-Unit('W')

(5.4)

or approximately 1.07 * 1026 W.

 

The total luminosity of the Sun at all wavelengths (bolometric luminosity) is taken to be approximately 3.862 * 1026 W., as given above.

``

 

Solution (C) Luminosity, Flux, and Temperature

Luminous flux is related to temperature by the formula:

 

F := sigma*T^4

 

where σ is the Stefan-Boltzmann constant (5.67 * 10-8 W m-2 K-4). Therefore, luminosity may be expressed as:

 

L := 4*Pi*r^2*sigma*T^4

 

Knowing the luminosity and radius of the Sun, we can calculate its effective surface temperature, as follows:

 

with(ScientificConstants)

``

evalf(Constant(sigma, units))

0.5670398973e-7*Units:-Unit(('kg')/(('s')^3*('K')^4))

(6.1)

L := 3.862*10^26*Unit('W')

0.3862000000e27*Units:-Unit('W')

(6.2)

r := 6.95508*10^8*Unit('m')

695508000.0*Units:-Unit('m')

(6.3)

``

solve(L = 4*Pi*r^2*evalf(Constant(sigma, units))*T^4, T)

5785.569728*(Units:-Unit('W')*Units:-Unit('m')^2*Units:-Unit(('kg')/(('s')^3*('K')^4))^3)^(1/4)/(Units:-Unit('m')*Units:-Unit(('kg')/(('s')^3*('K')^4))), (5785.569728*I)*(Units:-Unit('W')*Units:-Unit('m')^2*Units:-Unit(('kg')/(('s')^3*('K')^4))^3)^(1/4)/(Units:-Unit('m')*Units:-Unit(('kg')/(('s')^3*('K')^4))), -5785.569728*(Units:-Unit('W')*Units:-Unit('m')^2*Units:-Unit(('kg')/(('s')^3*('K')^4))^3)^(1/4)/(Units:-Unit('m')*Units:-Unit(('kg')/(('s')^3*('K')^4))), -(5785.569728*I)*(Units:-Unit('W')*Units:-Unit('m')^2*Units:-Unit(('kg')/(('s')^3*('K')^4))^3)^(1/4)/(Units:-Unit('m')*Units:-Unit(('kg')/(('s')^3*('K')^4)))

(6.4)

``

We take the first solution, rejecting the negative and imaginary solutions:

 

simplify(evalf(5785.569728*(Unit('W')*Unit('m')^2*Unit('kg'/('s'^3*'K'^4))^3)^(1/4)/(Unit('m')*Unit('kg'/('s'^3*'K'^4)))))

5785.569728*Units:-Unit(K)

(6.5)

``

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References

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Cox, A. (Ed.) (2000). Allen's Astrophysical Quantities, 4th ed. New York: AIP/Springer.

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