(adsbygoogle = window.adsbygoogle || []).push({}); 1. The problem statement, all variables and given/known data

I have made a solution to an exercise and I need some to check it and please notify me if I have made any misstakes.

Exercise

Calculate the radiation flux density of the surface of Earth from the sunlight that is reflected against the surface of Mars. As seen from the Earth, Mars will be in the opposite direction in comparison with the Sun. The Bond albedo of Mars is 0.15. Assume that Mars reflects about 4 times more in the direction towards the Earth in comparison with the sunlight that is reflected isotropic from the whole surface of Mars.

Calculate the apparent magitude that Mars will have in this case

3. The attempt at a solution

If we know the distance r between Mars and the Sun vi can express the radiation flux density at the surface of Mars as

[tex]F_{p}=F_{\odot}\left(\frac{R_{\odot}}{r}\right)^{2}[/tex]

where [tex]F_{\odot}[/tex] and [tex]R_{\odot}[/tex] are the radiation density flux and radius of the sun.

The total flux on the surface of Mars will be

[tex]L=F_{p}\cdot\pi\cdot R^{2}[/tex]

whereRis the radius of Mars

The flux reflected by Mars is

[tex]L'=L\cdot A[/tex]

whereAis the Bond albedo of Mars

Mars is observed at the distancedfrom the Earth and if the sunlight is reflected isotropic from Mars the radiation flux intensity at the surface of Earth will be

[tex]F=\frac{L\text{'}}{4\pi d^{2}}[/tex]

According to the exercise, 4 times as much sunlight will be reflected in comparison with the isotropic reflection, and then we find

[tex]F=\frac{L'}{\pi d^{2}}[/tex]

The expression for the radition flux density at the surface of Earth will be

[tex]F=4\cdot A\cdot F_{\odot}\left(\frac{R}{d}\right)^{2}\left(\frac{R_{\odot}}{r}\right)^{2}[/tex]

For the numerical calculation I will use the mean distances for Sun-Earth and Mars-Earth, and I use the following numbers:

[tex]A=0.15[/tex]

[tex]F_{\odot}=1366\, W/m^{2}[/tex]

[tex]R=3.386\cdot 10^{6}m[/tex]

[tex]d=1.49\cdot 10^{11}m[/tex]

[tex]R_{\odot}=6.955\cdot 10^{8}m[/tex]

[tex]r=2.29\cdot 10^{11}m[/tex]

By using this numerical values I get the result

[tex]F\approx3.9\cdot10^{-12}W/m^{2}[/tex]

Now I will continue with the solution for the apparent magnitude, which can be calculated by following equation:

[tex]m=-2.5\log\left(\frac{F}{F_{0}}\right)[/tex]

But the reference of radiation flux density [tex]F_{0}[/tex] is unknown. My idea was to determine it by using the apparent magnitude of the Sun and its radiation flux density at the surface of Earth, because they can be found:

[tex]m_{\odot}=-26.74[/tex]

[tex]F_{\odot}=1366\, W/m^{2}[/tex]

Then I found the reference of radiation flux density to be

[tex]F_{0}=2.7\cdot10^{-8}W/m^{2}[/tex]

I can now determine the apparent magnitude for Mars in this case:

[tex]m=-2.5\log\left(\frac{3.9\cdot10^{-12}}{2.7\cdot10^{-8}}\right)\approx9.6[/tex]

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# Astronomy - Correct solution? Part 1

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