# Homework Help: Astronomy - Correct solution? Part 1

1. Sep 14, 2011

### tosv

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
$$F_{p}=F_{\odot}\left(\frac{R_{\odot}}{r}\right)^{2}$$
where $$F_{\odot}$$ and $$R_{\odot}$$ are the radiation density flux and radius of the sun.

The total flux on the surface of Mars will be
$$L=F_{p}\cdot\pi\cdot R^{2}$$
where R is the radius of Mars

The flux reflected by Mars is
$$L'=L\cdot A$$
where A is the Bond albedo of Mars

Mars is observed at the distance d from the Earth and if the sunlight is reflected isotropic from Mars the radiation flux intensity at the surface of Earth will be
$$F=\frac{L\text{'}}{4\pi d^{2}}$$

According to the exercise, 4 times as much sunlight will be reflected in comparison with the isotropic reflection, and then we find
$$F=\frac{L'}{\pi d^{2}}$$

The expression for the radition flux density at the surface of Earth will be
$$F=4\cdot A\cdot F_{\odot}\left(\frac{R}{d}\right)^{2}\left(\frac{R_{\odot}}{r}\right)^{2}$$

For the numerical calculation I will use the mean distances for Sun-Earth and Mars-Earth, and I use the following numbers:
$$A=0.15$$
$$F_{\odot}=1366\, W/m^{2}$$
$$R=3.386\cdot 10^{6}m$$
$$d=1.49\cdot 10^{11}m$$
$$R_{\odot}=6.955\cdot 10^{8}m$$
$$r=2.29\cdot 10^{11}m$$

By using this numerical values I get the result
$$F\approx3.9\cdot10^{-12}W/m^{2}$$

Now I will continue with the solution for the apparent magnitude, which can be calculated by following equation:
$$m=-2.5\log\left(\frac{F}{F_{0}}\right)$$

But the reference of radiation flux density $$F_{0}$$ 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:
$$m_{\odot}=-26.74$$
$$F_{\odot}=1366\, W/m^{2}$$

Then I found the reference of radiation flux density to be
$$F_{0}=2.7\cdot10^{-8}W/m^{2}$$

I can now determine the apparent magnitude for Mars in this case:
$$m=-2.5\log\left(\frac{3.9\cdot10^{-12}}{2.7\cdot10^{-8}}\right)\approx9.6$$

2. Sep 14, 2011

### cepheid

Staff Emeritus

Normally what I'm using to seeing done is to say that the surface area of Mars that reflects sunlight is just a circle of area pi*R^2 where R is the radius of Mars, and that circular area is oriented perpendicular to the incident radiation.. The reason is that this is the cross-section that actually intercepts incident radiation from the sun. You assume that all the light is reflected normal to this surface and none in any other direction.

But I guess they want you to do it their way. Even so, what you've done can't be correct -- can you see why? Simply multiplying by 4 can't be right, because then you're saying that Mars somehow reflects MORE radiation than is incident upon it (by a factor of 4)! Instead, what I think they're asking you to assume is that the incident radiation is somehow divided into 5 parts, and 4 parts of that is reflected back towards the source, with the remaining 1 part being spread out over all other directions. In that case, the scaling factor would be 4/5. Does that make sense?

Another thing is that it is not really necessary to explicitly include F0. You can just use the relation for a relative magnitude difference between two sources:

m1 - m2 = -2.5log(F1/F2)

where in this case source 1 can be the sun and source 2 can be Mars, or vice versa. Since you know both the apparent magnitude and flux of the sun, and you know the flux of Mars, you can solve for the apparent magnitude of Mars.

3. Sep 14, 2011

### tosv

I think it makes more sense than my own interpretation, because the surface cannot reflect more light than it receive from the source. It is more probable that the scaling factor would be 4/5.

Instead of my previous expression I should use
$$F=\frac{4}{5}\cdot \frac{L\text{'}}{4\pi d^{2}}$$