I was challenged to present the odds of the object observed during the Apollo 11 mission to be a SLA panel, travelling in the same direction that the
CSM-LM on their way to the moon, and showing a close to constant apparent magnitude (not flashing).
If you didn't read part I, you can find it here.
The model below, is a simple but representative model for the light reflection of the panel.
Calculus of luminous power, apparent magnitude and a first set of odds.
1. Let's agree that the convex reflective surface has an albedo of 0.8 and the concave surface of 0.1.
2. Let's agree that there is no diffuse light in space. This means that the sun light is reflected by the object and strikes directly the eye of the
observer, or the object will not be seen.
3. Let's agree that visible light (visible spectrum) corresponds to 40% of the total irradiance of the sun.
4. Lets agree that the position of the object in relation to the observer is a random event (the reflecting surfaces could be facing any
The luminous power of the Sun reflected by the object (visible spectrum), is given by the equation:
Pr = (h x w) x P x Al, where Pr is the luminous power reflected, h=6.4 m, w=4.7 m, P is the solar constant and, for the visible spectrum is equal to
1.361 kw/m2 x 0.4 = 0.544 kw/m2, and Al is the albedo of the surface.
For Al=0.8, Pr= 13,090 w, and for Al=0.1, Pr=1,636 w.
The luminous power flux (Pf), luminous power by surface unit, at a distance L between the object and the observer is given by the equation:
Pf=Pr / (h x (Pi x L/2)), where Pi= 3.1415.
For L= 22,000 m and Al=0.8, Pf=0.0592 w/m2. For Al=0.1, Pf=0.0074 w/m2.
For L= 10,000 m and Al=0.8, Pf=0.1304 w/m2.For Al=0.1, Pf=0.016 w/m2 .
For L= 500 m and Al=0.8, Pf=2.604 w/m2 . For Al=0.1, Pf=0.325 w/m2 .
For comparison, if you applied this same equation to the moon (Al=0.12, diameter 3474 km, distance Earth- Moon of 380,000 km), you'll find luminous
power flux on Earth, during the full Moon, of 0.000683 w/m2. This tell us that, even in the worst scenario (Al=0.1 and L=22 km), the object would be
so bright and the power flux so high that would be sufficient to do a precision handwork.
Through this analysis, we see that even in the worst scenario, if it was the panel, the brightness should be the first descriptive feature, and
looking to it through the telescope would be a painful experience.
First set of odds
The probability associated to the observer being in the path of the reflected light.
The imaginary illuminated surface at a distance L between the observer and the object is given by the equation:
S=h x (Pi x L/2), where S is the area of the imaginary surface, Pi= 3.1415, and h=6.4m.
For L= 22 km, S=0,221 km2 .
For L= 10 km, S=0,101 km2 .
For L= 22 km, S=0,005 km2 .
The surface area of a sphere of radius R, with R=L is given by the equation:
A=4 x Pi x R2, where A is the area of the surface of the sphere. Because of the fact that we have 2 reflecting surfaces, pointed in opposite
directions, we'll take in account only the surface area of half of the sphere:
Ah=2 x Pi x L2, where Ah is the surface area of half of the sphere, Pi=3.1415.
For L= 22 km, Ah= 3041.062 km2 .
For L= 10 km, Ah= 628.319 km2 .
For L= 0.5 km, Ah= 1.571 km2 .
The probability of the observer to be in the path of the reflected light at distances L= 22 km, L=10 km and L=0.5 km is given by the equation:
P=S/Ah, where P is the probability.
The analysis show that the probability that the "non flashing object" was not the panel, is an interval of P=99,682% and P=99,993%. If we associate
this to the fact that the light would be extremily bright reassures my opinion that it was not the panel ( ... the further away you consider the
object the higher the probability that it was not the panel).
In the making of this analysis I came up with a theory that would help the panel cause (but coming from me would be highly speculative): NASA would
need just one word (and then a lot of math and justifications), and even then, the odds would still be low. Until then, if ever, the object, in my
opinion, was not the panel.
Next topic? If necessary, the odds associated with the strange motion of the panel can be analysed.
Thank you for reading.