Apparent magnitude of a spherical body with specular, rather than diffuse reflectivity? How bright were Sputnik 1 and Vanguard 2?

Question

How bright are geostationary satellites due to reflected sunlight? and Did Sputnik 1 tell us more than “beep”? What science was improved by information gained from its orbiting the Earth? now have me wondering how bright and visible Sputnik 1 would have been.

Sputnik 1 was a smooth, shiny, reflective metal sphere 58 cm in diameter with an initial periapsis and apoapsis of about 200 and 900 km altitude.

Vanguard 2 was similar, with a 51 cm diameter and a 560 x 2,953 orbit.

Normally we treat asteroids with an albedo and some model for diffuse reflectivity based on phase angle. See answers to the questions linked below.

But in this case it's probably a good approximation to use geometrical optics and assume specular reflection. The problem is that I don't know how to do that!

Question: Apparent magnitude of a spherical body with specular, rather than diffuse reflectivity? How bright were Sputnik 1 and Vanguard 2?

• Calculating the apparent magnitude of a satellite
• Why is Enceladus's albedo greater than 1?
• What is the difference between albedo, absolute magnitude or aparent magnitude?
• When would we detect a tiny meter size natural satellite in a geostationary orbits?
• Venus' magnitude during inferior conjunction

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Screen shots from the PeriscopeFilm video "Science in Space" Early 1960's Space Exploration Film Sputnik & Explorer Vanguard Rocket 12494 The photographic negative contains an image of Vanguard 2 from a tracking camera. Click for full size:

Answer 1

This once appeared in an IAO 2003 A,B 1 problem (the first problem on the first page).

The solution here.

A summary:

Let:
Albedo of moon be $\alpha_m$
Albedo of sputnik be $\alpha_s$
Distance between object and observer: $d$
Radius of objects: $R$
Solar constant, $F_\odot$

Flux of moon to observer: $F_m = \frac{F_\odot \alpha_{m} \pi R_{m}^2}{2 \pi d_{m}^2}$
Flux of Sputnik to observer: $F_S = \frac{F_\odot \alpha_{S} \pi R_{S}^2}{4 \pi d_{S}^2}$

\begin{align*} \Delta m = m_{S} - m_{m} &= -2.5 \log_{10} \frac{F_S}{F_m} \\ &= -2.5 \log_{10} \frac{1}{2}\frac{\alpha_{S}}{\alpha_{m}}\frac{R_{S}^2}{R_{m}^2}\frac{d_{m}^2}{d_{S}^2}\\ &= -2.5 \log_{10} \frac{1}{2}\frac{1}{0.14}\left ( \frac{0.58 \ \mathrm{m}}{3475000 \ \mathrm{m}}\frac{378000 \ \mathrm{km}}{200 \ \mathrm{km}} \right ) ^2 \\ & = 16^m \end{align*}

Therefore, apparent magnitude of Sputnik is $16^m - 12.7^m \approx 3.3^m$, well below $6^m$.

However, I do not understand the calculation of the flux and the associated comments in the document given above:

We do not take into account effects related to diagrams of scattering and random/mirror scattering.... difference between the 2 and the 4 in $2 \pi d_{m}^2$ and $4 \pi d_{S}^2$

Any ideas?

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Edit:

Maybe it's because assuming sun-object-observer angle to be small (e.g. as in full moon)

• moon reflects light into a hemisphere facing observer only, hence $2 \pi d_{m}^2$ (diffuse reflection?)
• Sputnik 1, when $\theta > 45^{\circ}$, the light rays are reflected into the hemisphere away from the observer as well, hence light is distributed into the whole sphere (at least approximately), hence $4 \pi d_{S}^2$ (specular reflection?)

(from researchgate)