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?

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:

Screen shots from the PeriscopeFilm video "Science in Space" Early 1960's Space Exploration Film Sputnik & Explorer Vanguard Rocket 12494 Screen shots from the PeriscopeFilm video "Science in Space" Early 1960's Space Exploration Film Sputnik & Explorer Vanguard Rocket 12494

  • $\begingroup$ Your link to Sputnik 1 gives a magnitude of 6.0. No such luck for the Vanguard 2 satellite. $\endgroup$ Commented Sep 3, 2020 at 19:26
  • $\begingroup$ @WayfaringStranger Thanks! I'm pretty sure that the brightness will still depend on phase angle (as discussed in the first linked question) as well as 1/r^2. Does the link mention the phase angle and distance when it was magnitude 6? $\endgroup$
    – uhoh
    Commented Sep 4, 2020 at 0:01
  • 1
    $\begingroup$ Sure it'll depend on phase angle, and how far the satellite is from earth in its non-circular orbit. I'd guess that magnitude 6 was about as bright as it got. Apparently the final rocket stage made it to orbit too, and was deliberately made reflective so as to be a magnitude 1 object. $\endgroup$ Commented Sep 4, 2020 at 14:57
  • $\begingroup$ If the reflection model is truly specular then the reflection is just a delta function. Not sure that's what we're looking for here. Why the BRDF of specular reflection is infinite in the reflection direction?. Is there a meaningful way to define "specular-like"? $\endgroup$ Commented Feb 13, 2022 at 13:20
  • 1
    $\begingroup$ @uhoh The BDRF is a delta function. If we're happy here to assume a perfectly specular sphere with a specified diameter, then the problem ought to be solvable. This would, as you indicate, use the non-zero angular extent of the sun. The result will however be strongly a function of the geometry. $\endgroup$ Commented Feb 13, 2022 at 16:01

1 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:

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?


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?)
Reflection from a spherical mirror

(from researchgate)

  • $\begingroup$ "The solution here" (in a link) is not a good Stack Exchange answer, these are usually called "link-only" answers. Links break; something hosted on Github could disappear at any minute. Can you summarize the solution here in your answer post, and estimate an apparent magnitude for at least one of the two spacecraft mentioned in the title? Thanks! $\endgroup$
    – uhoh
    Commented Oct 3, 2022 at 4:22
  • 1
    $\begingroup$ @uhoh Sorry, I was in a rush . Let me fix it ;) $\endgroup$
    – Cheng
    Commented Oct 3, 2022 at 9:46

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .