Brightness of Sputnik 1

rkresken@esoc.esa.de
Wed, 3 Mar 1999 09:20:38 +0100

Hi!

I followed the discussion about the visibility of Sputnik 1 with great
interest .
When I thought about it, I realized that the magnitude of that satellite
should be
quite easy to calculate since it must have been close to that of an ideal
"christmas tree ball" of 58 cm diameter.
To calculate the magnitude difference to the sun, one can use the formula

0.4(m2-m1) = log(I1/I2)

I1/I2 is the ratio of the intensity of the sunlight and the light reflected
from an
object at certain distance.
Since the brighness of a "christmas tree ball" satellite does NOT depend on
the
phase angle, the entire visual light power it receives is distributed
evenly on the
surface of an imaginary sphere with a radius that equals the distance
satellite-
observer.  The Intensity I is power devided by area, and the received
visual power
is I1 multiplied by the projected surface of Sputnik
P1 = I1*r^2*Pi

This is the same as the total energy received by the imaginary sphere
P2 = I2*4*R^2*Pi

The intensity ratio is then I1/I2 = 4*(R/r)^2
This can be plugged into the first formula to get the magnitude difference.
At an overhead perigee pass with an altitude of 230 km, (m2-m1) gets 29.5.
With the sun being -26.8, the satellite magnitude is 4.2. At an apogee
overhead pass at 945 km, one gets 7.2.
Based on this calculation, I think one can say that Sputnik 1 was a
potential
faint naked eye object. I assume that perigee occured over the northern
hemisphere.
As a test case, I applied that calculation to Echo 2, another spherical
shiny
satellite. With a diameter of 41 m and a typical altitude of 1100 km, I get
-1.6m. That matches eyewitness account quite well.

Comments are welcome!
Has anybody observed a similar satellite? Observation reports could be
helpful to check the validity of that calculation.

Regards,
Rainer