4
$\begingroup$

On Nov. 25 at 2:30a.m. from southern California while viewing the moon with a 70mm D x 400mm F telescope a point of light was seen at about the 7 o'clock position below the moon. As viewed the point of light was clear and seemed to be tracking with the moon. As I tracked the moon the point of light stayed within the same distance and in view. If it was a planet or star wouldn't they track slightly different than the moon ?

$\endgroup$
3
  • $\begingroup$ I viewed it for about 35 minutes, I should have tagged this to the amateur astronomer, I just discovered this site and still learning the ropes. I thought perhaps any stars or planets would track away eventually. It seemed very odd. It was very bright, just enough to get my attention with the naked eye, then with binouclurs, then had to break out the telescope. If a satelilite was positioned between the moon and a observer would it be brighter due to reflected light of the moon ? $\endgroup$
    – jdcrlist
    Nov 26, 2015 at 21:17
  • $\begingroup$ It's clear it wasn't a geosynchronous satellite. One of those would have a fixed position relative to the background stars. Also, its apparent magnitude would be something like +10 and up. It could have been a non-geosynchronous satellite; however, every one of these I've seen has flown across the entire visible sky in a matter of minutes. Can you provide the year for this observation, and the latitude/longitude? What was the TFoV for the observation? $\endgroup$
    – Alphecca
    Aug 30, 2018 at 17:27
  • $\begingroup$ "geosynchronous satellite. One of those would have a fixed position relative to the background stars." A geostationary satellite is fixed relative to the observer. In a stationary telescope, it would appear as a fixed point of light with the stars drifting through the field of view. In a telescope tracking at siderial rate to follow the stars, a geostationary satellite drifts through the field of view. A geosychronous satellite would drift north to south, or vice versa, very slowly, but appear at the same azimuth and elevation from the observer at the same time of day each day. $\endgroup$
    – Iaxc
    Jan 5, 2019 at 14:47

1 Answer 1

2
$\begingroup$

The answer to the headline question is clearly: No.

The full moon has an apparent diameter of about 0.5°. Its apparent motion is roughly 360°/24h, neglecting the orbital motion of Earth and Moon. Hence it apparently moves about 24h*0.5°/360° = 0.033h = 2 min. Hence the moon moves one moon diameter in about 2 min. (The neglected orbital motion induces an error of about 3.65% (0.997d/27.3d), divide the siderial Earth day by a siderial moon orbit, hence an error of just a few seconds for the apparent motion of 0.5°.) In 35 minutes, Moon's apparent motion would be about 17 moon diameters.

The geosynchronous satellite, in contrast, remains staying at the same apparent position, by definition.

Planets or stars move apparently about the above error estimate different from the moon, hence roughly 27.3-times slower relative to the apparent moon than a geosynchronous satellite.

$\endgroup$
1
  • 1
    $\begingroup$ The 360°/24h apparent motion applies to everything in the sky (the Moon, the Sun, the stars, the planets and geostationary satellites). This is what a stationary observer on Earth experiences and is due simply to the Earth's rotation. You need to consider the proper motion, which is the orbital motion that you would observe if the Earth stopped spinning. Here, the Moon does 360°/27 days which is about 0.5°/hour. So in your 35 mins, the Moon would've moved about a half-diameter against the fixed stars. $\endgroup$ Aug 30, 2018 at 6:38

You must log in to answer this question.

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