How long does it take the moon to move through an angle equal to its own angular diameter (1/2 a degree) against the background of stars? Is this based on the synodic or sidereal month? Why?
2007-09-08 15:31:59 · 3 answers · asked by ChEMIsTrY ChICkiE 1 in Science & Mathematics ➔ Astronomy & Space
If viewed from the center of the earth, it would be about 54.6 minutes. The moon's sidereal period -- the period of rotation with respect to the stars -- is 27.3 days. Divide that by 720 (the fraction of a circle in 1/2 degree) and express as minutes to get 54.6.
But for an actual observer on the surface of the earth, it will take a bit longer than that due to the effects of parallax.
2007-09-08 18:40:42 · answer #1 · answered by Keith P 7 · 0⤊ 1⤋
The definition of sidereal month has to do with the motion of the moon against the background stars. This question is like asking why spinach tastes like a plant.
The synodic month is based on the phase of the moon which in turn depends on the relative positions of the Earth, Moon AND Sun. Because the whole Earth-Moon system is orbiting the Sun (in the same direction that the moon orbitrs the Earth), the moon has to travel a bit more than one full orbit around the Earth to return to its original phase. The sidereal month describes simply the motion of the moon around the Earth ignoring the position of the sun.
Another way to look at this is the synodic motion of the moon is based on its angular motion *relative to the sun* while the sidereal is based on its motion *relative to the background stars*.
2007-09-09 22:00:28 · answer #2 · answered by Mr. Quark 5 · 0⤊ 0⤋
Well...... It takes about 28 days for the Moon to go 360 degrees so
.5/360 = x/28 => about 56 minutes. (If you want it closer, get a better number for an orbital period âº)
Doug
2007-09-08 23:05:37 · answer #3 · answered by doug_donaghue 7 · 0⤊ 0⤋