Re: Visible change in the phase of the moon?

allnor@xxxxxxxxxxxx wrote:
<snip> I spent some time viewing it [the Moon] myself. This would
have been around 5 or 6 PM. A few hours later (11 PM) I had
another look at the moon. This thime I got the distinct impression
that the edge between lunar night and day had changed. I was not
prepared for any such effect when I first viewed the moon . . . .
I am just wondering, is there a visible change on the lunar edge at
this sort of time scale?

1) How fast does the terminator move in terms of selenographic lunar
longitude per hour?

The lunar synodic period (lunar month from Earth) is ~29.53 days long,
you have the terminator moving at an average of:

360 selenographic longitude degrees /(29.53 days * 24 hours )
= 0.5079 selenographic longitude degrees per hour
at the lunar equator
- or -
0.51 longitude degrees per hour * ( 3600 arcseconds / 1 degree) * ( 1 /
hour / 3600 seconds )
= 30 arcsecs / minute
= 0.5 arcsecs selenographic longitude / second

2) How big in linear size (kilometers) is 1 degree of lunar longitude
on the Moon's equator?

The equatorial diameter of the Moon is 3,476.2 km. Therefore, its
circumference is 3,476.2 * pi km or -

10920.8 km = (3476.2 * pi) km

There are 21600 arcmins in 360 degrees (60 mins/deg * 360 deg) of lunar
longitude, so, per degree and per arcmin of lunar longitude there are:

30.34 km / deg = 10920.8 km / 360 degrees
0.51 km / arcminute = 10920.8 km / 21600 arcminutes

3) How fast does the Moon's terminator move in terms of kilometers per
hour across the surface of the Moon?

The Moon's terminator crosses this distance of 30.34 km per
selenographic longitude degree at a rate of 0.51 degs per hour. So the
terminator moves at the rate of:

15.41 km per hour = 30.34 km/ selenolgraphic deg * 0.51 seleno deg /
hour

The terminator moves about 15.41 km per hour across the Moon's surface
at the lunar equator.

4) How big is this 15.41 km (per hour) on the Moon's equator when seen
from the Earth?

It depends in part on the varying Earth-Moon distance. The average is
384,400 km. We'll use that. So, using the half-angle and/or sine
small angle (image scaling) formulas, 15.41 km on the Moon's surface
one viewed from Earth has an angular size of:

8.3" celestial arcsecs =
206525 arcsecs per radian * ( 15.41 km / 384400 km )

To recap:

Every hour at the Moon's equator -
= 8.3" celestial arcsecs in the eyepiece
= 15.41 km
= 0.51 degs of lunar selenographic longitude

The above sizes are only good at the 0 deg N, 0 deg E lunar longitude -
about the center of the Moon's disk when the terminator is at 1st or
3rd quarter.

(Backchecking the physical terminator rate to the lunar cycle as seen
from Earth, note that a half-lunar cycle - new Moon or full Moon
involves the terminator moving across 10920.8 km / 2 kilometers or
5460.4 km of lunar circumference. At 15.41 km per hour, it takes 354.34
hours or 14.76 days. That's about 1/2 a 29.53 day lunar cycle.)

Bear in mind that the 5 most common crater types have the following
sizes:

Type Size range N
ALC 0.5-20 km 796
BIO 2-20 km 636
SOS 2-35 km 119
TRI 13-50 km 79
TYC 30-170 km 32

5) How big are objects on the Moon's surface at the equator at 0 deg E
lunar longitude - where the Moon's terminator crosses the lunar equator
at the 1st and 3rd phases?

For a general solution for celestial arcsecs in the eyepiece to a
physical distance on the Moon, we use the half-angle or the small angle
sine formulas -

- where A is the size of the object in arcsecs as seen in the eyepiece:

- where the Moon is about 384400 km distance

Objsize_on_Moon km = ( A / 206265 arcs per radian ) * Moon's distance
Objsize_on_Moon km = ( A / 206265 arcs per radian ) * 384,400 km

Objsize_on_Moon km = 1.864 km/arcsec * Object_size_arcsecs_in_eyepiece

Again, the above sizes are only good at the 0 deg N, 0 deg E lunar
longitude - about the center of the Moon's disk when the terminator is
at 1st or 3rd quarter.

6) How big are objects on the Moon's surface at the equator when the
craters are from 0 deg to 90 degs E or W lunar longitude?

There is a projection-of-view problem - the Moon is spherical and when
we look at it as a 2d projection. Although this is too much for your
4-year-old nephew, having an older child look at a basketball from 10
feet and then close up may help to explain it. Tape two circular
"craters" on the basketball - one near the apparent center, another
near the apparent side.

As we move progressively away from center of the Moon's disk staying at
the Moon's equator, in either east or west longitude, we see
progressively more degrees of lunar longitude.

In the eyepiece, this is seen in the foreshortening of craters.
Circular craters near 90 deg E or 90 deg West lunar longitude look like
thin ovals, foreshortened in longitude but not latitude. Because we
are looking at an angle, we see more kilometers per arcsec of longitude
in the eyepiece.

The cosine function is used to adjust the 1.864 arcsec image scale in
the eyepiece at 0 deg N, 0 deg E lunar longitude and along the lunar
equator between 0 and 90 degs E or W lunar longitude:

Objsize_on_Moon km = Object_size_arcsecs_in_eyepiece * ( 1.864
km-arcsecs / cos(Obj_longitude_EorW) )

For example, at 0 deg N, 0 deg E, a crater 1 arcsec in diameter is
1.864 km in diameter, but a 1 arc sec crater at 0 deg N, 80 deg E
longitude is:

Objsize_on_Moon km = 1 arcsec * ( 1.864 km-arcsecs / cos(Radians(80
degrees)))
= 10.73 km

So, a 1 arcsec object at 80 degs E lunar longitude is 10.73 km in size.


The same conversions can be done using the number of days before and
after the first quarter (when the terminator is at 0 deg E longitude):

Objsize_on_Moon km = Object_size_arcsecs_in_eyepiece * ( 1.864
km-arcsecs / cos( 360 * ( Days-after-1st-quarter / 29.53 ) )

Three days after 1st quarter the Moon is at ~ 36.5 degs E longitude. A
1 arcsec crater on the terminator at the lunar equator is:

2.32 km = 1 arcsec * ( 1.864 km-arcsecs / cos(Radians(36.5 deg)) )

Let's say that three days after the 1st quarter you see a crater that
is 43.1 arcsecs in size in the eyepiece. Then it's physical size is:

100 km = 43.1 celestial arcsecs * ( 1.864 km-arcsecs /
cos(Radians(36.5 deg)) )

7) How fast does the terminator move across a crater on the Moon's
equator at various degrees of E or W lunar longitude?

If the terminator moves 15.4 kilometers per hour across the Moon's
surface, it will take:

6.49 hours = 100 km / 15.4 km/hour

for the terminator to cross the crater at 80 degrees E lunar longitude.


Because of this projection-of-view problem, the Moon's terminator does
not move at uniform rate in terms of arcsecs per hour seen as seen in
the eyepiece or with the naked-eye. (The Moon's near-side face has an
apparent size of about 1800" and 1800"/8.3" does not work out to 14.76
days.) The terminator will appear to move faster near the center of
the disk, and slower as it gets nearer the 90 degree E or W edge of the
Moon's apparent disk.

The matter becomes more complicated as you take both E-W lunar
longitude and N-S declination into consideration.

To answer your base question, assuming that a crater was near the
terminator on Moon's the 1st or third quarters, in six hours the
terminator moved:

49.3" = 8.3" per hour * 6 hours
92.4 km = 15.4 km * 6 hours

Tycho is about 83km in diameter; Copernius 93km; Plato about 95km.

These changes are easily seen in the eyepiece at 10x or above and with
the naked-eye and more difficulty across 8-9 hours.

Hope that's not too much info or too far off your question. Thanks for
a good question.

- Peace - Canopus56

P.S. - Corrections to the above by advanced lurkers would be
appreciated.

More reading:

Astronomy Dept. of the U. of Michigan. 2000. Lunar Features and
Mountain Heights. (Webpage)
http://www.astro.lsa.umich.edu/Course/Labs/lunar_features/lunar_intro.html

.

Relevant Pages

  • Re: First Views with an Older C90
    ... The edge of the light was beautiful but the rest of the moon ... it's a fundamental fact of lunar observing. ... Features show most clearly along the terminator, ... Far from the terminator, the Sun is more ...
    (sci.astro.amateur)
  • Re: moon atlas; questions about moon
    ... and six craters near it. ... At the time of your post, the lunar terminator was at approx. ... contains photos of the Moon by lunar age. ...
    (sci.astro.amateur)
  • Re: Colourful spring sunset
    ... recognize the term "terminator". ... hires high-power imaging involving the moon is accomplished around the ...
    (sci.astro.amateur)
  • Re: Worlds first computer may be even older than thought
    ... the 2nd C BC when this thing is supposedly built. ... to work but one of the rivals for finding longitude in the 18th C was ... enough to use lunar distances until the 18th century. ... The neolithic sailors used the stars, ...
    (sci.military.naval)
  • Re: Worlds first computer may be even older than thought
    ... the 2nd C BC when this thing is supposedly built. ... to work but one of the rivals for finding longitude in the 18th C was ... enough to use lunar distances until the 18th century. ...
    (sci.military.naval)