PDA

View Full Version : duration of Full Moon viewed from Earth



buczas
2008-Nov-12, 09:17 PM
I know that the "Full Moon duration" question has been asked and aswered many times BUT I find the "official" answer not satisfying.
To add complexity to the question I would like you to take into account that we (observers on Earth) do not see exactly the 50% part of the Moon. So - having that in mind - how long do we see that part fully lit during a Full Moon?

To visualise the concept think of an observer on a rocket launching from Earth - he sees more and more of the Earth's surface, but he'll never see the entire hemisphere, no matter how far he gets. Apply that to an observer on the Earth looking at the Moon.

antoniseb
2008-Nov-12, 09:30 PM
Do you mean how long does an Earthbound observer get to see the entire exposed face of the Moon in direct Sunlight? If that is the question, the answer will be some range of times depending on where the Earth and Moon are in the respective orbits.

Just using back of the envelope calculations, My guess is that it is about an hour. Is this a coincidence that it is also roughly the duration of a total lunar eclipse? Maybe, I haven't thought it through.

buczas
2008-Nov-12, 09:40 PM
Yes, I mean exactly that.
The "official" answer idealises the perspective and assumes that we see the entire Moon hemisphere (since it is so far away).
Good point noticing the difference between apogee and perigee in calculations.

01101001
2008-Nov-12, 09:41 PM
Welcome to BAUT Forum.


I know that the "Full Moon duration" question has been asked and aswered many times BUT I find the "official" answer not satisfying.

I'm afraid it sounds to me like you want to nitpick the usual astronomical definition of "full moon" and turn it into something ultraliteral, private, and not of use for communication with other people.

Why don't you skip what seems to be an ingenuous question and just tell us what you think the new "official" meaning of the phrase "full moon" should be and why you think it will serve us better?

Or, might I be mistaken and do you have a real question?

buczas
2008-Nov-12, 09:49 PM
I'm not familiar with the new definition, looking it up right now.
I'm not asking about the astronomy definitions at all nor want to nitpick any of them. My question is more about perspective and spatial geometry, sorry if I didn't make myself clear.

ngc3314
2008-Nov-12, 10:04 PM
Because the Earth subtends a larger angle in the lunar sky than does the Sun, this perspective definition not only gives no full Moon in most months, but gives a full Moon from only some parts of the Earth when it does happen.

01101001
2008-Nov-12, 10:11 PM
I'm not familiar with the new definition [...]

I was speaking of a new definition that you might have wanted to provide to us, different from what was already in use. I guess now that's not the case.

If you are satisfied with the definition(s) that people already use, maybe you can cite one you agree with, so that someone might have a chance to answer your question to your satisfaction?

pghnative
2008-Nov-12, 10:13 PM
Wouldn't a truly full moon ("50% lit") happen only when the sun was opposite the viewer --- ie, during an eclipse?

I'm not advocating that as a definition for a full moon (just the opposite, in fact), just commenting that the ideal of 50% lit seems impossible to me, and would only occur for a brief instant in time.

buczas
2008-Nov-12, 10:44 PM
Ok, let me rephrase that question and clarify.
I'm ok with definitions we all use and they are not my concern. What I would like all of us to note is that we do not see the entire hemisphere of any round object, no matter how far away it is from us (the farther we go the more we see, but never 100% of a hemisphere). Given that, I wonder how long exactly it takes between one terminator dissapearing on one edge of the Moon and another appearing on the other for an observer located on Earth.

Hornblower
2008-Nov-13, 01:37 AM
What do you mean by an "official" answer? By whose authority would such an answer be considered "official"?

Anyone with the orbital elements and the necessary mathematical skill can calculate the interval during which the entire visible portion of an idealized smooth sphere would be receiving light from at least part of the Sun's disk.

PraedSt
2008-Nov-13, 02:28 AM
Given that, I wonder how long exactly it takes between one terminator dissapearing on one edge of the Moon and another appearing on the other for an observer located on Earth.

The moon is in constant motion. So the answer to your question is: the lit area of the moon, during what we call a 'full moon'....that area is maximum for an instant, and only an instant.

Our full moons last longer, ~1 night, because we're not so picky! :)

hhEb09'1
2008-Nov-13, 02:41 AM
The moon is in constant motion. So the answer to your question is: the lit area of the moon, during what we call a 'full moon'....that area is maximum for an instant, and only an instant.And, since the path of the moon can deviate five degrees from the ecliptic, and the earth is only two degrees in the lunar sky, many times, no observer on earth might see a "full" moon. I think that's what pghnative was getting at.

PraedSt
2008-Nov-13, 02:51 AM
just commenting that the ideal of 50% lit seems impossible to me, and would only occur for a brief instant in time.

And, since the path of the moon can deviate five degrees from the ecliptic, and the earth is only two degrees in the lunar sky, many times, no observer on earth might see a "full" moon. I think that's what pghnative was getting at.
Yes, definitely, my bad. I got distracted by the OPs following post:

Ok, let me rephrase that question and clarify.
So that's the question done isn't it? No-one sees 50%; and whatever anyone does see, only lasts an instant.

Veeger
2008-Nov-13, 05:15 AM
So that's the question done isn't it? No-one sees 50%; and whatever anyone does see, only lasts an instant.

Initially I would agree - but - perhaps now, I am becoming caught up in the OP and doing my own nitpicking.

Since we only see a portion of the hemisphere, (the amount dependends on the current distance to the moon and the position in the sky relative to the observer) the duration of the full moon would be the time it takes the entire visible surface to be lit until, the trailing terminator just enters the visible area. It would be longer than an instant but the time would depend on the geometry of time and place so there is no simple answer in my opinion.

:)

tony873004
2008-Nov-13, 06:33 AM
I watched a full moon in a telescope that was headed for eclipse. An hour before 1st penumbral contact, I could see shadows on one edge. The cool thing was that the shadowed limb flipped nearly 180 degrees, through the south pole just prior to penumbral contact. I fell asleep before penumbral contact, but I was very convinced that a shadowed limb would never disappear from view. Hence, there is no such thing as a full moon.

I've always wanted to try this again. As others have pointed out, a full moon at perigee might actually become truly full. Next time we have clear skies during a lunar eclipse, I intend to find out.

PraedSt
2008-Nov-13, 07:22 AM
Initially I would agree - but - perhaps now, I am becoming caught up in the OP and doing my own nitpicking...It would be longer than an instant but the time would depend on the geometry of time and place so there is no simple answer in my opinion.:)
Good nitpick. You've made me work out a formula.

Ok, a simple model:
1. The Sun is very, very, far behind your head.
2. The Moon rotates around your head every 28 days.
3. The inclination of the Moon's orbit is 0 deg.
4. The Moon's orbital radius is constant.

Then let p = the proportion of the Moon's hemisphere we can see. At 'full Moon', all of p will be lit.
Now let T = the time, in days, p remains lit; which is what I think the OP is asking.

Then I get: T = 14*(1-p)

Feel free to diss/improve... :)

hhEb09'1
2008-Nov-13, 10:59 AM
Feel free to diss/improve... :)Well, that didn't get us very far, did it? :)

Simplifying assumptions (http://www.planetscapes.com/), ignoring inclination, oblateness, eccentricity, libation, our height and the distance between our two eyes.
1. The sun is 149600000 km away, with a radius of 695000 km
2. The moon is 384400 km away, with a radius of 1737.4 km
3. The lunar day is 29.53059 days, the lunar rotation period is 27.32166 days
4. The earth radius is 6378.14 km, with a rotation period of 23.9345 hours

When the earth, moon, and sun are aligned, a line tangent to both moon and sun would be perpendicular to each radius. So, the lit surface of the moon would be 180 degrees plus 2 x arcsin ( (695000 - 1737.4) / (149600000 + 384400) ), or 180.53 degrees (OK, 180.5296705859504540827400437424514 according to usoft).

In order to avoid eclipse, the moon must be a half degree away from the ecliptic at "full" moon, so someone on earth should be able to see that extra quarter degree, like tony873004 says.

buczas
2008-Nov-13, 07:35 PM
Since we only see a portion of the hemisphere, (the amount dependends on the current distance to the moon and the position in the sky relative to the observer) the duration of the full moon would be the time it takes the entire visible surface to be lit until, the trailing terminator just enters the visible area. It would be longer than an instant but the time would depend on the geometry of time and place so there is no simple answer in my opinion.


Yes, that's exactly my point. We can as well apply my question to an imaginary rotating sphere coloured black (left hemisphere) and white (right hemisphere). As the sphere rotates we see the colours change but due to previously mentioned factors, there is a delay between one terminator dissapearing and another one appearing on the opposite side. My question - what is that delay, assuming that the sphere is the size of the Moon, and just as far from the observer. The fact noticed by hhEb09'1 (about there being more than 50% of the Moon lit by the Sun) can be included later in this thinking, but let's keep things simple for now.

hhEb09'1
2008-Nov-13, 08:12 PM
For a person standing on the earth directly below the moon, the reduction in angle of the view of the surface of the moon from 180 degrees is 2 x arcsin ( 1737.4 / ( 384400 - 6378.14) ) or 0.53 degrees (0.52666815505425298379446469379299). So, the sun illuminates .53 more degrees, but we see .53 + .53 less than that, or a little more than a full degree. The coincidence of the .53s is the same coincidence that the sun in the sky is about the same size as the moon in the sky, both a little more than a half degree.

buczas
2008-Nov-13, 08:58 PM
So, with a lunar day being about 708 hours long the angular speed of the terminator is about 0.5 deg/h. It then gives two hours of visible surface without any terminators visible (correct me if I'm wrong, I'm no mathematician). That's close to antoniseb's calculations and this answer suits me fine.
Thanks!

mugaliens
2008-Nov-13, 11:04 PM
I know that the "Full Moon duration" question has been asked and aswered many times BUT I find the "official" answer not satisfying.
To add complexity to the question I would like you to take into account that we (observers on Earth) do not see exactly the 50% part of the Moon. So - having that in mind - how long do we see that part fully lit during a Full Moon?

To visualise the concept think of an observer on a rocket launching from Earth - he sees more and more of the Earth's surface, but he'll never see the entire hemisphere, no matter how far he gets. Apply that to an observer on the Earth looking at the Moon.

The Great Carnac:

Answer: "The numbers 363,104 km, 405,696 km, 1,738.14 km, 1,735.97 km, 27.321582 days, 24 hours, 152,097,701 km, and 147,098,074 km."

Question: "The measurements required to compute the duration of a full Moon as observed from Earth, not withstanding atmospheric distortion."

PraedSt
2008-Nov-14, 01:01 AM
For a person standing on the earth directly below the moon, the reduction in angle of the view of the surface of the moon from 180 degrees is 2 x arcsin ( 1737.4 / ( 384400 - 6378.14) ) or 0.53 degrees (0.52666815505425298379446469379299). So, the sun illuminates .53 more degrees, but we see .53 + .53 less than that, or a little more than a full degree. The coincidence of the .53s is the same coincidence that the sun in the sky is about the same size as the moon in the sky, both a little more than a half degree.
Good figures hhEb09'1 :)
Ok, so we see a bit less than one full degree from 180? Is that right? Sorry, am feeling dense this week/month. So ~179 deg.
Plugging that into my simple formula, that makes p = 0.9941
Which makes T = 14*(1-p) = 0.0824 days or ~ 2 hours. Phew!
Thanks hhEb09'1!

hhEb09'1
2008-Nov-14, 09:45 AM
Good figures hhEb09'1 :)
Ok, so we see a bit less than one full degree from 180? Is that right? Sorry, am feeling dense this week/month. So ~179 deg.
Plugging that into my simple formula, that makes p = 0.9941
Which makes T = 14*(1-p) = 0.0824 days or ~ 2 hours. Phew!Almost! we see a half degree less, but the sun illuminates a half degree more. (To convince anyone that this is the same coincidence as the full eclipse coincidence, think of the sun perfectly eclipsed by the moon--the dark part of the moon we'd see would line up exactly at the edge of the part illuminated on the other side by the sun. So, the amount that we don't see of 180 degrees is matched by the amount the sun illuminates past 180.) I think that final answer is the same though, matching buczas (http://www.bautforum.com/questions-answers/81205-duration-full-moon-viewed-earth.html#post1364933)'s. However, because the Earth subtends an angle of two degrees in the lunar sky, whenever the moon is not in full eclipse, it will be away from that geometry and many if not most people on Earth would see beyond the 180.53 line, maybe from the "bottom". That was tony873004 point.
Thanks hhEb09'1!yw :)

Jeff Root
2008-Nov-14, 11:57 AM
To convince anyone that this is the same coincidence as the full eclipse
coincidence, think of the sun perfectly eclipsed by the moon--the dark
part of the moon we'd see would line up exactly at the edge of the part
illuminated on the other side by the sun.
Very nice!

-- Jeff, in Minneapolis

PraedSt
2008-Nov-14, 12:45 PM
Yeah, I thought that was pretty nifty too. Good one hhEb09'1. You'd write a good textbook! :)

hhEb09'1
2008-Nov-15, 06:22 PM
Very nice!Thanks Jeff, Praed. Of course, the example here is with the moon on the opposite side of the earth from the sun, so the geometry is not exact--but then, it's not exact period, what with the eccentricity of the orbits.

I also noticed that the diagrams that I was drawing to calculate the deficit from 180 degree that we see of the moon are the same diagrams I use to calculate the apparent diameter of the moon. In other words, the reason we see .53 degrees less of the moon is because the moon looks like it's .53 degrees in our field of view.