Moon brightness question...

Here's one I was discussing with a friend, wondered if anyone here could
"cast any light"...

If you view the full moon, it doesn't look shaded, like a sphere. That is,
there is no discernible decrease in brightness towards the limb, even though
the glancing angle the sun's light makes with the surface is shallower and
shallower, so presumably less light should be reflected back to the viewer.

You don't see this with asteroids, so what is it about the moon's surface
that gives it this property?

Thanks!

Glass spherules?

Pierre MK-UK

The Moon has no atmosphere. With no atmosphere to reflect light from the Sun
around it's surface the boundary between night and day is very abrupt.
Either full exposure to the Sun or complete darkness is seen. If you do see
any detail on the night side of the Moon you are viewing it with Earthshine,
the light reflected off the Earth to the Moon.

James


"MichaelJP" wrote in message
...
Here's one I was discussing with a friend, wondered if anyone here could
"cast any light"...

If you view the full moon, it doesn't look shaded, like a sphere. That is,
there is no discernible decrease in brightness towards the limb, even
though the glancing angle the sun's light makes with the surface is
shallower and shallower, so presumably less light should be reflected back
to the viewer.

You don't see this with asteroids, so what is it about the moon's surface
that gives it this property?

Thanks!

Hmmm. Maybe...

The parts of the lunar sphere away from the centre (as viewed from
earth) are "angled away" from you. The closer you get to the edge, the
more "angled away" it is.

Therefore - any "square arcsecond" viewed near the edge actually
contains more lunar surface area within it's boundaries than a square
arcsecond viewed at the centre of the lunar disc.

So although there IS reflected intensity dropoff per unit area on the
moons surface (as you move toward the edge), this is balanced by seeing
more "unit areas" of the moons surface within a given square arcsecond
of view.

So the brightness appears pretty much constant across the disc.

This may also be utter bullpucky - but it seems logical to me...

Cheers
Beats

"justbeats" wrote in message
oups.com...
Hmmm. Maybe...

The parts of the lunar sphere away from the centre (as viewed from
earth) are "angled away" from you. The closer you get to the edge, the
more "angled away" it is.

Therefore - any "square arcsecond" viewed near the edge actually
contains more lunar surface area within it's boundaries than a square
arcsecond viewed at the centre of the lunar disc.

So although there IS reflected intensity dropoff per unit area on the
moons surface (as you move toward the edge), this is balanced by seeing
more "unit areas" of the moons surface within a given square arcsecond
of view.

So the brightness appears pretty much constant across the disc.

This may also be utter bullpucky - but it seems logical to me...

Cheers
Beats


Hmm.. not sure about the calculation there, but wouldn't that apply to any
sphere illuminated fully? Certainly if you illuminate a model of the moon,
you will not see the same effect as the real moon.

It must be something to do with the properties of the lunar surface itself.
Most surfaces shade by a cosine rule, where full brightness falls off to
zero smoothly as the surface turns from face on to 90 degrees to the light
source. It seems that the lunar surface reflects 100% until very nearly 90
degrees, then suddenly reflects nothing back...

"MichaelJP" wrote in
:

"justbeats" wrote in message
oups.com...
Hmmm. Maybe...

The parts of the lunar sphere away from the centre (as viewed from
earth) are "angled away" from you. The closer you get to the edge, the
more "angled away" it is.

Therefore - any "square arcsecond" viewed near the edge actually
contains more lunar surface area within it's boundaries than a square
arcsecond viewed at the centre of the lunar disc.

So although there IS reflected intensity dropoff per unit area on the
moons surface (as you move toward the edge), this is balanced by seeing
more "unit areas" of the moons surface within a given square arcsecond
of view.

So the brightness appears pretty much constant across the disc.

This may also be utter bullpucky - but it seems logical to me...

Cheers
Beats


Hmm.. not sure about the calculation there, but wouldn't that apply to
any sphere illuminated fully? Certainly if you illuminate a model of the
moon, you will not see the same effect as the real moon.

It must be something to do with the properties of the lunar surface
itself. Most surfaces shade by a cosine rule, where full brightness
falls off to zero smoothly as the surface turns from face on to 90
degrees to the light source. It seems that the lunar surface reflects
100% until very nearly 90 degrees, then suddenly reflects nothing
back...


Reflection is probably not the right term. It may depend on the way the
rough surface of the moon scatters the incident light.

I found the following article indicating that the moon does show
significant limb darkening at radio wavelengths.

http://www.df5ai.net/ArticlesDL/EMEP...MEscatter.html

Klazmon.

MichaelJP wrote:
If you view the full moon, it doesn't look shaded, like a sphere. That
is, there is no discernible decrease in brightness towards the limb,

Moon is not a smooth sphere. It has a surface that can be considered to
be almost randomly orientated reflectors. Additionally, the apparent
(from our perspective) increased concentration of these towards the limb
tends to compensate for the "almost" in the previous sentence.


Best,
Stephen

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"Stephen Tonkin" wrote in message
...
MichaelJP wrote:
If you view the full moon, it doesn't look shaded, like a sphere. That is,
there is no discernible decrease in brightness towards the limb,

Moon is not a smooth sphere. It has a surface that can be considered to be
almost randomly orientated reflectors. Additionally, the apparent (from
our perspective) increased concentration of these towards the limb tends
to compensate for the "almost" in the previous sentence.


Best,
Stephen

Interesting - I suppose that this random scattering effect means that the
normal shading rules don't apply then.

Also I guess that the similar effect seen on planets such as Venus is due to
random scattering as well, but this time from a gaseous surface?

Wasn't it justbeats who wrote:
Hmmm. Maybe...

The parts of the lunar sphere away from the centre (as viewed from
earth) are "angled away" from you. The closer you get to the edge, the
more "angled away" it is.

Therefore - any "square arcsecond" viewed near the edge actually
contains more lunar surface area within it's boundaries than a square
arcsecond viewed at the centre of the lunar disc.

So although there IS reflected intensity dropoff per unit area on the
moons surface (as you move toward the edge), this is balanced by seeing
more "unit areas" of the moons surface within a given square arcsecond
of view.

So the brightness appears pretty much constant across the disc.

This may also be utter bullpucky - but it seems logical to me...

I've started checking every sufficiently curved object I can find. I
position it in full sunlight and look at it from an angle as close as
possible to a full moon configuration - so that the shadow of my head
just misses it.

So far everything I've checked exhibits limb darkening.

--
Mike Williams
Gentleman of Leisure