Apparent magnitude of Earth from moon?

[Evil Dr Ganymede]

We know the apparent magnitude of a full moon from Earth is about -13.

But what's the apparent magnitude of a full Earth as seen from the moon? I know it'd be brighter, but I've never seen a number for this...

What would be the apparent magnitude of a "half-full Earth" from the moon too, for that matter?

[tony873004]

Just a guess....
The Earth's radius is 4 times larger than the moon's radius, so it's surface area of its visual disk will be 16x greater than the Moon disk from Earth.

The Moon is a dark almost charcoal black color. The Earth has bright white cloulds, snowy plains and continents, dark land, and dark water except where the Sun reflects. That part is bright.

And sometimes its cloudier than other times, sometimes the Sun's reflective point is over land or clouds rather than open ocean. If you ever look at Earthshine on a crescent moon, you'll notice that some days it's brighter than other days, and can even change from one hour to the next.

So my guess .... 16x + some varying amount.

A better way to tell... Get a camera with a zoom lens large enough that the moon easily fills the spot meter. When there's a crescent moon, Center the dark side of the moon where the earthshine is. Becareful not to let any of the bright sunlit side or darkness of space enter the spot meter. Record the shutter speed that your light meter wants for a given apatuer and ISO setting. Then wait for a couple weeks for a large Gibbous moon. Take your camera, with the same settings for apateure and ISO and meter the ground you're standing on. Since the Moon is a dark color, find some dark ground, maybe asphalt. There's your difference.

There's an experiment I've always wanted to do but it would require 2 people in different parts of the world. Use Celestia to track the Earth from the Moon, and pay attention to the Sun's reflective point. Find an opportunity where this reflective point is on land, but then crosses a large lake or sea. Have somebody in that part of the world report on weather conditions at the time to make sure it's not cloudy (maybe a weather report or satellite pic would work as well) Have somebody in the part of the Earth that can see the crescent Moon illuminated with Earthshine and see if there's a big temporary jump in the intensity of Earthshine on the moon during the time that the shiny spot crosses the water.

[Evil Dr Ganymede]

So you reckon it'd be less than x100... x100 is a difference of 5 magnitudes isn't it?

I wonder what the apparent magnitude of the earthshine seen on a new moon itself is...

Actually, I'll crosspost what I was asking about in the Users board here, because there are some astronomical questions here:
-------------------------------------------------------------

Right. That's intentional. Your planet orbits A with a semimajor axis of 0.4, so when B is 11au, A is still 27.5 closer. The 1/R^2 attenuation of the electromagnetic force means that B will be a dimmer by a factor of about 1/750. Also, A is intrinsically over 3.5 times brighter than B, so A is going to appear over 2500 times brighter than B. Celestia adjusts the apparent light source brightnesses with a power function; with the exponent I've chosen, light from B should be just barely visible. I could adjust the exponent to bring out the light from B more, but I'm not sure that this would be more realistic.

OK, just for ease of reference, here's the Prometheus CEL URL I'm referring to all the time. You'll need the code from the thread on the Bugs board too.
cel://PhaseLock/Rigel%20Kentaurus%20A:Prometheus/Rigel%20Kentaurus%20B/2366-02-26T19:54:24.16683?x=AI5ZmJkqPKaB7+b//////w&y=oEJZ4FCPrA8sndL//////w&z=qtRYzZxDdzTDiCo&ow=-0.367940&ox=-0.121596&oy=0.674201&oz=-0.628719&select=Rigel%20Kentaurus%20B&fov=20.765589&ts=1.000000<d=0&rf=38711&lm=71

Now, the apparent magnitude of B over the range of 11 to 30 AU (or whatever the extremes are) is from -18 to -20. The full moon has a magnitude of only about -13. So it's what, at least 100 times bright than the full moon, even when it's furthest from the planet... and yet it provides no visible illumination? As it is the full moon casts noticeable, visible shadows when it's up. In this case B should certainly do the same if you were standing on the surface.

Because B isn't illuminating the planet at all though, I presume that B wouldn't cast ring shadows if Prometheus had a ring either. I could believe that it'd be hard to see the shadow while A was above the horizon, but you would surely see a shadow cast by B if it was the sole illuminator given what I just said.

Look at the crescent moon in the sky, and you can clearly see earthshine. Wouldn't B's illumination look similar to that? I wonder what the apparent magnitude of the full Earth is as seen from the moon?

Though the thing with earthshine is that you can see it when the moon is new/crescent, because the Earth is full as viewed from that hemisphere on the moon. Do we not see earthshine when the moon is "half full" because the illuminated side is too bright for us to see the dark side? Or is it because the Earth would also be "half full" as seen from the moon at that time, and that lowers the apparent magnitude of Earth enough that it doesn't illuminate the moon's surface enough to make it visible anymore? I suspect it's more the latter than the former...

[selden]

Oh Evil One,

I'm confused. What I think you're saying you're seeing is not at all what I'm seeing when using Celestia v1.4.0pre3.

I substituted a surface texture of "astar.*" so that Prometheus is white, making it easier to see the illumination. (Omitting the surface texture does result in an incorrect illumination.)

Here's what I see when using "Render path: Basic"
with Ambient Light set to None.



The planet is illuminated on both sides, both from the nearby star and from the more distant one.
(I'm pretty sure the shadow band is a bug, though. I don't think Chris has had a chance to update the algorithm used for terminator shading.)

I did upgrade my graphics card last week, but my previous tests convinced me that the lower Render paths are drawn exactly the same.

System:
256MB 500MHz P3, Win XP Pro SP2
128MB FX5700LE, ForceWare v66.81 (beta)
Celestia v1.4.0pre3

[selden]

Oh, Evil One,

Adding rings was easy: I just inserted these lines in the definition of Prometheus:

Code: Select all

   Rings {
      Inner    7450
      Outer  14022
      Texture "saturn-rings.png"
   }


They're just Saturn's rings with 1/10 the radii (I lopped off the trailing 0s).

Here's what Celestia v1.4.0pre3 draws for the Render paths available on my system.

Render path: Basic = no ring shadows.

Render path: Multitexture = planet shadow on rings, no ring shadows on planet.

Render path: OpenGL Vertex program = buggy shadow

Render path: OpenGL Vertex program + Nvidia combiners = buggy shadow

Render path: NVIDIA GeForce FX = brightest only

Render path: OpenGL 2.0 = both shadows OK.

I believe that there is a bug in the two OpenGL vertex shaders paths, since the shadows are being cast through the planet onto the "dark" side.

The others demonstrate limitatations in what's available.

Note that in the v2.0 path, you can see shadows of the rings cast onto the planet by both stars. If you look closely, you can see both shadows of the planet on the rings, too.

[chris]

The planet is illuminated on both sides, both from the nearby star and from the more distant one.
(I'm pretty sure the shadow band is a bug, though. I don't think Chris has had a chance to update the algorithm used for terminator shading.)

The shadow band is accurate--at least, it's as accurate as the assumption that the surface is Lambertian. The dark band is the area where the light from both stars arrives at a direction almost normal to the surface.

--Chris

[Fafers]

We know the apparent magnitude of a full moon from Earth is about -13.

But what's the apparent magnitude of a full Earth as seen from the moon? I know it'd be brighter, but I've never seen a number for this...

Just a guess:

The area of Earth's disk as seen from the Moon is 13.5 times the area of the full Moon as seen from Earth. So:

ratio=13.5*(albedoE/albedoM)
ratio=13.5*(0.35/0.07)=67.5 times
(some 4.6 magnitudes)

[Evil Dr Ganymede]

The planet is illuminated on both sides, both from the nearby star and from the more distant one.
(I'm pretty sure the shadow band is a bug, though. I don't think Chris has had a chance to update the algorithm used for terminator shading.)

The shadow band is accurate--at least, it's as accurate as the assumption that the surface is Lambertian. The dark band is the area where the light from both stars arrives at a direction almost normal to the surface.

--Chris

I haven't had a chance to check what Selden's done on my own PC, but the only time I got that shadowband was when I had the planet orbiting the Barycentre - but that's not how my system looks. Selden, did you use the exact orbital configuration that I did - ie did you put the planet around A?

[selden]

Oh, most Evil One,

Yes, it's orbiting A.

*However*, I just tried it on my system at work, which has an FX5200, and it shows just one side being illuminated!

I think the major difference is the driver version. At home I'm running ForceWare v66.81 beta, downloaded from Nvidia's Web site. I haven't had time to upgrade the system I use at work, which is still running v61.76.

It's lunchtime now, so I'll do a system backup and try installing the beta drivers.

Chris,

Are you allowed to say what version of drivers you've been using for development?

[Evil Dr Ganymede]

Hrm. I'm using driver version 61.77...

[selden]

I was sure I'd posted the results of my upgrade here, but apparently not. The single-sided illumination on my system at work was caused by a typo in the STC catalog file which prevented the second star from being considered part of the same system. It was not caused by the version of the graphics driver.