Earth From Mars Options

[paxdan]

they should post in advance the time they are doing this so we can go outside and wave.


what are the chances that MSL will be able to resolve the moon too?

[Guest_Richard Trigaux_*]

what are the chances that MSL will be able to resolve the moon too?

Difficult to say on this image!!

The Moon shows 13 time less surface than Earth, and its Albedo is 7% against 35% for Earth. This means that it is 65 times less luminous.

On the picture Earth occupies 3-4 pixels large, but I doubt it is its real dimention, rather it is smearing from high exposure or optical imperfections.

So we should expect that the Moon is visible as a 1 pixel spot, white or at least clearer than the background.

More, Oppy is on the equator of Mars, so that the Moon ellipse is expected to be near the vertical. But I have no idea of its size.

The worse difficulty is in fact the JPEG noise of the image. Increasing contrast with a graphic processor makes appear many jpeg spots, but some possible candidates also appear:



Best candidates:
65 pixels at right and 125 pixel above
78 pixels at left and 51 pixels lower.
Lesser candidates (more likely JPEG noise)
8 pixels at left and 26 above
22 pixels at left an 6 under

That makes two candidates, but one or both of them could be a star, or just jpeg noise.

Jpeg images are bad for this kind of sport, as they add much speckles which can be misinterpreted as bright spots, or they can even make disappear a real spot. PNG images are better, and they can be made so what not to take much more download time.

All this makes Earth very bright seen from Mars, much more than many stars!!

Attached thumbnail(s)

 

[ElkGroveDan]

The worse difficulty is in fact the JPEG noise of the image. Increasing contrast with a graphic processor makes appear many jpeg spots, but some possible candidates also appear:
Best candidates:
65 pixels at right and 125 pixel above
78 pixels at left and 51 pixels lower.
Lesser candidates (more likely JPEG noise)
8 pixels at left and 26 above
22 pixels at left an 6 under

That makes two candidates, but one or both of them could be a star, or just jpeg noise.

According to the JPL caption this color image is created from three photos (as you would expect). I suggest separating the layers and looking for a dsitinctive speck common to all three.

[dilo]

This is a sum of L4+L5+L6 images probably used to generate Nasa image:

Obviously, there isn't compensation for motion and Earth appear as a sequence of 3 aligned colored points slightly to the left of the image center.
Based on image time, I simulated the view with StarryNight software:

Considering that each pixel of PanCam is 58 arcsec wide, real Earth image should be a fraction of pixel and Earth-Moon apparent distance is only 5 pixel!
So none of Richard Moon "candidates" satisfy calculated orientation/distance... moreover, consider that first image show a lot of "false stars" (comparable to Earth in luminosity but visible only in one image) which should be cosmic ray hits and/or hot pixels...

[Bubbinski]

I wish I were on Mars with a small telescope so I could look at Earth from that vantage point. Mars is right now about magnitude -0.2 in the morning sky on Earth...I suspect Earth would be brighter since it's larger. After all, Venus is practically the same size as Earth and is at almost -4 magnitude in our evening sky.

1 AU = 150 million km or 92 million miles. Right now, on the Solar System Live website http://www.fourmilab.ch/cgi-bin/uncgi/Solar - Venus is 1.623 AU from Earth and Earth is 1.162 AU from Mars. So that translates into Venus being 149,316,000 miles from Earth and Earth being 106,904,000 miles from Mars as of this morning. Also keep in mind, however, that Venus has a very bright, very reflective cloud cover and Earth wouldn't reflect as much sunlight back out into space. So I'd guess on about a -3, maybe -3.5 magnitude for Earth from Oppy's vantage point on Mars.

Anyway, if the rovers ever became immobile but were still operational otherwise, couldn't the Pancam be programmed to track Earth with the precision of a telescope's clock drive and observe it that way?

Bubbinski (owner of a 90 mm Orion Mak telescope)

[dilo]

I wish I were on Mars with a small telescope so I could look at Earth from that vantage point.  Mars is right now about magnitude -0.2 in the morning sky on Earth...[]  So I'd guess on about a -3, maybe -3.5 magnitude for Earth from Oppy's vantage point on Mars.  

Anyway, if the rovers ever became immobile but were still operational otherwise, couldn't the Pancam be programmed to track Earth with the precision of a telescope's clock drive and observe it that way?

Bubbinski (owner of a 90 mm Orion Mak telescope)

You aren't too far from right: as you can see from my simulation, actual magnitude of Earth from Mars is -2.9... unfortunately, moon is 1/40 of Earth luminosity (magnitude +1.2) and the strong luminosity contrast, joined to the small apparent distance, makes it invisible.
This became even more clear when I tryed to make a super-resoultion image of Earth, summing enlarged and precisely shifted 9 images taken with different filters (in a time window of less than 5 min); result is compared with simulation with the same scale:

Earth image is slightly elongated by motion during exposure but is clear that, even if the PanCam could compensate for sky motion (and this shouldn't possible) resolution would be too low for any useful astronomical observation of Earth...

Regards.
Marco.

[Guest_Richard Trigaux_*]

Good work Dilo,

I must admit that none of my candidates was right.

I think my candidates were white (decomposing in RGB would not have eliminated them).

This is an evidence that jpeg images are really tricky with small details.

Worse a bright point like here is often surrounded with a zone of still poorest rendering.
Is it possible that the original record in a non-compressed format may show the Moon? About resolution, Earth and Moon must de separated. About luminosity, if you consider that a white pixel has a value of max 256, and if you divide by 40, this still makes a brightness of minimum 6. So in order to the Moon being visible, we need a very dark sky, or at least the noise must be weaker than 6. Fortunatelly the Earth is overexposed, and thus the luminosity of the Moon must be much larger than 6, ans so I think it could be visible on the original non-compressed format. Where the compression takes place? If it is aboard Opportunity, there is no hope.

[helvick]

If it is aboard Opportunity, there is no hope.

Nice work all round folks.

One question for any knowledgeble astronomer types- Given that Mars is further out than earth would it be possible to take a shot of earth nearer earth/mars conjunction when the earth would be a small enough crescent not to completely drown out the moon which could be half illuminated provided it was timed right? Or does conjunction geometry mean that Sun's position means that these would be daylight shots which would almost certainly drown out the moon.

This Venus Crescent shows the sort of thing I'm talking about.
Venus Crescent

[Tman]

Based on image time, I simulated the view with StarryNight software:

Hi dilo, great work, but could it be that your StarryNight pic shows the view mirror-inverted like through a telescope? Because of Earth is still behind Mars by orbit sun (left of the sun on Mars). For example take a look at: http://deepimpact.umd.edu/amateur/where_is.shtml

[garybeau]

I tried the same thing with Starry Night.
Date = 5/29 1 hour after sunset. (12:12 UT)
Location = 1° 58.8' S 5° 56.4' W Meridiani Planum
Pan Cam FOV 16.8 deg x 16.8 deg

The moon came out at 11 o'clock position.

Did I do something wrong?

http://i6.photobucket.com/albums/y207/gary...itledimage1.jpg

http://i6.photobucket.com/albums/y207/gary...itledimage2.jpg

http://i6.photobucket.com/albums/y207/gary...itledimage3.jpg

http://i6.photobucket.com/albums/y207/gary...itledimage4.jpg

http://i6.photobucket.com/albums/y207/gary...itledimage5.jpg



Gary

Edit: Correct location 1° 57' S 5° 31.8' W

[4th rock from th...]

Nice work all round folks.

One question for any knowledgeble astronomer types- Given that Mars is further out than earth would it be possible to take a shot of earth nearer earth/mars conjunction when the earth would be a small enough crescent not to completely drown out the moon which could be half illuminated provided it was timed right? Or does conjunction geometry mean that Sun's position means that these would be daylight shots which would almost certainly drown out the moon.

This Venus Crescent shows the sort of thing I'm talking about.
Venus Crescent

Let's see... When Mars is at opposition, it's closer to the Earth and fully illuminated by the Sun (as seen from Earth). At this time, if you were on Mars, the Earth would be between the Sun and Mars. So it would be like New Moon and invisible, unless it transited the Sun. So, from the Martian point of view, Earth is just like Venus for us.
The Moon very close to the Earth and would show the same phases as Earth.
So if Earth is at 1st quarter (when seen from Mars) the Moon whould show the same phase.

[Guest_Richard Trigaux_*]

As "4th rock from the sun" says, and countrarily to helvick hope, Earth and Moon, seen from mars, have alway the same phase, so their luminosity vary accordingly, and the contrast is constant.

However we observe, with Venus seen from Earth, that there is a moment, say 1 month before and after conjuction, where the luminosity is maximum. We can expect a similar phenomenon to take place with Earth seen from Mars, and in this case the Moon is brighter, and thus more visible.

So what I think is that a human on Mars would see the Moon, at least near the conjunction, both separation and luminosity being good enough, provided that there is no large town drowning the sky with its light. We could too imagine what mythology and legends the Earth's companion would have inspired to a supposed Mars civilization...

[dilo]

Sorry for late reply, at certain point yesterday my attention was deviated by great news about Opportunity status...
Gary/Tman, you are right... when I simulated with StarryNight, I was interested only to apparent distance and magnitudes, so I didn't used exact Opportunity location and I also rotated image in order to obtain illumination from low. This my (wrong) guess generated lot of misunderstanding and I apologize for this, but, I repeat, initially I wasn't interested to orientation!
Richard, about luminosity of pixel, real Moon value should be about half of your estimation, because sky (background) level is close to 120 in this region and we must subtract this contribute from Earth luminosity... now, a luminosity increase of 3 is comparable to the typical noise fluctuation of background, even in the less noisy image (1p168067125esf55dip2682l1m1, see enlarged detail below):

This make impossible to identify the moon, also because, based on correct simulation, long exposure caused a partial overlap of Earth and Moon trails...
Yes, part of the noise is due to jpeg artifacts which can be surely reduced using original images. In fact, also MER normally transmit compressed images to us, but it's compression algorythms are more efficient than jpeg and contains with less artifacts... most important, original images have a 12-bit precision, and this should make more easy do detect weak details!
I hope someone in this Forum can access to original 12-bit images... anyway, probably they will be published here: http://anserver1.eprsl.wustl.edu/
Thanks to all contributors.

[edstrick]

Richard Triquaux said: As "4th rock from the sun" says, and countrarily to helvick hope, Earth and Moon, seen from mars, have alway the same phase, so their luminosity vary accordingly, and the contrast is constant.

That only assumes their phase functions are identical. Earth's atmosphere is forwared scattering, while the relatively dark powder of the lunar regolith is highly backscattering. At low phase angle, say a full Earth and full Moon seen from Venus, the moon will be relatively bright, though still much darker than Earth. At high phase angle, a crescent Earth and Moon seen from Mars, the moon will be much darker in proportion than the Earth.

[dvandorn]

they should post in advance the time they are doing this so we can go outside and wave.


what are the chances that MSL will be able to resolve the moon too?

I have a very important reason for quoting the original post, pic and all.

You see, the amount of time it takes to make an exposure like this means the picture just isn't going to come out well.

The problem here is obvious. The reason the Earth isn't as distinct as it could be is obvious.

Somebody moved.

-the other Doug