# Where in the Pale Blue Dot image should the Earth's moon be?

In the description of this image it is claimed that Moon should be detectable in famous "Pale Blue Dot" image.

Detailed analysis also suggests that Voyager detected the moon as well, but it is too faint to be seen without special processing.

https://photojournal.jpl.nasa.gov/catalog/PIA00450

And recently NASA released reprocessed version of it by Kevin M. Gill to bring out the details using latest image processing techniques. https://photojournal.jpl.nasa.gov/catalog/PIA23645

But there isn't any mention of Moon or where it should be in the image with respect to Earth so I was hoping someone can work out where Moon is supposed to be, perhaps image processing masters have coaxed it out unknowingly and we are just not paying attention!

• I heard Carl Sagan on TV saying "the Earth was even less than a single pixel". So there would be only two pixels separated by a few dark pixels. – Uwe Feb 13 '20 at 21:50
• "Earth was a crescent only 0.12 pixel in size". This would mean the moon should be (384000/6371)*0.12 ~ 7.2 pixels from the earth. Suffice to say, I haven't found anything like that in either image, but I didn't get to look at the raw data. – Polygnome Feb 14 '20 at 8:48
• Moons diameter is only 0.2727 of the Earths diameter, so Moon was only 0.032 pixel in size. Was the camera able to detect such a tiny and dark spot? – Uwe Feb 14 '20 at 12:11
• My WAG (wild -A -guess) is that the "special processing" involves many images over time & a data-fitting analysis done to find positional perturbations or occultations. Similar to our indirect detection of planets around other stars. – Carl Witthoft Feb 14 '20 at 13:14
• @Uwe what you forget is that the object size , for all objects less than a pixel, is irrelevant. The camera pixel limits resolution, but it's the object's brightness vs. the camera SNR which determines whether that pixel goes "bright" – Carl Witthoft Feb 14 '20 at 13:16

# A bit below and to the left:

Note that the size of the red dots is not representative of the size of the Earth or Moon.

### Where was Voyager 1 when the photo was taken?

According to the catalog page for PIA23645 (also linked above). The photograph was taken 1990-02-14 04:48 GMT. This is an abomination; GMT hasn't existed since the 70s; one presumes UTC was intended instead. Converting this (see SOFA iauDtdb(...), currently using the Fairhead and Bretagnon model) produces 04:48:57.1850933 TDB (we could go with :49, given that the original precision was to the nearest minute; this definitely doesn't matter that much). Using the JPL Horizons data API on this time tells me that at that instant, Voyager 1 was at coordinates (in AU referred to the ICRF/J2000 reference frame) of:$$\begin{bmatrix} X\\Y\\Z \end{bmatrix} \approx \begin{bmatrix} -15.74203225317424\\ -30.05725969411413\\ +21.85720068189130 \end{bmatrix}$$

### Generating a simulation

We can now simply take this position, put a virtual camera in an orrery there, and try to match the contextual image given in PIA00450 (also linked above).

I used my own orrery, which is based off of more JPL Horizons data. This orrery correctly takes into account the offset of the Earth from the Earth–Moon barycenter and the Sun from the solar system barycenter (although in both cases there is little difference at the time of interest).

I measured the location of Earth, Venus, and the center of the Sun on PIA00450, and then manually tried to adjust the virtual camera's field of view, direction, and rotation to get a match. I wasn't able to get all three points to line up perfectly (although I could get them very close). I attributed this to the necessarily high amount of error in trying to estimate the Sun's center from the huge washed out lens flare in the image. So, I fine-tuned the result using the positions of the Earth and Venus alone.

Taking Earth and Venus to be centered in PIA00450 at respective pixel coordinates $$<480.8,1413.2>$$ and $$<1162.8,1721.2>$$ (from the top left), I was able to match the orrery's rendered output to less than a pixel using a field of view of $$0.7071^\circ$$, a "center" position of $$<-0.3389,0.4785,0.0>$$ (again, AU), and a rotation around the camera's axis of $$5.22^\circ$$ clockwise.

A better matching, or even the actual positioning data, may be possible to achieve. However, it would likely not change the result noticeably.

### Validation

After writing this answer, I encountered JPL's online solar system simulator. Although it is not quite as flexible, I was able to quickly generate the result with approximately the right parameters:

The camera parameters are not comparable, and the time is slightly different, but this clearly shows that my result is substantively correct. This simulator is a great tool to answer these sorts of questions!

• Though starting with "abominable data" you've generated some pretty wonderful results! – uhoh Apr 26 '20 at 5:47
• 👍 Who's now going to do the statistical analysis if the slightly brighter pixel in about the right place is significant above noise? – asdfex Apr 26 '20 at 14:34