Fun with shadows - again! The Taurus-Littrow fly-by

[onebigmonkey]

I am continuing with my studies of Apollo shadows on the moon and I skipped ahead of my current focus (Apollo 15) to Apollo 17 and its stupendous fly by of Taurus Littrow.

Seemed to me that, in the presence of bright sunshine, there might just be a shadow of a command module to be seen somewhere.

A likely candidate can be seen in AS17-147-22465, a 49Mb TIFF of which you can get here

https://archive.org/details/as17-147-22465

and you can see what I think is the CSM shadow here:



That little blob on the mound is not visible in the other images in this sequence, neither does it appear to be anything on the lens.

So, is it the CSM shadow?

Well, we have to do some educated guess work here. We know from the mission times that the sun angle at this point is 14.75 degrees.

The mission information suggests an altitude for the LM at this point of around 10000 metres. This would mean that the shadow of the CSM would be roughly 25 km from the point directly beneath the CSM.

What would this actually look like?

Google Moon allows you to position a marker at a known height above ground. We can also draw a line from the point where the shadow falls 25 km in length, and position the CSM 10 km above the end point. This is what I've done in the image below.

The black line represents a 25km distance from the location of the shadow. The CSM figure is 10km above the end of that. The red line represents the path followed by the CSM, which coincidentally is pretty much with the sun behind it. The yellow line represents the path of light from the sun, based on a position determined by Stellarium and the path of shadows on the ground.



Thoughts?

I also think I've spotted where the LM's shadow is:



You can also find what looks like the CSM shadow half way up the south massif in https://archive.org/details/as17-147-22466

What I need to know is are the numbers right and are my conclusions reasonable?

For anyone who decides to check the 16mm footage, most of the dark blobs that appear to trail the movement of the CSM & LM seem more likely to what we in the trade call 'bits of crap'.

[Allan F]

The size of the shadow - the CSM covered an area of around 44 square meters - would that be visible fjrom 10 km?

[bknight]

obm, I don't want to rain on your parade, but the image you have posted appears to 22465 and has a "splotch" on the peak, but this image taken, from ALSJ:

Gene took this photo on the orbit before final descent. On the final approach, Gene flew Challenger down over the Sculptured Hills (low, knobby hills in sunlight just right of frame center) and out over the valley floor before landing a little north of the Trident group of craters. The CSM with Ron Evans can be seen in the left distance with the South Massif in the background. Henry is the furthermost of the three largish craters at the foot of the North Massif to the right. The labeled version of this frame picks out a few other features for orientation. Compare this view with the CSM Pan Camera frame and note the distance across the valley from Henry to Bear Mountain and the outcrop hill beyond (in shadow here).

22466 looks like this

with the CSM further " up" the mountain.

Rev 12, CSM, Apollo 17 Landing Site.

EDIT:
In the second image 22466, it would seem that a shadow would be about mid-way up the South Mastif. but I don't see any dark spots.  It may be just me though.

[ka9q]

I can't imagine either Apollo spacecraft casting a shadow on the moon from that altitude. Consider how large (small, actually) it would appear from the surface compared with the half degree diameter of the sun.

To cast a shadow you have to cover most or all of the sun. Most of it gives you a penumbra, all of it gives you an umbra.

[bknight]

You could be right and another thought why was the LM above the CSM while preparing to land?  Maybe Bob or Jay might be able to answer thagt one.

[onebigmonkey]

This is why I like to run things by people here - a fresh perspective is always helpful.

One thing I'm conscious of is that I very much want a shadow to be visible in the image, and while I've found what I believe are good candidates there is every possibility that they are just blemishes on the image from one source or another. I'm trying to avoid pareidolia!

While I agree that the CSM & LM shadows would be small, one thing in favour of them being visible on the ground is that there is no atmosphere to scatter sunlight and diffuse out the shadow. It's certainly possible for high altitude objects to cast a discernible shadow here on Earth (like, um..clouds ) the question is how small can an object be to cast a noticeable shadow from high up!

To answer bknight's question this was dealt with on another thread: the LM initially moved to a position above the CSM so that it could be inspected against the black background of space. The CSM moved to a higher orbit later.

[onebigmonkey]

Looking at the descent footage itself at http://apollo17.org/ the altitude is called out as 1000 feet just as the LM shadow appears over the rim of Camelot crater.

It's a very small shadow on there, so my original claims may well be a false positive.

Ah well - I have at least got something I can work with there

[12oh2alarm]

Time for some math!

Let the CSM be a disk of radius 5m. At what distance would the disk cover the whole sun (alpha = 0.5 degrees)?

Trigonometry says radius/distance = tan alpha = alpha (for small alpha).
So 5m/distance = 0.5 * pi/180 <=> distance = 2*5m*180/pi = 573m.

If the CSM is further than that, it will be an "annular" solar eclipse and there will be nothing more than a small dip in the light curve on the ground. At more than 5km distance it will look more like a Venus or Mercury transit.

That puts the shadow hypothesis squarely into the "nope" field.

[ka9q]

You could be right and another thought why was the LM above the CSM while preparing to land? 

Assuming this shot was taken shortly after LM/CSM separation, it's because the CSM executes a downward separation burn. This keeps the period of the two spacecraft the same, but they will appear to move around each other within the orbit plane once per orbit. You may have seen slow flyarounds of the ISS taken from the shuttle; this is essentially how it was done.

Caveat: I know Apollo 11's CSM executed the downward separation maneuver; the procedure may have been different for Apollo 17.

[bknight]

Assuming this shot was taken shortly after LM/CSM separation, it's because the CSM executes a downward separation burn. This keeps the period of the two spacecraft the same, but they will appear to move around each other within the orbit plane once per orbit. You may have seen slow flyarounds of the ISS taken from the shuttle; this is essentially how it was done.

Caveat: I know Apollo 11's CSM executed the downward separation maneuver; the procedure may have been different for Apollo 17.

The images were both taken during the orbit 12 after the separation and circularization maneuver taking place at GET 109:17:29.  I didn't realize the CSM reduced altitude while separating.  Good things come by being in this forum.

[12oh2alarm]

Trigonometry says radius/distance = tan alpha = alpha (for small alpha).
So 5m/distance = 0.5 * pi/180 <=> distance = 2*5m*180/pi = 573m.

Factor of 2 missing, since I use the radius instead of the diameter, I should use (alpha/2) instead:
distance = radius/tan(alpha/2).
The maximum distance for the disk to completely cover the sun is now 1145m.
Further than this distance the CSM casts no more umbra, only penumbra, which gets quickly unnoticeable on film as distance increases.

[bknight]

At what angle value does this equation revert back to tan(angle) = d/D form?

[12oh2alarm]

At what angle value does this equation revert back to tan(angle) = d/D form?

I'm not sure I understand the question. We can use two slightly different formulae:

tan(alpha/2) = radius/distance    (eqn. 1)
tan(alpha)   = diameter/distance  (eqn. 2)

You could ask for the error (difference in computed distance) for both equations. Let's do that!
If you use the tan x = x approximation the distance is exactly the same: 1145.915m
If you don't, the results are

5m/tan(.25deg) = 1145.908m   (using eqn. 1)
10m/tan(.5deg) = 1145.887m   (using eqn. 2)

All three values are within 28mm (a good inch), a relative error of 2E-5 or 2 parts in 100,000.
For estimating umbral shadow existence, this is completely negligible and the approximation justified.

You might also ask, at what angle does the error for the tan x = x approximation become larger than some given threshold. Let's compute that as well. Say, 1 percent relative error, or 0.01:
We need to solve (tan x - x)/x >= 0.01
A numerical solution yields x > 0.172, or 9.86 degrees, at which angle the aproximation error exceeds one percent.

[bknight]

Umm, the definition that I saw:
http://astro.physics.uiowa.edu/ITU/glossary/small-angle-formula/

Since these angular diameters are often small, we can use the small angle approximation which will give us:
tan theta ~ theta

That was the size of the angle I was asking, not sure if this answers the question.

[12oh2alarm]

Umm, the definition that I saw:
http://astro.physics.uiowa.edu/ITU/glossary/small-angle-formula/

Since these angular diameters are often small, we can use the small angle approximation which will give us:
tan theta ~ theta

That was the size of the angle I was asking, not sure if this answers the question.

That's the same formula as my eqn. 2 and there is no specific angle at which the approximation "reverts" to the exact formula. The error is a function of the angle which is always nonzero, except for the trivial case of theta = 0. That's why I provided the one percent error example.