Moon crosses Earth: a remarkable sequence

There will be no readers’ wildlife today: I’m saving up the photos for when I’m in Poland. Instead, have a gander at this photo:

Screen Shot 2016-07-27 at 6.57.58 AM

This still photo of the moon interposed between a satellite and Earth was going around yesterday, and some who saw it automatically cried “Photoshop!” (I don’t think I’ve ever posted an amazing photo that hasn’t aroused that cry.) Now, it’s okay to say, “I’m dubious,” but a proper skeptic should say, “I better investigate further.” And if you did, you’d see that this photo was genuine, as described by NASA.

First, a time-lapse taken from the NASA site. You’re seeing here what you never see from Earth: the “dark side” of the moon. The bit below tells you why we never see the moon’s bum, which you need to know:

dscovrepicmoontransitfull

Part of NASA’s explanation:

A NASA camera aboard the Deep Space Climate Observatory (DSCOVR) satellite captured a unique view of the moon as it moved in front of the sunlit side of Earth last month. The series of test images shows the fully illuminated “dark side” of the moon that is never visible from Earth.

The images were captured by NASA’s Earth Polychromatic Imaging Camera (EPIC), a four megapixel CCD camera and telescope on the DSCOVR satellite orbiting 1 million miles from Earth. From its position between the sun and Earth, DSCOVR conducts its primary mission of real-time solar wind monitoring for the National Oceanic and Atmospheric Administration (NOAA).

. . . A NASA camera aboard the Deep Space Climate Observatory (DSCOVR) satellite captured a unique view of the moon as it moved in front of the sunlit side of Earth last month. The series of test images shows the fully illuminated “dark side” of the moon that is never visible from Earth.

The far side of the moon was not seen until 1959 when the Soviet Luna 3 spacecraft returned the first images. Since then, several NASA missions have imaged the lunar far side in great detail. The same side of the moon always faces an earthbound observer because the moon is tidally locked to Earth. That means its orbital period is the same as its rotation around its axis.

I don’t fully understand “tidal locking,” but it has to do with the Earth’s huge mass acting to slow down the Moon’s originally faster rate of rotation, and with the gravitational pull of the Moon, which causes Earth’s tides. The slowing apparently stops when the rotational speeds of the Earth and Moon are identical. A physics maven might explain the phenomenon further in the comments. What is clear is that this locking is absolutely precise: over centuries the rate of the Moon’s rotation has exactly matched that of Earth’s—to the extent that even our distant forebears didn’t see the Moon’s bum.

Actually, we can see a bit more of the Moon’s hidden side, as it wobbles a bit on its axis, a phenomenon called lunar libration. This allows us to see, in toto, 59% of the Moon’s surface. And, of course, the “dark side of the Moon” isn’t really dark, for it’s illuminated by the sun when that side is between the Earth and the Sun. We just don’t see the “dark side” because of the precise locking of rotational speed.

h/t: Alan G

46 Comments

  1. Posted July 27, 2016 at 7:34 am | Permalink

    Reblogged this on The Logical Place.

  2. Posted July 27, 2016 at 7:44 am | Permalink

    Re tidal locking:

    If the Moon’s rotation were faster than its orbit there would be a “tidal bulge” sweeping around it (just like the tides on Earth, though Earth’s gravitational effect on the Moon is much bigger than the Moon’s on Earth, since the Earth is more massive).

    Since the bulge is in rock, that would dissipate a huge amount of energy; thus the lowest-energy configuration is when the rotation and orbital periods are the same, so that the tidal bulge stays in the same place on the moon. So that’s the configuration the system heads for.

    Alternatively, consider that, if the moon were rotating faster, the tidal bulge would rotate to ahead of the Earth-Moon line of centres. Thus Earth’s gravity would be pulling backwards on that tidal bulge, so reducing the rotation rate of the Moon. That proceeds until the rotation and orbital periods are the same, at which point the tidal bulge is along the line of centres, and the system is then tidally locked.

    • Chris G
      Posted July 27, 2016 at 8:00 am | Permalink

      Very clear explanation, thanks Coel.
      Easy to forget (actually, it never really occurred to me) that the moon has a tidal-bulge created by the gravitational pull of the earth. But once that’s understood the idea of it acting as a ‘brake’ until the locking occurs makes perfect sense,
      Chris G.

    • Randall Schenck
      Posted July 27, 2016 at 8:58 am | Permalink

      So if the earth rotated as the moon, in sinc with it’s orbit, we would not be here to talk about it.

      • Posted July 29, 2016 at 3:03 pm | Permalink

        One day, it will.

        If the expanding sun doesn’t obliterate the Earth and Moon first …

        /@

    • Posted July 27, 2016 at 9:17 am | Permalink

      As an aside, the theory of tidal locking was first worked out by George Darwin, notable scientist, mathematician and astronomer, and son of an even more notable scientist named Darwin.

    • Gregory Kusnick
      Posted July 27, 2016 at 10:23 am | Permalink

      Thus Earth’s gravity would be pulling backwards on that tidal bulge, so reducing the rotation rate of the Moon.

      This backwards pull also has a small forward component relative to the Moon’s orbital motion. So the Moon’s orbital radius gradually increases as its rotation slows, and angular momentum is conserved.

      This works both ways: the Earth is still spinning within its tidal bulge of air and water, losing rotational energy to tidal friction, and transferring angular momentum to the Moon’s orbital motion. Eventually Earth and Moon will be mutually tide-locked (if the expanding Sun doesn’t engulf them both first).

      • dabertini
        Posted July 27, 2016 at 11:49 am | Permalink

        Is the Moon not drifting away ie. increasing its orbital radius because of this?

        • Gregory Kusnick
          Posted July 27, 2016 at 12:21 pm | Permalink

          Yes it is; it formed much closer to the Earth and has been moving away ever since. So the fact that the Moon and Sun appear the same angular size in the sky is a temporary state of affairs that we happen to be lucky enough to witness.

          Given enough time, the outward spiral will cease when the Earth’s rotation becomes locked to the Moon. Then they’ll very gradually spiral back in toward each other as the combined Earth-Moon system sheds angular momentum due to (much weaker) tidal interaction with the Sun.

    • gravelinspector-Aidan
      Posted July 27, 2016 at 5:44 pm | Permalink

      Not disagreeing with Coel and Gregory’s comments, but one further point. Though the Moon has locked to it’s orbit around the Earth, the Earth continues to transfer angular momentum to the Moon via the oceanic bulges being dragged ahead of the Earth-Moon line by the Earth’s location. This puts a torque on the Moon.
      Detailed studies of rocks deposited in tidal channels in the past reveal different combinations of cycles of frequent oscillations of current flow (interpreted as daily tides) and slower oscillations of speed of current flow (interpreted as the difference between spring and neap tides. But instead of having a day:month ratio of 1:28.5, in the past this ratio was higher – up to 1:31.5 according to this review. (I can send you a PDF if necessary). We do have geological confirmation of these theoretical considerations. but we also have enough evidence to see that the changes are not simple pattern. When you consider the “roughness” of the ocean bed and the changes of continent-ocean arrangement through time, you’d be less likely to expect a simple, stately progression.

    • Michael Waterhouse
      Posted July 27, 2016 at 8:41 pm | Permalink

      The ‘tidal bulge’.
      That is caused by the different gravitational strengths between the near side and the far side, due to the square of the distance reduction, isn’t it?

      • Gregory Kusnick
        Posted July 27, 2016 at 8:50 pm | Permalink

        Yes, that’s one way to look at it.

        Another way is to observe that orbital period increases with orbital radius. The rock on the near side is tracing out a smaller orbit with the same orbital period as the Moon’s center. It’s not going fast enough for that smaller orbit, and so falls toward the Earth.

        Similarly, the rock on the far side is tracing out a larger orbit with the same period. It’s going too fast, and falls away by centrifugal force.

        • Gregory Kusnick
          Posted July 27, 2016 at 8:53 pm | Permalink

          (Technically, period increases with radius to the 3/2, if I’m remembering right. But that doesn’t alter the argument.)

        • Michael Waterhouse
          Posted July 28, 2016 at 7:21 pm | Permalink

          Thanks, I hadn’t looked at it that way before.
          Makes sense.

  3. Posted July 27, 2016 at 7:55 am | Permalink

    Celestial mechanics…wonderful!

  4. ThyroidPlanet
    Posted July 27, 2016 at 8:04 am | Permalink

    Is that a lunar Green Flash?

  5. rickflick
    Posted July 27, 2016 at 8:59 am | Permalink

    One reason this looks “Photoshopped” is the earth is bright and colorful while the moon is dull, dark and colorless. A quick glance makes you think you’re looking at a black and white photo combined with a color photograph – something a clumsy Photoshopper might do, rather than something achievable only with advanced space technology.

  6. Posted July 27, 2016 at 9:26 am | Permalink

    I notice these great photos were taken with a 4 megapixel camera. So why are we forking out dollars (or euros) for cameras with detectors 5 times that?

    • Posted July 27, 2016 at 12:12 pm | Permalink

      In high-end cameras, the image sensor area is usually a better indicator of the image quality than the pixel count. A decade ago, before more economical CMOS sensors began to compete with CCD sensors, the pixel count was a more accurate measure of image quality and even now there is still some correlation.

      Nowadays, there are a lot more kinds of trade-offs that can be made in image sensors so judging their quality is trickier than it used to be. Some high pixel count cameras have noisier low-quality pixels.

      • Posted July 27, 2016 at 1:05 pm | Permalink

        Update that may be relevant to image sensor size. I was reading the wikipedia article and saw that DISCOVR was put into storage for over a decade. Reading between the lines, politics may have been involved, but I’m not sure. So maybe the sensor is older than would be expected. https://en.wikipedia.org/wiki/Deep_Space_Climate_Observatory#History

        • rickflick
          Posted July 27, 2016 at 1:19 pm | Permalink

          Yes, that could be an issue. Remember Voyager 1 was launched in 1977, when Radio Shack released it TRS-80. V1 and V2 are still alive.

    • Thanny
      Posted July 27, 2016 at 5:28 pm | Permalink

      Because people don’t understand and don’t bother to ask the question you just did.

      Most of the pixels in these consumer-grade super-duper-multi-megapixel cameras are effectively noise. The sensor is too small, and the lens much too small.

      The camera that took this picture, called EPIC, is a 12″ diameter telescope with a 15 micrometer per pixel sensor. That’s ten times the pixel size of a typical phone camera, and yields a full sensor size of at least 3cm square. On top of that, it’s a 12-bit sensor, so there are 4096 levels for each pixel, compared to the 256 levels per pixel you see on your screen. While I’m not positive, I believe it’s a monochromatic sensor, so each color is shot at full resolution through a filter, then combined.

      That makes for a whole lot of signal and very little noise, which is going to clobber pretty much every consumer digital camera on the market. You’d have to get the top end professional models to even compare.

      That’s the background info. The actual answer can be encapsulated with another question. Can you imagine anyone being able to sell a camera with information like the description above, to a member of the general public? Wouldn’t it just be easier to come up with a single number, the label of which begins with “mega”?

      • rickflick
        Posted July 27, 2016 at 10:03 pm | Permalink

        The sale of these simple cameras is actually quite an amazing thing. For the most part, people get what they want and need in a simple inexpensive system. Just look on Flickr, Facebook, and Youtube. Not many are asking any questions. They’re content.

        • Posted July 29, 2016 at 8:12 am | Permalink

          Yes, basic snapshots are what 98+% of the general public (the people you need to sell to to make money on a consumer product) are after.

          As I like to summarize it:

          For the vast majority of people, a “good” photo is one that:
          1. Has a person or people in it
          2. The person or people can be recognized

          That really is about it (have a look around social media; though food counts somewhat now as a subject).

          So, the 2MPx fixed-focus, digital zoom only, camera on most “smart” phones is just the ticket for most people. And I can’t blame them at all.

  7. Eric Shumard
    Posted July 27, 2016 at 9:29 am | Permalink

    The astronomer and mathematician George Howard Darwin, Charles Darwin’s son, is credited as the first to work out the theory of tidal locking.

  8. Tom Czarny
    Posted July 27, 2016 at 9:55 am | Permalink

    A most welcome respite from the grief and horror here at ground level.

  9. John Harshman
    Posted July 27, 2016 at 10:08 am | Permalink

    I had no idea there were satellites that far away from earth. But I think I can see my house from there, or at least where my house ought to be.

    The far side of the moon is way different from the near side: hardly any maria. I forget why that is.

    • Gregory Kusnick
      Posted July 27, 2016 at 10:31 am | Permalink

      DSCOVR isn’t in Earth orbit; it orbits the L1 equilibrium point between and Sun.

      As I recall there are various theories about the maria. I favor the idea that when the Moon became tide-locked, it stopped spinning with the heavier side facing downhill.

      • gravelinspector-Aidan
        Posted July 27, 2016 at 4:39 pm | Permalink

        it stopped spinning with the heavier side facing downhill.

        That raises the question of why the maria are concentrated on one face of the Moon. Which is a good question – the incoming impactors would be expected to come in fairly evenly distributed around the ecliptic.
        The centre of mass of the Moon is, IIRC, some 60km from the centre of figure, which rather implies that the distinction dates back to the construction and differentiation of the Moon. But I’d have to go back and read up on that.
        (I thought I mailed Jerry about this a couple of weeks ago, including a discourse on the magnificent recovery of SOHO, another denizen of L1.)

      • cherrybombsim
        Posted July 27, 2016 at 5:55 pm | Permalink

        When the moon formed, it was a lot closer to Earth than it is now, and became tidally locked really, really fast, very soon after it formed. The mare were erupted much later, when the moon was already tidally locked.

        • Gregory Kusnick
          Posted July 27, 2016 at 7:03 pm | Permalink

          Fair enough, but it still seems likely that if there’s any mass asymmetry, the most stable orientation is with the heavier side facing Earth. So now the question becomes, why did the maria erupt on the heavier side?

          • jeremyp
            Posted July 28, 2016 at 6:38 am | Permalink

            The near side of the Moon is protected from collisions by the Earth.

            • Gregory Kusnick
              Posted July 28, 2016 at 12:48 pm | Permalink

              I presume you don’t mean that it’s protected by the solid body of the Earth, since that obstructs less than a tenth of a percent (if I’m doing the math right) of the nearside sky.

              So the claim is presumably that the Earth’s gravity deflects potential impactors away from the Moon. I’m willing to be convinced, but naïvely it seems like it would have the opposite effect of magnifying the Moon’s collisional cross-section, much as optical gravitational lensing magnifies the images of distant galaxies.

              On further research, it seems the prevailing theories involve either a thinner crust on the near side, or a higher concentration of magma-inducing radioactivity. But in either case it just pushes the question back a level.

  10. John Harshman
    Posted July 27, 2016 at 10:10 am | Permalink

    Hey, what’s that brownish smudge in the center of the earth’s image? It stays in the center as the earth rotates. Is that the moon’s reflection?

    • Gregory Kusnick
      Posted July 27, 2016 at 10:32 am | Permalink

      The Sun’s reflection, I think.

  11. bPer
    Posted July 27, 2016 at 11:48 am | Permalink

    Prof. Coyne, I think you understand the phenomenon but you messed up your description in the penultimate paragraph. The rotation speed of the Earth is not relevant to this issue. “The slowing apparently stops when the rotational speeds of the Earth and Moon are identical” should read “The slowing apparently stops when the rotational speed of the Moon and its orbital period are identical”.

    Also, regarding lunar libration, it’s not that the Moon “wobbles” on its axis; it’s that the lunar orbit is not circular but slightly elliptical. The Moon rotates at a constant rate, but travels faster in its orbit at perigee and slower at apogee. This makes it look to us like we’re viewing the Moon at slightly different angles at various parts of the orbit.

    • gravelinspector-Aidan
      Posted July 27, 2016 at 4:44 pm | Permalink

      The fact that the rotation poles of the Earth, Moon, and the Moon’s orbit around the Earth are not parallel also contributes to liberation. Plus the Earth is not a point location, but subtends around 2 degrees in the Moon’s sky.

  12. Posted July 27, 2016 at 11:48 am | Permalink

    So its an earth eclipse to the satellite?

  13. Gnu Atheist
    Posted July 27, 2016 at 12:00 pm | Permalink

    “There is no dark side of the moon, really. As a matter of fact, it’s all dark.”

    • stuartcoyle
      Posted July 28, 2016 at 4:10 pm | Permalink

      There is a “Far Side”.

      “Far Side of the Moon” is not a bad title for a prog rock concept album.

  14. Heather Hastie
    Posted July 27, 2016 at 12:48 pm | Permalink

    This is so cool, and even better with all the explanations of how and why it happened. Thank you all! 🙂

  15. Posted July 27, 2016 at 4:08 pm | Permalink

    One reason it looks faked to me is that there is no shadow of the moon visible crossing the earth. Evidently the sun is opposite the earth (directly behind the camera), so the moon’s shadow is always hidden under it.

    Doesn’t this mean the photo had to be taken during a solar eclipse?

    • Gregory Kusnick
      Posted July 27, 2016 at 4:14 pm | Permalink

      Yes, we’re seeing a solar eclipse from the Sun’s POV. But it looks like you had to be in Hawaii to observe it from Earth.

    • gravelinspector-Aidan
      Posted July 27, 2016 at 4:50 pm | Permalink

      There would have been a solar eclipse IF the lunar umbra had touched the surface of the Earth. But because the lunar orbit is not a circle, the Earth-Moon distance varies (recall the definition of a circle), and some times the shadow touches down, and other times it doesn’t.
      It is possible that there was an eclipse while this series was taken, but not necessary.
      Incidentally, most partial eclipses of less than 90% are not noticed by “people on the streets”.

  16. JohnnieCanuck
    Posted July 28, 2016 at 8:50 am | Permalink

    The tidal bulge explains how the Moon lost its rotational energy, but not what keeps it locked in one orientation.

    If the moon is longer or more dense through a particular axis, then one end will end up pointing towards the Earth.

    If we replace the Moon with two satellite masses in circular orbits, one where the nearest point was, and the other 3500 km away at the furthest point, they will have different orbital periods. The furthest will take longer than the Moon does and the closer will take less. A tether between them would force them to orbit at the same time and would be in tension. The tether would align with the Earth.

    The Moon has sufficient gravitational force to overcome this tension. The Roche limit describes how deep it could be in the Earth’s gravitational well before the tension pulled it apart.

  17. Posted July 29, 2016 at 7:59 am | Permalink

    “some who saw it automatically cried ‘Photoshop!’ (I don’t think I’ve ever posted an amazing photo that hasn’t aroused that cry.)”

    Almost all the “Amazing!” photos I’ve seen online have, in fact, been photo-shopped. It’s very easy ti do. And it’s become so ubiquitous that most people have stopped noticing (if they ever did).

    My intent, where I point it [photo-shopping] out in comments, is to explain exactly what I see that leads me to that conclusion. If I fail to do that, I apologize and will try to do better.


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