In the first chapter, while talking about how Aristotle was able to conclude that the Earth is spherical, Hawking says that had the Earth been a flat disk the shadow of the on the Earth on the moon would have been

elongated and elliptical unless the eclipse always occurred at a time when the sun was directly under the center of the disk

I do not understand how the Earth will cast a shadow on the moon if the Earth were flat. And, therefore, the claim later in the sentence is something I am confused about. Any illustration of this idea would help.

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    $\begingroup$ "I do not understand how the Earth will cast a shadow on the moon if the Earth were flat" A flat piece of paper still casts a large shadow in one direction, no? $\endgroup$ – JMac Jul 23 '19 at 15:27
  • $\begingroup$ Yes but if it did, you would have to set up the moon, the earth, and the sun in a very odd fashion that won't make the sun and moon visable. Wouldn't it? $\endgroup$ – Sal_99 Jul 23 '19 at 15:32
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    $\begingroup$ @JMac "how" here means like how it looks a certain way not how it can happen at all $\endgroup$ – user234190 Jul 23 '19 at 16:13
  • $\begingroup$ "have"have to set up the moon, the earth, and the sun in a very odd fashion" and that's exactly the argument here... $\endgroup$ – Jasper Jul 23 '19 at 16:37
  • $\begingroup$ A flat disk earth must be at an angle to the plane containing the sun and moon, to explain the position of the sun and moon in the sky. Hawking is assuming people thought the earth was a circular flat disk - if it was square for example, the shadow in an eclipse would obviously look a different shape. $\endgroup$ – alephzero Jul 23 '19 at 16:51

So to have a lunar eclipse you first need to have the Earth directly between the Sun and the Moon -- in other words you need to have a full moon as seen from Earth.

Here is a rough diagram of what a full moon might look like when seen from a disk Earth. I have drawn a disk of land and oceans and another disk of atmosphere above it, but one might have imagined the atmosphere instead as a dome or similar:A sun, an earth disk, and a moon lined up diagonally. The sun is beneath the earth-disk indicating nighttime, the moon is therefore above it, visible low in the horizon.

Notice that the Sun does not need to be aligned right above or beneath the plane of the Earth in order to be opposite from the Moon; in this case it hits the earth mostly from the side rather than from the bottom. To see this diagram the Moon needs to be both (a) full and (b) low in the sky; the full moon is necessary for any eclipse while the low-in-the-sky observations require a certain degree of luck in the timing of these particular eclipses that will involve non-circular shadows.

Now, this disk is still able to cast shadows on things, as is any piece of paper that you cut into a circle. Since the light source is coming from beneath the disk (it is night-time!) we would draw its shadow up and out of the disk-Earth like so:

The same image but dotted lines have been drawn between the boundaries of the Earth and the boundaries of the disk with which to indicate both umbra and penumbra of a shadow that has been filled in in the remaining space. The moon is still visible in the foreground of that shadow.

Now the shadow here is not engulfing the Moon and that is no accident; not every full moon generates an eclipse but rather the moon can be displaced from the Sun, in this case in the third dimension (it is a little closer to you than the plane of the Earth and Sun and therefore the shadow misses it).

You will notice that there is a fuzzy edge to any shadow called the penumbra, and it comes from the light source having nonzero size -- if it were a point source (or infinitely far) then the shadow would have sharp edges; if it is a larger source (or closer) then the shadow has more blurry edges. In theory this could hide the entire effect, but in practice this blur is very visible during a lunar eclipse as the light which makes it through the atmosphere actually gets all of its blue color sucked out of it, so the boundary of the shadow on a lunar eclipse is seen to be a thin red line in every eclipse. So that basically tells you that this diagram is a little inaccurate in that the Sun should be much further away and the shadow being cast should be much sharper. In practice, you can make out the shape of the umbra when you are looking at a lunar eclipse, and you do not need to worry too much about the penumbra around it.

The only remaining factor is what that shape actually looks like. At these sorts of sharp angles where the eclipse is seen low in the sky, you would expect the eclipse to be a narrow ellipse. The lower in the sky the narrower the shadow should be, until you get an eclipse right at sunrise or sunset, when the moon is as low as can be across the horizon and the disk becomes a thin black line across the moon (plus a red penumbra on one or both sides).

So it might be hard to observe based on timing and coincidence, but sometimes you should be able to see the eclipse as narrower than the Moon is wide, so that you would have a band of darkness across the moon but with both sides of that band still illuminated. It would happen for eclipses just barely visible at sunrise or sunset in the opposite direction of the Sun.

But, this is not what is observed: lunar eclipses nearer sunrise/sunset when the moon is necessarily low in the sky, look the same as eclipses near midnight when the moon is straight above.

That is what Hawking is saying that the Greeks noticed and understood. The Earth must be "blown up" to a sphere or spheroid in this up-down dimension you are seeing, so that the Sun cannot hit it "edge-on" and thus cast a very narrow shadow onto the Moon during a lunar eclipse.

  • $\begingroup$ Would you say a ellipse considering the edge of the disk and the atmosphere only being on one side? $\endgroup$ – user234190 Jul 25 '19 at 6:45
  • $\begingroup$ Apologies for the late follow up, but when you mention sum setting and rising close to the eclipse, you mean that the sun sets/rises in the opposite end of the earth yea? Since that would be necessary for the Earth to get in the way of the sun's light $\endgroup$ – Sal_99 Jul 29 '19 at 5:23
  • $\begingroup$ @Sal_99 yes, to see the most visible implications of a flat earth you would want to see a lunar eclipse right before sunrise or right after sunset, when the moon would have to be very low in the sky. (It doesn't have to be that way usually, it's just that this is an eclipse we're talking about and during an eclipse they have to be opposite.) And at those times the sunrise and sunset would indeed have to be behind you while you are looking at the Moon. $\endgroup$ – CR Drost Jul 29 '19 at 15:18

The question comes down to how a sphere's shadow compares to a disks at angles not directly above. You don't have to translationally move either to see the difference, you can just rotate them. If you rotate a sphere under a light the shadow looks the same, if you rotate a circular disk under a light it will turn into something like an ellipse (not exactly) then like a straight line and back (which represents the case we are looking at here with the projection on the moon revolving around the earth like the sun). Though it will actually be a squeezed ellipse rather than an elongated one so hawking's wording is wrong in that case. Like if the flat earth blocked the sun while the moon was rising and the sun was setting you would see something like a line on the moon, but the line wouldn't be longer than the width of an eclipse from directly above. How visible or blurry the line is depends on the thickness and width of the disk.

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    $\begingroup$ As far as I know, if you allow for more than just rotation, it should allow for the shadow to elongate as well. Either way, as far as what it looks like goes, I'm not sure how you would tell apart a squeezed ellipse from a elongated one. You will always have a vertex and co-vertex, regardless of if a circle were squeezed or stretched to make the ellipse. $\endgroup$ – JMac Jul 23 '19 at 16:45
  • $\begingroup$ Elongated or squeezed depends on the the surface and angle you are looking at it from. Here it is the moon rotating around the earth and looking from earth presumably so it won't look elongated since people will see the size of the circle from when it's straight above. They can compare to that to say squeezed or elongated. $\endgroup$ – user234190 Jul 23 '19 at 16:53
  • $\begingroup$ Is this ellipse squeezed, or elongated? upload.wikimedia.org/wikipedia/commons/9/96/Ellipse-def0.svg $\endgroup$ – JMac Jul 23 '19 at 16:55
  • $\begingroup$ It depends on the original circle it came from if it were like the case of what we are talking about here. Hawking implies that the shadow looks longer in a certain direction than it was when it was a circle, that's wrong. $\endgroup$ – user234190 Jul 23 '19 at 16:58

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