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Message Subject
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Calling Astro...Please come in and explain how this works???..It's freaking me out.....
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The issue is this. A point on the Earth is travelling at 15 degrees per hour due to the rotation of the Earth, that is its angular speed. The moon in its orbit around the earth travels at 0.6 degrees per hour, again its angular speed. So if the Earth did not spin, the moon's shadow would certainly move from West to East, but since the Earth rotates through many more degrees per unit time than the moon in its orbit around the Earth, the Earth wins the race and therefore spins ahead of the moon's shadow, and therefore it will appear that the moon's shadow is traveling from East to West.
Quoting: leader But the Moon doesn't have to "win the race". It just has to go by the Sun. The Sun isn't that big and if you think about your field of view when looking up at the sky, the Sun doesn't take up many degrees of your vertical field of view. It actually takes about 0.5 degrees. Now if the moon goes roughly 0.54 degrees of its orbit per hour it doesn't take long to pass by the Sun. You can actually do the same calculation as I did before with kilometers but with degrees (distance 0.5 * 2 divided by speed 0.54) and it's pretty close to the same answer as before. You can see how going by the Sun in two hours or so would be much faster than the speed at which the Earth is rotating, right? Wrong. Do this. Draw two concentric circles. The inner circle represents the earth's surface circle; think of it as the equator circle. And the outer circle is the moons path. And think of it as looking down at it from the north pole direction. Now draw a dot outside both of the circles to represent the sun. Now draw a line that connects the center of the two concentric circles with the dot that represents the sun. That line will intersect the Earth circle where the shadow begins and the outer circle where the moon is initially located. Now from that initial line draw two more lines that both go threw the center of the concentric circles with one line 0.6 degrees counter clockwise from the initial line and the other that is 15 degrees counter clockwise away from the initial line. At the beginning of the eclipse, the shadow of the moon is where the initial line that you drew intersects the Earth's circle as I have previously said. In one hour, that point on the Earth's surface has traveled to where the 15 degree line intersects the Earth's circle, while the moon is located where the 0.6 degree line intersects the moon's orbit circle. Now draw a line that goes through where the 0.6 degree line intersects the moon's orbit circle and the sun, and continue that line down to the Earth's surface. Notice that that line intersects the Earth's circle to the West of the place on the Earth's circle where the shadow started. Therefore the shadow has traveled from East to West. You didn't count the fact that the Sun is pretty far away and that makes the Moon's shadow travel in relation to Sun's position. In your concentric circles example you would have to draw a dot pretty far away from the circles representing the Sun and draw the shadow of the Moon by starting from the far away Sun, then going through the Moon's orbit circle position and on to the Earth's surface circle. Then the shadow will go faster than any point on Earth's surface. The light source being far away is irrelevant. The fact that the object causing the shadow, the moon, is close is what matters. Face the fact, I just proved that for an eclipse where the shadow travels along the equator the shadow will travel from East to West. And I'm sure the same can be proven for all other eclipses. But if it isn't true for all other eclipses, then at least for many others it should be true for. But all recorded eclipses from what I am aware of are west to east. We got some trouble in paradise folks.
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