Operation Rahu Episode IX
How Far and How Big is the Moon?
Total lunar eclipse on Saturday 28 July 2018
There are 2 total lunar eclipse in year 2018 the first was 31 January and second was 28 July that was the longest eclipse of the 21st century because moon was passing almost at the center of Earth's umbra so it took 103 minutes
Geometric of total lunar eclipse
Total lunar eclipse on 27 - 28 July 2018 can be observed in many countries from Australia Asia Eastern Europe and Africa
Total lunar eclipse 28 July 29018 was the longest in 21st century (103 minutes) beacuse moon was passing almost at the center of Earth's umbra
Total lunar eclipse on 31 January 2018 took lesser time because moon was passing at the side of Earth's umbra
Total lunar eclipse (Blood moon) and Mars (Cr: National Astronomical Research Institute of Thailand)
Photo of partial lunar eclipse 28 July 2018 local time 04:53 by Songkla Obsevatory Thailand
Operation Rahu Episode IX
Operation Rahu have been conducting for 8 episodes and this one is the episode IX to measure distance Earth - Moon by using just one photo of partial lunar eclipse on 28 July 2018. Operation Rahu's methodology is modified from the ancient Greek mathematics invented by Aristarchus of Samon 3rd BC.
Logo of Operation Rahu Episode I - IX
Cr: website http://home.strw.leidenuniv.nl/~pvdwerf/teaching/awt2/aristarchus.html
How do we begin to measure the distance from the earth to the moon? One obvious thought is to measure the angle to the moon from two cities far apart at the same time, and construct a similar triangle, like Thales measuring the distance of the ship at sea. Unfortunately, the angle difference from two points a few hundred miles apart was too small to be measurable by the techniques in use at the time, so that method wouldn't work.
Nevertheless, Greek astronomers, beginning with Aristarchus of Samos (310-230 B.C., approximately) came up with a clever method of finding the moon's distance, by careful observation of a lunar eclipse, which happens when the earth shields the moon from the sun's light.
The ancient Greek astronomer "Aristarchus of Samos 3rd BC." was the first who measured how far the moon
To better visualize a lunar eclipse, just imagine holding up a quarter (diameter one inch approximately) at the distance where it just blocks out the sun's rays from one eye. Of course you shouldn't try this---you'll damage your eye! You can try it with the full moon, which happens to be the same apparent size in the sky as the sun. It turns out that the right distance is about nine feet away, or 108 inches. If the quarter is further away than that, it is not big enough to block out all the sunlight. If it is closer than 108 inches, it will totally block the sunlight from some small circular area, which gradually increases in size moving towards the quarter. Thus the part of space where the sunlight is totally blocked is conical, like a long slowly tapering icecream cone, with the point 108 inches behind the quarter. Of course, this is surrounded by a fuzzier area, called the "penumbra", where the sunlight is partially blocked. The fully shaded area is called the "umbra". (This is Latin for shadow. Umbrella means little shadow in Italian.) If you tape a quarter to the end of a thin stick, and hold it in the sun appropriately, you can see these different shadow areas
Question: If you used a dime instead of a quarter, how far from your eye would you have to hold it to just block the full moonlight from that eye? How do the different distances relate to the relative sizes of the dime and the quarter? Draw a diagram showing the two conical shadows.
Now imagine you're out in space, some distance from the earth, looking at the earth's shadow. (Of course, you could only really see it if you shot out a cloud of tiny particles and watched which of them glistened in the sunlight, and which were in the dark.) Clearly, the earth's shadow must be conical, just like that from the quarter. And it must also be similar to the quarter's in the technical sense---it must be 108 earth diameters long! That is because the point of the cone is the furthest point at which the earth can block all the sunlight, and the ratio of that distance to the diameter is determined by the angular size of the sun being blocked. This means the cone is 108 earth diameters long, the far point 864,000 miles from earth.
Now, during a total lunar eclipse the moon moves into this cone of darkness. Even when the moon is completely inside the shadow, it can still be dimly seen, because of light scattered by the earth's atmosphere. By observing the moon carefully during the eclipse, and seeing how the earth's shadow fell on it, the Greeks found that the diameter of the earth's conical shadow at the distance of the moon was about two-and-a-half times the moon's own diameter.
Question: At this point the Greeks knew the size of the earth (approximately a sphere 8,000 miles in diameter) and therefore the size of the earth's conical shadow (length 108 times 8,000 miles). They knew that when the moon passed through the shadow, the shadow diameter at that distance was two and a half times the moon's diameter. Was that enough information to figure out how far away the moon was?
Well, it did tell them the moon was no further away than 108x8,000 = 864,000 miles, otherwise the moon wouldn't pass through the earth's shadow at all! But from what we've said so far, it could be a tiny moon almost 864,000 miles away, passing through that last bit of shadow near the point. However, such a tiny moon could never cause a solareclipse. In fact, as the Greeks well knew, the moon is the same apparent size in the sky as the sun. This is the crucial extra fact they used to nail down the moon's distance from earth.
They solved the problem using geometry, constructing the figure below. In this figure, the fact that the moon and the sun have the same apparent size in the sky means that the angle ECD is the same as the angle EAF. Notice now that the length FE is the diameter of the earth's shadow at the distance of the moon, and the length ED is the diameter of the moon. The Greeks found by observation of the lunar eclipse that the ratio of FE to ED was 2.5 to 1, so looking at the similar isosceles triangles FAE and DCE, we deduce that AE is 2.5 times as long as EC, from which AC is 3.5 times as long as EC. But they knew that AC must be 108 earth diameters in length, and taking the earth's diameter to be 8,000 miles, the furthest point of the conical shadow, A, is 864,000 miles from earth. From the above argument, this is 3.5 times further away than the moon is, so the distance to the moon is 864,000/3.5 miles, about 240,000 miles. This is within a few percent of the right figure. The biggest source of error is likely the estimate of the ratio of the moon's size to that of the earth's shadow as it passes through
Moderninized formular modified from Aristarchus of Samos
After carefully looking at the ancient Greek mathematics I came to the conclusion that Earth - Moon distance can be calculated by measuring the proportional of diameter of Earth's umbra and diameter of the moon (FE : ED) with the help of one clear photo of partial eclipse as shown in the diagram.
Since the Earth's circumference and diameter were determined under "Operation Eratosthenes 21 Mar 2012" Phupek Temple Sakon Nakhon Thailand - Bayon Temple Seam Reap Cambodia (http://www.yclsakhon.com/index.php?mode=preview&lay=show&ac=article&Id=539365797)
The modernized formular of measuring Earth - Moon modified from the ancient Greek mathematic
Earth's circumference was found 37,823 Km under "Operation Eratosthenes Mar 21, 2012" (Phupek temple Sakon Nakhon Thailand - Bayon temple Seam Reap Cambodia)
how far and how big is the moon?
Thanks to the clear photo of partial eclipse taken by Songkla Observatory Thailand, I can easily determine the size of Earth's umbra by drawing a circle to fit the moon's eclipse as well as finding moon's diameter in the same time. Since Earth's diameter was known by Operation Eratosthenes 21 Mar 2012 so here we go!
Take one clear photo of the partial lunar beclipse and download into PowerPoint
Making picture of "Earth's umbra" by draw a circle that exactly fitted to the curvation of eclipse and draw another circle to create the completed size of moon
Measure the proportional of diameter of Earth's umbra and diameter of moon, here we got 11: 4.55
Using the modernized formular to find how far is the moon and we have 380,299 Km with 5.56% error on comparision with 402,704 Km of the Starry Night Astronomical Program
Size (diameter) of the moon can be determined by 380,299 / 108 = 3,521 Km (considering the moon's umbra can reach Earth on solar eclipse that means the conical length of moon's umbra must be 108 time of her diameter)
Photo of partial lunar eclipse taken at Songkla Observatory Thailand
Modernized formular to measure distance Earth - Moon with one photo of patial lunar eclipse
Distance Erath - Moon can be measured by the modernized formular as show in the diagram and we have 380,299 Km compared to the 402,704 Km from the Starry Night Astronomical Program (error 5.56%)
The distance Earth - Moon calculated by the Starry Night Astronomical Program = 402,704 Km
According to the updated version of Operation Eratosthenes on 23 Sep 2013 at Prasat Phupek Sakon Nakhon Thailand, yielded the new Earth's circumference at 39,628 Km (Earth's diameter 12,609 Km) with very small error 0.95%. It is recommended to apply this new figure for all of Operation Rahu.
Here we got the updated Distance Earth - Moon (viewed from Songkla Observatory) 398,458 Km with 1.05% error from the actual figure 402,704 Km