Telescopic images of moon seen from Selsey and Athens p is the parallax shift of the moon, measured to be 0.03°
Where the longitude of Selsey and Athens are φ and φ’ respectively, and the latitude of Selsey λ λ’, and the radius of the Earth is r.
Giving us the value of UG to be 2360 km., dividing by two gives the length of UE to be 1180 km. The distance length EM can then be calculated:
The actual lunar distance is 384000 km, our answer has an error of 17%. It could have been more accurate had we subtracted AE and the distance of E to arc UE from our answer.
Another way of measuring the lunar parallax is using a solar eclipse. During a solar eclipse, the earth and the moon occupy the same line of longitude in the sky at some point. Observing the eclipse from different latitudes along the Earth, the sun and moon would appear at different points along this longitude. A lunar parallax can be obtained by comparing the angle between the sun and the moon as observed from different points on Earth. Consider the following example: using the solar eclipse that occurred at August 11, 1999. At the time, according to naval data tables, the solar declination 11 was 15.34°. The lunar declination was 15.89°, so the angular difference between the sun and the moon to be 0.55°.
Fig. 12 shows how the sun and moon would look
from an imaginary viewpoint in the center of the Earth. There is no visible eclipse. However, from London, an eclipse can be seen.
Fig. 13: From London, at the climax of the eclipse, the moon and the sun appeared to have interchanged positions, and were a maximum distance of Φ from each other, which was measured to be 0.03°. Seen from London, the moon had been “lowered” in the sky. Assuming
Figure 12: lunar and solar declination
11 Angle of the sun from the plane of the earth’s equator
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