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Astronomical question and answer 219

 

Frank L. Preuss

 

How long is then now such a wandering of the shadow?

 

In the answer 207 we saw the picture of the eclipse in Novosibirsk and there the eclipse lasted 36 x 3 = 108 minutes.

In the answer 211 it says: "The cone shadow of the moon wanders with a speed of about 28 km per minute over the surface of the earth (at the equator)."

From the time and the speed one can now calculate the distance. When something moves along 108 minutes with a speed of 28 km/min, then the distance is = 108 min x 28 km/min = 3024 km.

When one looks at a globe, then the diameter there is 12 756 km long, and 3024 km is about a quarter of that.

Now follows a picture, which shows such footpaths of eclipses:

Sonnenfinsternisse2

"B Sichtbarkeit der Sonnenfinsternisse von 1968-1990
totale Sonnenfinsternis
ringförmige Sonnenfinsternis
ringförmig-totale Sonnenfinsternis"

"B Visibility of eclipses from 1968-1990
total eclipse of the sun
circular eclipse of the sun
circular total eclipse of the sun"

The footpath of the Novosibirsk eclipse of the sun of the 01.08.2008 has probably been approximately there, where, on the above picture, the eclipse of the 31.VII.1981 and of the 22.VII.1990 is indicated. Therefore not far from the North Pole, the centre of the picture, a little above of it.

 

This is the end of "Astronomical question and answer 219"
To the German version of this chapter: Astronomische Frage und Antwort 219

 

 

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