- #1
eldorado
- 2
- 0
Hello :)
When I read about Tsar Bomba, I find that all sites say "The fireball touched the ground, reached nearly as high as the altitude of the release plane and was seen and felt almost 1,000 kilometres (620 mi) from ground zero."
Diameter of the fireball is 8 km.
Now , there are some premises :
1 Distance between the explosion & point that fireball seen is 1,000 km.
2 Circumference of the Earth is about 40000 km.
3 Angle between the two lines which passes through center of the Earth & ( ground zero & 1000 km point ) = (1000/40000)*360 = 9 degrees.
4 Sight line of 1000 km point is the tangent of the earth.
5 To see the explosion from 1000 km point , fire must be at least 79.5 km high.
this conclusion came from:
hight of the fire to be seen 1000 km far = ( Radius of the Earth / Cos (9 degrees ) ) - Radius of the earth.so , how could the explosion seen from this distance when the hight of the fire is only 8 km ?
thanks for help :)
When I read about Tsar Bomba, I find that all sites say "The fireball touched the ground, reached nearly as high as the altitude of the release plane and was seen and felt almost 1,000 kilometres (620 mi) from ground zero."
Diameter of the fireball is 8 km.
Now , there are some premises :
1 Distance between the explosion & point that fireball seen is 1,000 km.
2 Circumference of the Earth is about 40000 km.
3 Angle between the two lines which passes through center of the Earth & ( ground zero & 1000 km point ) = (1000/40000)*360 = 9 degrees.
4 Sight line of 1000 km point is the tangent of the earth.
5 To see the explosion from 1000 km point , fire must be at least 79.5 km high.
this conclusion came from:
hight of the fire to be seen 1000 km far = ( Radius of the Earth / Cos (9 degrees ) ) - Radius of the earth.so , how could the explosion seen from this distance when the hight of the fire is only 8 km ?
thanks for help :)