- #1
My_name_is_Peter
- 2
- 0
The Earth rotates around its own axis within 24 hours. Theoretically, if at the equator perform a tower with a height - H, ignoring its effect on the slowdown of the Earth's rotation (we assume that the material of low density), objects and other complications, at a certain height, the linear velocities at the top of the tower would be very large compared to the stationary observer in relation to the axis of rotation of the Earth and was in the outer space at the height of the top. You can consider the view of the observers on the Earth on the equator and the one on the top of the tower, and in space next to the top of the tower passing him. You can increase the H value and try to calculate the linear velocity of the top of the tower relative to theoretical observer in space next to the top of the tower passing him.
The second thing can be if the tower is non-zero width along the meridian and its height would be theoretically 24 light hours, would it cover You some of the space view surrounding the equator by the width of the tower? Will You see the tower spreading from east to west? Would photones go from space be swept away by the tower if it did not reflect light? What would see the observer in space next to the top of the tower passing him? Will he see the meridian on the Earth?
You can also create a new questions, about it.
Thank You so much :)
and wish You Fascinating considerations :)
The second thing can be if the tower is non-zero width along the meridian and its height would be theoretically 24 light hours, would it cover You some of the space view surrounding the equator by the width of the tower? Will You see the tower spreading from east to west? Would photones go from space be swept away by the tower if it did not reflect light? What would see the observer in space next to the top of the tower passing him? Will he see the meridian on the Earth?
You can also create a new questions, about it.
Thank You so much :)
and wish You Fascinating considerations :)