- #1
fog37
- 1,568
- 108
Hello Forum,
We are on planet Earth that is rotating at an angular speed of omega = 7.27 × 10-5 rad/s. The Earth atmosphere is also rotating with Earth at the same speed (that is why it is at rest relative to the Earth surface).
Let's say I have a coin in my hand that is 1 meter above the Earth surface. That coin has the same angular velocity as anything that is on the Earth surface but a different and higher tangential speed v which is higher by omega*r = omega (1 meter). Point that are further away from the Earth center must have a linearly larger tangential speed v.
That said, If I launch the coin in the air and that coin reaches at height of 2 meter, I believe the coin will fall down and not drop exactly on my hand where it was initially. Explanation: the speed need at h=2 meter is higher than the speed at h=1 meter. When the coin reaches h=2m its speed is not sufficient to stay align with the speed of my hand at the lower height h=1 meter...
This effect may be so negligible that we may state that the coin fall back on my hand on the same spot where it took off... but if the height difference is significant we should notice this effect.
The same goes it we drop a coin from a very tall tower where the tangential speed v2 is higher than the tangential speed at lower altitudes v1. The coin will fall slightly ahead of the position from which it was dropped because its speed v2 will allow the coin to cover a longer distance compared to the distance covered by object traveling at speed v1...
A helicopter that raises straight up vertically from the surface of the Earth should progressively see the point from which it took off move away from directly below...
Is my thinking correct?
thank,
fog37
We are on planet Earth that is rotating at an angular speed of omega = 7.27 × 10-5 rad/s. The Earth atmosphere is also rotating with Earth at the same speed (that is why it is at rest relative to the Earth surface).
Let's say I have a coin in my hand that is 1 meter above the Earth surface. That coin has the same angular velocity as anything that is on the Earth surface but a different and higher tangential speed v which is higher by omega*r = omega (1 meter). Point that are further away from the Earth center must have a linearly larger tangential speed v.
That said, If I launch the coin in the air and that coin reaches at height of 2 meter, I believe the coin will fall down and not drop exactly on my hand where it was initially. Explanation: the speed need at h=2 meter is higher than the speed at h=1 meter. When the coin reaches h=2m its speed is not sufficient to stay align with the speed of my hand at the lower height h=1 meter...
This effect may be so negligible that we may state that the coin fall back on my hand on the same spot where it took off... but if the height difference is significant we should notice this effect.
The same goes it we drop a coin from a very tall tower where the tangential speed v2 is higher than the tangential speed at lower altitudes v1. The coin will fall slightly ahead of the position from which it was dropped because its speed v2 will allow the coin to cover a longer distance compared to the distance covered by object traveling at speed v1...
A helicopter that raises straight up vertically from the surface of the Earth should progressively see the point from which it took off move away from directly below...
Is my thinking correct?
thank,
fog37