when an helium filled balloon rises (in a windless scenario) it would just rise straight up, 90°, right? So I know it has the same speed as the spot on earth it was let go. But with anything spinning, it spins faster the further you get away from the center! So how does it pick up speed to stay over the same spot it has left the earth? From where does it get the force to spin faster? Or if it does not need it – why?
In: Earth Science
>But with anything spinning, it spins faster the further you get away from the center!
On planetary scales the balloon would have to go really high up to see this. The earth’s radius is 3,959 miles. If you calculate the circumference from that, you get 24,875 miles.
Adding one mile to the radius (i.e. if the balloon is one mile higher) gives you a circumference of 24,881 miles. So if the balloon is one mile above you, its path length is only 6 miles longer than your path length (on the surface).
That means that over the course of one entire rotation of 24,881 miles, it will be only 6 miles behind you. We complete one rotation in 24 h, so your rotational (angular) “speed” is 1,036.458 mph while the balloon would need to be moving at 1,036.708 mph to keep up.
**So for a balloon 1 mile up, it would only need to move 0.25 mph faster than you to keep up.**
A hot air balloon floats immersed in the air, similar to the way a fish might float immersed in the ocean. The fish doesn’t see the seafloor below it zoom off at 1000 miles per hour because it is inside the ocean which is rotating with the Earth. Similarly, a hot air balloon rests inside the atmosphere which is rotating with the Earth. The ocean and the atmosphere both rotate with the Earth for the same reason: friction with the solid Earth would prevent them both from moving (on average) across the surface.
Explanation Here:
https://physics.stackexchange.com/questions/525002/why-does-hot-air-balloon-rotate-with-the-earth
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