Notro

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If you're standing on Earth and jump straight up, you already have the same eastward velocity as the ground beneath you due to Earth's rotation. So when you're in the air, you continue moving east at almost the same speed as the surface below you.

The key effect that can make you land somewhere slightly different is the Coriolis effect, which causes a tiny horizontal deflection while you're airborne.

Let's estimate how high you'd need to jump to land more than 6 inches (0.1524 m) from where you started (the radius of a 1-foot-diameter circle).

For a vertical jump to height hhh, the time in the air is:

t=22hgt = 2\sqrt{\frac{2h}{g}}t=2g2h??

The Coriolis deflection for a vertical jump at latitude ?\phi? is approximately:

d˜43 O cos?? (2h)3/2gd \approx \frac{4}{3}\,\Omega\,\cos\phi\,\frac{(2h)^{3/2}}{\sqrt{g}}d˜34?Ocos?g?(2h)3/2?

where:

O=7.292×10-5 rad/s\Omega = 7.292\times10^{-5}\ \text{rad/s}O=7.292×10-5 rad/s (Earth's rotation rate)

g=9.81 m/s2g = 9.81\ \text{m/s}^2g=9.81 m/s2

At a mid-latitude of about 45°45^\circ45°:

d˜6.2×10-5h3/2d \approx 6.2\times10^{-5} h^{3/2}d˜6.2×10-5h3/2

(with ddd and hhh in meters).

Setting d=0.1524d = 0.1524d=0.1524 m and solving:

h˜180 mh \approx 180\ \text{m}h˜180 m

or about 590 feet.

The airtime would be:

t˜12 secondst \approx 12\ \text{seconds}t˜12 seconds

which is far beyond any human jump.

So, due solely to Earth's rotation, you'd need to jump roughly 600 feet straight up before the Coriolis deflection would carry you outside a 1-foot-diameter circle. The exact number varies with latitude (it's larger near the equator and smaller near the poles), but it's on the order of hundreds of feet, not inches or meters.

Interestingly, if you could somehow leap that high, you'd drift only a few inches despite being airborne for over 10 seconds because both you and the ground are already moving east together at nearly the same speed.

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Could be Lupis

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