re: How high would you have to jump in order to land in a different spot?

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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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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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On the equator or on the tip of one of the poles?

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On the equator or on the tip of one of the poles?

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Depending on your exact latitude, Earth spins at up to 1,000 mph (1,600 km/h).

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Earth's Rotation: Depending on your exact latitude, Earth spins at up to 1,000 mph (1,600 km/h). Orbiting the Sun: Our planet races around the Sun at about 67,000 mph (107,000 km/h). Orbiting the Milky Way: The entire Solar System speeds through our galaxy at roughly 514,000 mph (828,000 km/h). Galactic Motion: Relative to the cosmic microwave background radiation (the thermal leftover from the Big Bang), the Milky Way itself moves at 1.3 million mph (2.1 million km/h) toward the constellation Leo. We are fast as frick, boy!

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Oh! Scruffy knows this one! A bike!

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