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## New answers tagged earth

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The above answers do not take into account the vector nature of $\omega$. I venture to provide an answer that hopefully should satisfy the questioner. We Consider a point at colatitude $\lambda$ on the Earth's surface in the Northern hemisphere. We draw a local coordinate system at this point such that the x-axis points south, the y-axis points east, and the ...

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The point is if all effect was taken into account. Math would be summed up that effect of more mass under your feet still less than effect of distance from the center of mass Another view is. At equator there are bulge near you. But from all other side of earth the bulge is far from you. Compare to the pole that all bulge is equally far from you, that ...

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Now perfect balance between the centrifugal force of orbital rotation and sun's gravity is impossible so the earth's orbit should either be slowly decaying inwards or expanding outwards due to difference in magnitude of those opposing forces. This assumption is incorrect. We could make the same argument about a weight suspended from a spring. ...

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Very generally speaking, a crater is about 10 times the diameter of the meteor, with a direct hit. so at some 950 km in diameter, we can guesstimate a crater roughly covering 9,500 km, which is 1/4 the way around the Earth. If we give an impact speed of slightly greater than escape velocity of 12-13 km/s, it would take over a minute to complete it's ...

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It's rarer than that, more like 1 in 6000. Because each molecule of water has two atoms of hydrogen, then about every 2 in 6000 (1 in 3000) has a single atom of deuterium (DHO). And would be closer to 1 in every 6000$^2$ for a molecule of D2O. The linked question Deuterium density in seawater gives sources that show deuterium is well-mixed in the ocean. ...

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Let's just say, for the sake of argument, that the equator has a linear eastern velocity of 1,000 mph (actually 1,039). If you release a ball (presumably fired from a cannon) southward from a latitude of 70° north, your ball has an eastern velocity of about 342 mph (cosine of 70° = 0.342). As your ball travels over 45° north, the ground beneath it will have ...

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The direction of deflection of the ball should be independent where the observer is. Left or right deflection is with respect to an observer facing in the direction the ball is moving. Are you asking about a ball being thrown from the northern to the southern hemisphere? In that case the deflection would change from deflecting right to deflecting left as ...

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The vector $\mathbf r$ does change, even though its magnitude is nearly constant. Most importantly, the component of $\mathbf r$ which is perpendicular to the rotation axis is decreasing in this example. This fact leads to the explanation for which you are searching. Another way of seeing this would be to use this definition of angular momentum, which ...

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The error is just to consider an average speed $h\omega$. When the particle is at height $z$, its horizontal (relative to the Earth) speed is $v=2z\omega$. The time of of flight is $$t=\sqrt{\frac{2z}{g}}.$$ Differentiating this expression we get the time taken by the particle to move a distance $dz$, $$dt=\frac{dz}{\sqrt{2gz}}.$$ The horizontal distance ...

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But you can see the motion of the earth in it's orbit; it's called aberration of starlight, first measured by James Bradley in 1729. Even earlier, parallax of stars had been detected, by 1680. But you have to take detailed observations at widely separated times for the effects to be even the least bit obvious.

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We can detect it, but it takes very precise measurements. Stellar parallax, i.e. the relative displacement of close-by stars against the background of far away ones can be detected, but it's a very difficult measurement to make because the "motion" is very small (usually fractions of an arc second): Recently we have learned to build satellites that can ...

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