# Tag Info

## Hot answers tagged earth

3

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 ...

2

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.

2

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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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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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. ...

1

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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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 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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