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No. The Earth's atmosphere is held in place entirely by the Earth's gravity. There is no barrier at the top of the atmosphere keeping it from escaping. Running a straw from the ground to deep space would not change the gravitational force holding the air molecules down; it would, in fact, change nothing.

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A lens deflects light rays, bringing them to a focus. A gravitational lens is typically a galaxy or cluster of galaxies. A galaxy typically has trillions of stars. The sun deflects light rays a little. See How the Sun Warps Starlight, or Gravitational deflection of light. A ray that skims its surface is deflected by about 1.8 arcsec. These rays would come to ...

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Your error is assuming that sunrise and sunset occur at $\theta = 0$, this is incorrect for two reasons. The first being the sun's disk has a diameter of $0.5^{\circ}$, and the second is atmospheric refraction causes distortion which change the sun's apparent location on the horizon. Altogether, at sea level this leads to an angle of correction of about $-0.... 2 The system planet+people is isolated and no external torques act on it, so its angular momentum$L_{tot}$is conserved. Let's write $$L_{tot} = L_{pl} + L_{earth} \, ,$$ where$L_{pl}$is a complex function of all the positions and velocities of the world population and cars, while$L_{eath}$is the angular momentum of a solid sphere (but you can also ... 2 Back of the envelope calculation: The earth is a solid sphere with an average density several times that of the human body. Even billions of people will only be a thin layer if spread around the whole surface of the earth. So the mass of billions of people is minute compared to the mass of the earth. The earth rotates once in$24$hours. The equator is about ... 2 Okay, if we have to calculate the speed with which each person needs to walk/run to stop the earth we may try to conserve the angular momentum. Lets say that the world population is$P$and average mass of each person is$m$. Then the total mass of humanity would be$Pm$. Say everyone walks with a speed$v$. Then on conserving the angular momentum we get:$\$(...

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The moon also will feel tidal forces from the earth (we don't see this because there is no water on the moon!) even though the moon technically is falling toward the earth as the earth is falling toward the sun. Tidal forces are caused because of difference in how the forces are felt at different points across the orbiting object. Even though the Sun is much ...

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This contrasts to e.g., the ISS with reduced / microgravity only. I'm not sure how the ISS contrasts. Both raindrops and the ISS experience gravity and accelerate toward the earth. The only difference is that the raindrop will also experience drag and in short order will strike the surface. I speculate the inner core could equally move slowly / «sink» ...

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The core experiences centrifugal forces as well as centripetal forces, just as the rest of the Earth does from its orbit around the Sun, and from the Earth - Moon orbit. While the center of gravity and the center of mass are not always the same, given the Earth's mass, rotational speed, and distance from the Sun, the changes to the core would be negligible. ...

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The entirety of all humans is still negligible to the mass of Earth. However if everyone were to run in the same direction at once, the minute change would be canceled when they stopped, as stopping would transfer their momentum back to the Earth. Every action and reaction on the surface of Earth equal out.

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I include pictures from moon phases, you find them in the Internet quiet easily, so one has almost straight lines only at half moon. Perhaps you tell us which picture is the one you saw? Or draw a sketch yourself? From your question it is hard to imagine what you saw different from other times.

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