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You can't add forces on different objects to get an acceleration of one object. All the forces should act on the center of the sphere to have effect on breaking the sphere. All the three forces: the tidal force $F_{T}$, the gravitational force $F_{g}$ and the Tension force because of the rope pulling the sphere $T$ act at the center of the sphere for the ...


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The acceleration of the points $B$ and $C$ will be same assuming that the Earth is not breaking apart, i.e., the acceleration of all points on the Earth will be same. If you think of the Earth as a very strong rod, only the Tension in the rod varies so that the entire rod accelerates with the same acceleration as that of the center of mass of the rod. So, ...


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Fermi transport will be the equations that describe how each polarisation component changes direction, as the ray passes through the gravitational field (or rather, curved spacetime). So for circular polarisation to become eliptical, you need to check whether othogonal components cease to be orthogonal (when Fermi-Walker transported along an arbitrary ...


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Between points A and C The earth can be considered as a wall and thus to break off the 'wall' a relative acceleration with respect to the wall is a must , If the tensile strength is 10N and mass 10 kg then the rope must move at a relative acceleration of 1m/s^2 or Acceleration(a) - Acceleration(c) = 1


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The number of tides stays just about the same, as the earth turns under the tidal bulges. On a water earth we would have two tides per day. Local landforms impact that greatly, varying from place to place. The tidal force is the difference of the moon's gravity on the near and far sides of the earth. It falls off as $\frac 1{r^3}$, so the if the moon ...


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The tides will get smaller as the moon moves away having less gravitational effect. As long as there are tides there will be two per day because this is based on the Earths rotation not the moons distance. As the Earth spins the tides rise on the side facing the moon and the opposite side because of gravity. Its these tidal forces along with the Earths spin ...


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Just imagine there would be no moon at all. Obviously there would be no tides due to the moon. As Dr. Chuck pointed out in the comments, there would still be tides due to the sun, but the tidal forces of the Sun are only about 46% of the moons tidal forces (taken from german Wikipedia about Tides: https://de.wikipedia.org/wiki/Gezeiten). So the farer the ...


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No. Tides are caused by the gradient in the gravitational field. As you get further from the moon, the field drops as $\frac{1}{r^2}$ and the gradient changes as $-\frac{1}{r^3}$. If there is a gradient, then objects closer to the moon will accelerate towards it more rapidly than objects further away from it. The effect of this is nicely illustrated in an ...


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Gravitational waves can be emitted from a rotating object, but the object must not be axisymmetric. For example, a perfect sphere will not radiate gravitational waves, but a sphere with some sort of bulge may. We can calculate the radiated energy from such a source (see for instance this paper). However, for gravitational waves to have effects on the same ...



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