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Imagine a U tube full of water where the Moon's gravity pulls upward on one arm, but not the other. Water would rise in that arm. Water will not flow from one arm to the other if the pressure is the same at the bottom of each arm. $\rho(g_{Earth} -g_{Moon})(h+\Delta h) = \rho g_{Earth} h$ From this you can show $\Delta h = h \frac{g_{Moon}}{g_{Earth}... 0 For a wave, amplitude is crest to mean or trough to mean and range is crest to trough. Amplitude = 1/2 range, so the answer to your question is +60 to -60 or +54 to -54. Not 30 to -30 or 27 to -27. Source. I imagine you read the Wikipedia explanation on how much tides vary. That 54cm would require no sun, but still liquid oceans and the Moon to ... 1 It is all question of scale. If we could harness tidal energy at large scale and more effectively it would slow down the earth faster but at the scales practical now it is not even comparable to topographic changes due to erosion of the continental shelves which brake the tides. Shifting of tectonic plates and mountain ranges could affect the balance of ... 19 Ultimately, the short answer is yes. The moon is slowly moving away from the Earth, at a rate of 4 centimeters per year. This is due to it being in a slightly higher orbit than equilibrium. At the same time, the Earth, being a satellite of the sun, is having its rotation rate slowed, until it will eventually be tidally locked with the sun. Both of ... 34 The moon does work on the Earth because the Moon orbits around the Earth at a different speed than the Earth's rotation. So, as time passes, different parts of the Earth feel the maximum strength of the Moon's pull, leading to oscillations in the height of the Earth's surface, which can be exploited for energy extraction. The Earth has always been ... 0 To formalize @GrahamReid's comments, Tides get their energy from the tidal forces exerted on the Earth by the Moon. Since the moon orbits around the earth much slower than the earth rotates, the moon drags water and rock around the earth in the direction opposite its rotation. In this manner angular momentum and energy is transferred from the rotation of ... 1 What you need to pull an object off the surface of another, gravitationally, is for the tidal acceleration$F_{\rm tidal}$from the "external" body at the surface of the Earth (or whatever other body) to exceed the gravitational acceleration$g$of Earth on the surface. This comes with the caveat that if the tidal forces are too strong, they will start to ... 2 Assuming that the gravity well acts as an attracting mass whose force follows Newton's law of gravitation, your well will suck the object from Earth's surface when the force from the well is larger than the force from the earth. That is,$GM_{earth}m/R_{earth}^2<GM_{well}m/r_{well}^2$, where$r_{well}\$ is the distance between object and well. Earth ...