Can a spinning object with an unbalanced mass cause “tidal acceleration” on an orbiting satellite, without tidal mass displacement involved? The sun spins and it is not 100% balanced in mass. Look at the sun from North Pole and dissect it into 4 quadrants A B C D, each of those quadrants is not exactly equal to 25% of the mass. By the quadrants not being equal is the point that gravity is attracting to not located directly in the center of spin? Is the point where gravity is attracting to orbiting the center of spin the cause of tidal acceleration on orbiting satellites?
Tides are caused by the difference in gravitational attraction at different points on the attracted object. The Moon would cause tides on Earth even if it was a perfect, uniform sphere that was not rotating on its axis. This is because the force with which the Moon pulls on the water on the near side of the Earth is greater than the force of its pull on the far side.
100% I agree that the force is greater on the closer side of the earth than the opposite side. But is the main cause not that the force is directional to the moon than that the force is stronger on one side than the other. Even on the far side the water still wants to move in the same direction as the closer side? What im asking is when the tide rise or fall does the center of gravity point change in the earth because the mass balance has changed? If the center of gravity changes its not located at the center of spin, is the offset of the center of gravity and center of spin the cause of tidal acceleration? Because the center of gravity would start to circle/orbit the center of spin within the earth or satellite host.
Well, it's both. On the far side the water wants to move in the OPPOSITE direction as the closer side - i.e. away from the Moon and away from the center of the Earth. This is due to orbital mechanics. Picture three masses in orbit around the Moon. Put them on a line extending from the center of the Moon, and put them 4000 miles from each other, so that one is closest to the Moon and one is farthest. The center one is in a perfectly circular stable orbit. If all are going the same speed, the closest one will fall towards the Moon - because it is going slower than the ideal orbital speed. The farthest one will move away from the moon - because it is going faster than ideal orbital speed. In the case of the three masses, they will eventually stabilize in new elliptical orbits. But water on Earth isn't free to seek a new orbit, so it stays here - but moves slightly in response to those forces. That's what a tide is. Nope; you get two bulges, one on either side, so no net change in CoG. Keep in mind that there are two CoG's - one of the Earth, and another of the Earth/Moon system.
no and yes caveat, it seems most likely that the atmospheric tides are primarily governed by the sun.
On a perfect globe of water they would be equal. However, variations in coastlines, ocean depths, changes in temperature, atmospheric pressure etc conspire to make them less than equal in almost all cases. Yes, although since there's no easily defined "edge" of the atmosphere it's harder to quantify.
Not to mention confounding factors, such as the fact that the atmo expands and contracts as it interacts with the sun's radiation and winds.
