a couple Qs if you don't mind, because you obv know a lot about this.
I'm just a guy who was interested enough to Google and throw together some calculations.
so I guess if the barycenter is not in the middle of earth, then the earth wobbles as the moon goes around. Is this what causes tides, it's essentially the sloshing of the ocean as the earth wobbles? I always knew that "the moon causes tides", but I never understood the mechanism.
The gravitational forces between the earth and moon are major components of tides. However the barycenter doesn't seem to contribute directly. Moving the earth and moon farther apart (and thus moving the barycenter further away from the center of the earth) actually causes the tides to become weaker. This actually happens regularly as the moon gets closer and then farther from the earth in its orbit (the moon's orbit is not perfectly circular, but slightly elliptical). When the moon is at its closest, the tides are barely higher, and at its furthest the tides are barely lower.
I guess a second question would be, is there a certain distance at which the moon would escape earth's gravity? I wonder what it is, esp compared to the current distance away? would it be 2x, or 10% or 10x?
Gravity accelerates two objects toward one another, based on their mass and their distance. It works the same whether the two objects are initially moving toward each other or away from each other (away from each other, we usually call "decelerating", but there's no difference in the math).
Escape technically occurs when the two objects are moving away from each other, but the deceleration due to gravity will never be enough to overcome their initial velocity at their initial distance from each other. Gravity diminishes as the objects move farther apart, which results in less deceleration over time. In the case of escape, the velocity will never reach zero.
The answer is "yes"... assuming the moon magically appeared at that new farther distance but traveling at the same velocity as it is currently. According to the Wikipedia link on escape velocity:
The escape velocity at a given height is (square root of 2) times the speed in a circular orbit at the same height
Also, the orbital velocity of an object decreases as its distance increases. So increasing the distance of the moon would decrease how fast it would need to be going to stay in orbit.
But remember that the orbital distance suddenly increased but the orbital velocity did not change...
Let's say the moon is orbiting at distance R with orbital velocity V. Thus, all we need to do is figure out at what new distance R2 the new orbital velocity V2 = V * (1 / square root of 2).
This page contains the formula we need. Solving for r, r is proportional to 1/(v^2). So R2 is proportional to 1 / (V2^2), and substituting the equation above we find that R2 is proportional to 2 / (V^2), which equals 2 * (1/V^2), which equals 2 * R. R2 = 2*R.
Thus the answer is "2 times the original orbital distance, 478,000 miles".