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This is all fine, but in local space also the unit of distance (let's call it "rho") is different from "r" of reference coordinates - and not just time. Of course, considering rho as the distance from the origin would not make sense (at r_s it would tend to infinity), but using its first and second derivatives will be correct! (I mean d^2 ...

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Your intuitive guess is indeed correct. To find the velocity, we first use conservation of energy. The energy per unit mass of a particle is given by $$e = -g_{tt} \dot{t} = \left(1 - \frac{r_s}{r}\right) \dot{t}.$$ When the particle is at rest at $r_1$, its four velocity $u^\mu = (\dot{t}, \dot{r}, \dot{\theta}, \dot{\phi})$ has only a $t$-component, and ...

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Since you're a math major, the way to think about this is that Newton's Second Law is an ODE. If $h$ is the distance between the object and the center of the Earth, the force on the object is (as you have shown) $$F = - m g \frac{r^2}{h^2}$$ where $r$ is the radius of the Earth. (Note the minus sign, since the force is attractive. Also, $g$ is more ...

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Summary: Very deep - on the order of 1-10 kilometers. Analysis: Lets make some assumptions. First, we are dealing with air at standard temperature, pressure, etc. Second, lets assume that the walls of the pit are smooth, parallel, and acoustically hard so we don't have to worry about funky reflections. Third, assume that the strike leads to an impulse ...

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The denser end has less bouyancy and will start tilting downwards eventually in a precessing motion. when damped it will point straight down. In a vacuum, the entire rod will have no bouyancy, so it will drop straight down without experiencing any torque.

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