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In terms of milliseconds, how long would it take an object in a vacuum to freefall from its starting point, to the ground a meter away, accelerating at precisely 1g?

Like, if NASA were to make a vacuum chamber, not a single millimeter above or below sea level, suck all the air out of it, and, using a mechanical hand, drop some kind of pellet to the ground that is precisely 1,000 millimeters beneath it, how many milliseconds later would it hit the ground?

I'm pretty sure that this requires a more precise definition of 1g besides simply 9.8 meters per second squared. I read that the actual acceleration of 1g is in fact 9806.65 millimeters per second squared, so this requires some very precise and delicate math.

Oh, by the way, when I said "not a millimeter above or below sea level," what I mean was, the 1,000mm-high vacuum chamber is exactly 500mm above sea level and 500mm below sea level. As much as is physically possible, the acceleration is precisely 1g. After all, your velocity is increasing at a linear rate, due to your acceleration in a vacuum, but at the same time, according to Newton's gravity equation, the acceleration that dictates the linear rate at which your velocity increases will itself increase exponentially as you get closer to the gravity source!

Yeah, I know, this is every logistics experts' wet dream! heehee!

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In terms of milliseconds, how long would it take an object in a vacuum to freefall from its starting point, to the ground a meter away, accelerating at precisely 1g?

Like, if NASA were to make a vacuum chamber, not a single millimeter above or below sea level, suck all the air out of it, and, using a mechanical hand, drop some kind of pellet to the ground that is precisely 1,000 millimeters beneath it, how many milliseconds later would it hit the ground?

I'm pretty sure that this requires a more precise definition of 1g besides simply 9.8 meters per second squared. I read that the actual acceleration of 1g is in fact 9806.65 millimeters per second squared, so this requires some very precise and delicate math.

Oh, by the way, when I said "not a millimeter above or below sea level," what I mean was, the 1,000mm-high vacuum chamber is exactly 500mm above sea level and 500mm below sea level. As much as is physically possible, the acceleration is precisely 1g.

Position of an initially stationary object with respect to time for constant acceleration is:

$x=\frac{1}{2}at^2$

a in this case is 1g, so you wind up with:

$t=\sqrt{\frac{2}{g}\times 1m}$

Or about 0.456+/-0.004 seconds. (456 +/- 4 ms)

If you wanted to be more precise for a changing acceleration you use a differential equation:

After all, your velocity is increasing at a linear rate, due to your acceleration in a vacuum, but at the same time, according to Newton's gravity equation, the acceleration that dictates the linear rate at which your velocity increases will itself increase exponentially as you get closer to the gravity source!

Yeah, I know, this is every logistics experts' wet dream! heehee!

Umm, it's not really exponential, it's inversely proportional to the square.

$\frac{d^2x}{dt^2} = \frac{F}{m} = \frac{GM}{x^2}$

Where x=0 represents the centre of mass of earth.

Solving this is a little bit complicated, but there are a number of reasons why we can consider something at or near the earth's surface to have a constant acceleration (and thus we don't have to).

Firstly, earth is spinning, so when you are at the equator some of the gravity is 'used up' keeping you moving around in circles.

On top of this earth isn't exactly spherical, and even where it is there are fluctuations in its density.

These factors add up to 'exactly 1 g' not really being overly precise. Wiki says it's $9.80\pm0.02 ms^{-2}$

Or varying by about 0.2%

Earth is about 7000km in radius, or 7000000m.

If you fall 1 metre then you are only changing in height by 1/70000%. So the change due to a little change in altitude is going to be insignificant compared to doing your experiment where there are iron deposits, or moving north or south a bit.

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