4 ms·

You're correct, and my napkin guess is that it's going about 1.5mph faster than sea level. But the ISS is going a whopping 16100 mph faster than sea level [0].

You're correct, and my napkin guess is that it's going about 1.5mph faster than sea level. But the ISS is going a whopping 16100 mph faster than sea level [0]. Popping that into MS Mathematics using the Lorentz equation, we see that gamma now is 0.2 billionths above unity. Which is super small, but definitely detectable. In contrast, 1.5mph doesn't yield anything - I'm guessing (1.5/670616629)^2 is nearing the floating point epsilon (I guess MS Math is using floats?) [1]. [0] http://www.wolframalpha.com/input/?i=orbital+velocity+of+ISS http://www.wolframalpha.com/input/?i=orbital+velocity+of+ISS [1] http://en.wikipedia.org/wiki/Machine_epsilon http://en.wikipedia.org/wiki/Machine_epsilon EDIT: Right, as noted elsewhere, this assumes Everest is at the equator. Which Wikipedia tells me is not a great approximation for where Nepal is. So assume that Nepal ended up in the wrong place for a little while, and the arguement holds. Otherwise, we've got a bit more calculating to do.

haberman 12y ago
Thanks for the info! What's the right answer if someone makes this counterargument: in the inertial reference frame of the Earth the Earth is not rotating, so neither the sea nor Mt. Everest are moving at all, so there is no relative velocity difference and no time dilation from velocity difference. Btw, I'm not sure about the phrase "I'm guessing (1.5/670616629)^2 is nearing the floating point epsilon" (unlike physics, I do have a fair amount of expertise in floating point formats). The way you put it suggests that the "machine epsilon" represents the smallest increments that floating point can represent, and therefore suggests that (1.5/670616629)^2 represented as float will underflow to zero or be so inaccurate as to be meaningless (sorry if I'm misreading you). I find the definition of "machine epsilon" given on that page somewhat confusing. I think it's more intuitive to think of floating point error in terms of percentages. That page says the "machine epsilon" for float is ~1e-7; an equivalent and IMO more intuitive way to say it is that float is accurate to ~0.00001%. The range of float goes far smaller than 1e-7; FLT_MIN is ~1e-38, and that's not even considering subnormal numbers. So float can very easily represent the results of (1.5/670616629)^2 ≈ 5e-18, and like any other representable float this is accurate to no worse than ~0.00001%, which is pretty decent.
- HCIdivision17 12y ago
  On the topic of the floating point mechanics, I'll readily defer to you. I'll be honest: I was grasping for an explanation of why the result was simply a solitary "1", and there's likely a much smarter explanation (like "After a number of decimal places, nobody gives a damn." - false, but useful typically). As for the counterargument: I don't think that works. And that's taking into account the harrowing liberties I'm willing to assume for the sake of a physics argument. In the inertial reference frame of the Earth, the Earth is rotating - as in, there's rotational inertia in that frame. This has measureable effects on stuff, from time dilation from velocity (very small) to time dilation from gravity warping spacetime (noticeably larger, but still small). Put another way, here's a thought experiment. The mantel of the planet is molten rock, and can be treated as a viscous fluid. If the planet were spinning, the fluid would bulge out at the latitudes where the planet is spinning fastest (centrifugal force stuff). Otherwise the planet would be a sphere. This is directly measurable, and - in fact - the planet's a sphere. Mostly. It bulges out a bit at the equator [0]. So if you're careful, you'll note it's never fair to say you're in a non-rotating inertial frame on earth. But in practice it almost never matters. Unless you're doing something crazy like measuring femtillionths of a second with one of the most sensitive devices we can build - then we start being a bit wrong. Or you're just trying to be accurate with your GPS satellites (which are way higher and faster than a mountain top). I rant a bit about this, as this sort of counterargument comes up a lot. I think it's due to the completely unreasonable mismatch of scales people are used to. Feynman ranted a bit on it about QM, and we're running into it here with relativity [1]. Here we're trying to talk about something reasonably, and the levels of precision are completely unreasonable: we're talking about a couple mile's difference over the span of thousands of miles to have an effect on the order of a few (bi|tri)llionths of a second; how do you keep such a scale in mind? It's like the silly analogies of hitting a baseball in NYC and nailing a bumble bee in San Fransico for precision. By the same token, we're hurtling through space, whipping around the sun and being wobbled so hard by the moon the ocean sloshes over our beachfronts. And that seems perfectly normal, even though it directly implies enormous forces at work. /rant (and sorry for that - I find this sort of thing facinating) TL;DR: I think that counterargument is simply wrong. But completely understandably so. [0] http://image.gsfc.nasa.gov/poetry/ask/a11818.html http://image.gsfc.nasa.gov/poetry/ask/a11818.html [1] http://bouman.chem.georgetown.edu/general/feynman.html http://bouman.chem.georgetown.edu/general/feynman.html

haberman 12y ago

Thanks for the info! What's the right answer if someone makes this counterargument: in the inertial reference frame of the Earth the Earth is not rotating, so neither the sea nor Mt. Everest are moving at all, so there is no relative velocity difference and no time dilation from velocity difference. Btw, I'm not sure about the phrase "I'm guessing (1.5/670616629)^2 is nearing the floating point epsilon" (unlike physics, I do have a fair amount of expertise in floating point formats). The way you put it suggests that the "machine epsilon" represents the smallest increments that floating point can represent, and therefore suggests that (1.5/670616629)^2 represented as float will underflow to zero or be so inaccurate as to be meaningless (sorry if I'm misreading you). I find the definition of "machine epsilon" given on that page somewhat confusing. I think it's more intuitive to think of floating point error in terms of percentages. That page says the "machine epsilon" for float is ~1e-7; an equivalent and IMO more intuitive way to say it is that float is accurate to ~0.00001%. The range of float goes far smaller than 1e-7; FLT_MIN is ~1e-38, and that's not even considering subnormal numbers. So float can very easily represent the results of (1.5/670616629)^2 ≈ 5e-18, and like any other representable float this is accurate to no worse than ~0.00001%, which is pretty decent.

HCIdivision17 12y ago
On the topic of the floating point mechanics, I'll readily defer to you. I'll be honest: I was grasping for an explanation of why the result was simply a solitary "1", and there's likely a much smarter explanation (like "After a number of decimal places, nobody gives a damn." - false, but useful typically). As for the counterargument: I don't think that works. And that's taking into account the harrowing liberties I'm willing to assume for the sake of a physics argument. In the inertial reference frame of the Earth, the Earth is rotating - as in, there's rotational inertia in that frame. This has measureable effects on stuff, from time dilation from velocity (very small) to time dilation from gravity warping spacetime (noticeably larger, but still small). Put another way, here's a thought experiment. The mantel of the planet is molten rock, and can be treated as a viscous fluid. If the planet were spinning, the fluid would bulge out at the latitudes where the planet is spinning fastest (centrifugal force stuff). Otherwise the planet would be a sphere. This is directly measurable, and - in fact - the planet's a sphere. Mostly. It bulges out a bit at the equator [0]. So if you're careful, you'll note it's never fair to say you're in a non-rotating inertial frame on earth. But in practice it almost never matters. Unless you're doing something crazy like measuring femtillionths of a second with one of the most sensitive devices we can build - then we start being a bit wrong. Or you're just trying to be accurate with your GPS satellites (which are way higher and faster than a mountain top). I rant a bit about this, as this sort of counterargument comes up a lot. I think it's due to the completely unreasonable mismatch of scales people are used to. Feynman ranted a bit on it about QM, and we're running into it here with relativity [1]. Here we're trying to talk about something reasonably, and the levels of precision are completely unreasonable: we're talking about a couple mile's difference over the span of thousands of miles to have an effect on the order of a few (bi|tri)llionths of a second; how do you keep such a scale in mind? It's like the silly analogies of hitting a baseball in NYC and nailing a bumble bee in San Fransico for precision. By the same token, we're hurtling through space, whipping around the sun and being wobbled so hard by the moon the ocean sloshes over our beachfronts. And that seems perfectly normal, even though it directly implies enormous forces at work. /rant (and sorry for that - I find this sort of thing facinating) TL;DR: I think that counterargument is simply wrong. But completely understandably so. [0] http://image.gsfc.nasa.gov/poetry/ask/a11818.html http://image.gsfc.nasa.gov/poetry/ask/a11818.html [1] http://bouman.chem.georgetown.edu/general/feynman.html http://bouman.chem.georgetown.edu/general/feynman.html