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| Physics Question - ma = mg |
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| TimFox:
One of the great 19th-century physics experiments was carried out by the Hungarian Baron Eötvös from 1885 to 1909. See https://en.wikipedia.org/wiki/E%C3%B6tv%C3%B6s_experiment His group showed that gravitational and inertial mass of different materials was equivalent to a sensitivity of about 10^-8. Note that g, the gravitational acceleration due to the earth, does change with altitude, and slightly due to variations in the density underfoot, but for most purposes (except for inertial-guidance navigation) it is a constant vector pointing down. Kater's pendulum is a classic way to measure g https://en.wikipedia.org/wiki/Kater%27s_pendulum In old laboratory textbooks, the period of the pendulum was measured by having it oscillate in front of a grandfather clock and comparing the frequency to that of the clock pendulum (illustration in above Wikipedia reference). |
| mathsquid:
Rather than going about defining different subscripted g's, I'd use the law of universal gravitation and G = 6.67408 x 10-11 m3 kg-1 s-2 $$ F= G \frac{m_1m_2}{r^2}$$ Plugging in mass and mean radius of the earth gives you $$ F = 6.67408 \cdot 10^{-11} \, \frac{5.971 \cdot 10^{24} m}{(6.371\cdot 10^6)^2} \approx 9.8m$$ meters/s2 |
| radiolistener:
--- Quote from: bostonman on May 13, 2021, 03:03:47 am ---Obviously ma = mg in this case. --- End quote --- This is not obvious. This is known as equivalence principle, which says that the gravity mass equals to the inertial mass. From experiments we know that the gravity mass is very close to the inertial mass, the difference is less than 10^-14 or something like that. But we don't know if they are equals exactly or there is a very small difference which we cannot catch due to insufficient measurement resolution. And there is no explanation why they are very close. Nobody knows that. So, this is very not obvious. :) Probably you will be interested to read the Chapter 28 "Electromagnetic mass" in The Feynman Lectures on Physics Volume 2. It is about the mass of electron. A little quote: --- Quote ---We only wish to emphasize here the following points: (1) the electromagnetic theory predicts the existence of an electromagnetic mass, but it also falls on its face in doing so, because it does not produce a consistent theory—and the same is true with the quantum modifications; (2) there is experimental evidence for the existence of electromagnetic mass; and (3) all these masses are roughly the same as the mass of an electron. So we come back again to the original idea of Lorentz—maybe all the mass of an electron is purely electromagnetic, maybe the whole 0.511 MeV is due to electrodynamics. Is it or isn’t it? We haven’t got a theory, so we cannot say. We must mention one more piece of information, which is the most annoying. There is another particle in the world called a muon—or μ-meson—which, so far as we can tell, differs in no way whatsoever from an electron except for its mass. It acts in every way like an electron: it interacts with neutrinos and with the electromagnetic field, and it has no nuclear forces. It does nothing different from what an electron does—at least, nothing which cannot be understood as merely a consequence of its higher mass (206.77 times the electron mass). Therefore, whenever someone finally gets the explanation of the mass of an electron, he will then have the puzzle of where a muon gets its mass. Why? Because whatever the electron does, the muon does the same—so the mass ought to come out the same. There are those who believe faithfully in the idea that the muon and the electron are the same particle and that, in the final theory of the mass, the formula for the mass will be a quadratic equation with two roots—one for each particle. There are also those who propose it will be a transcendental equation with an infinite number of roots, and who are engaged in guessing what the masses of the other particles in the series must be, and why these particles haven’t been discovered yet. --- End quote --- |
| TimFox:
Obviously, the gravitational and inertial mass can be defined as equal for one specific material, such as pure gold. Baron Eötvös showed that this was true (within very small experimental error) for a wide range of materials. My favorite statement (attributed to) Albert Einstein is "Raffiniert ist der Herrgott, aber boshaft ist er nicht,”. Surely only a demon would violate the equivalence principle by the teeny margin of 10^-14. |
| AntiProtonBoy:
--- Quote from: Brumby on May 13, 2021, 07:21:03 am ---But then it is spoiled by this: --- Quote ---Weight on any planet in kg: weightplanet = m * gplanet / 9.81 --- End quote --- For starters, kg is a unit of mass - not a weight. Secondly, what is that divide by 9.81 all about? --- End quote --- kg and unit of mass equivalence is only true on Earth. In other words 1 kg of weight equals to 1 unit of mass exerting exactly 9.81 newtons of force on earth; or alternatively we can also say that 1 kg of weight equals to 1 unit of mass accelerated by 9.81 m/s2. If you want to measure weight on another planet, then you will need to scale by the ratio gplanet / 9.81. Example: if you weight 100 kg on earth, then you will weigh 16.53 kg on the moon. Here is why: weightearth = m = 100 kg gmoon = 1.62 weightmoon = m * gmoon / 9.81 weightmoon = 100 * 1.62 / 9.81 weightmoon = 16.53 kg Wiki has a pretty good explanation on this: In scientific contexts, mass is the amount of "matter" in an object (though "matter" may be difficult to define), whereas weight is the force exerted on an object by gravity. In other words, an object with a mass of 1.0 kilogram weighs approximately 9.81 newtons on the surface of the Earth, which is its mass multiplied by the gravitational field strength. The object's weight is less on Mars, where gravity is weaker, and more on Saturn, and very small in space when far from any significant source of gravity, but it always has the same mass. https://en.wikipedia.org/wiki/Mass_versus_weight Also, how do I insert LaTeX formulas in this forum? |
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