Why don't they measure Earth's gravitational waves which has stronger effect holding us on planet and sounds to me maybe stronger instead of measuring far black hole mass created wave 1/1000th of proton size change in spacetime?
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$\begingroup$ Earth doesn't produce much in form of gravitational waves, but we have instruments that are extremely sensitive to changes in Earth's gravity. They are called absolute gravimeters and here is an example: ngs.noaa.gov/GRD/GRAVITY/ABSG.html. $\endgroup$– CuriousOneFeb 16, 2016 at 10:10
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$\begingroup$ Gravitational waves are to the gravitational field what electromagnetic waves are to the electric and/or magnetic field. see my answer here physics.stackexchange.com/questions/229213/… it needs accelerating or deccelerating gravitational fields (necessary but not sufficient) to get a gravitational wave $\endgroup$– anna vFeb 16, 2016 at 10:27
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3 Answers
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It's important to understand that gravitational radiation is generated for non-static systems only. The Earth, in isolation, is a static system (nothing is changing, it just sits there), as are most planets and stars, and so would not radiate at all. This makes sense if you think about it, because if it was radiating it would be losing energy, yet nothing is changing, so it can't be radiating.
You can consider the Earth-Moon system however, which isn't static, and which does radiate, like any pair of bodies in orbit about each other. There is a formula for computing the power radiated for two bodies in orbit about each other where the field is weak:
$$ P = \frac{32 G^4}{5 c^5}\frac{m_1^2 m_2^2 (m_1 + m_2)}{r^5} $$
Where the terms have the obvious meanings (see Wikipedia for more details). So we can compute this for the Earth-Moon system, and we get $P \approx 8\times 10^{-6}\mathrm{W}$: this is a really small power, radiated by the whole system, and detecting it would be absurdly hard (even by the standards of LIGO, which can detect absurdly tiny strains). The Sun-Earth system radiates about $200\mathrm{W}$ which is more but still an extremely tiny power when compared with the energies in the system. I guess the Sun-Jupiter system might be more, although it will still be tiny.
By comparison the event detected by LIGO in September was, at its peak, radiating more power than the power radiated by all the stars in the universe. That's why it was detectable.
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$\begingroup$ Just as a curiosity: what was the power radiated in the two black holes merging, revealed by ligo? $\endgroup$ Feb 27, 2016 at 10:34
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1$\begingroup$ @KimPeek: they quote a peak GW luminosity of $3.6\times 10^{49}\mathrm{W}$, which is about $200M_\odot/s$ -- it was converting about 200 solar masses to energy per second at peak. $\endgroup$– user107153Feb 27, 2016 at 11:28
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Not every gravitational field is a wave. The earth has an almost static gravitational potential. It's like trying to measure the electric potential of an electric charge with a microscope (a measurement device designed to measure light, electromagnetic waves).
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When you think about GW, think that they move at speed of light and pass through the objects on their way.
Earth's gravitational (field) on the other hand, moves along with earth. Therefore, if you try to measure that as a passing wave, you can not, because, the instrument will move along with earth at the same speed and it will never pass through the equipment due to same speed.
You can measure changing earth's gravitational field from space, if the equipment is not moving with earth. Passing of that field can be considered like a wave, but it will pass with same speed as that of earth, and not at speed of light. Therefore the moving field of earth is not a true gravitational wave.
A true GW gets detached from the source and travels at speed of light.
This explanation is independent of how GW are generated.
