I understand that gravitational waves pass quite freely through massive bodies.

Quoting http://www.ligo.org/science/GW-Potential.php:

Gravitational waves will change astronomy because the universe is nearly transparent to them: intervening matter and gravitational fields neither absorb nor reflect the gravitational waves to any significant degree.

Now, there must be some interaction between the Earth and gravitational waves, otherwise we wouldn't be able to detect them. I'd like to understand the magnitude of this interaction. If you had two infinitely sensitive detectors on different sides of the Earth, how much weaker would a gravitational wave be after it passed through Earth?


1 Answer 1


In a 1969 paper, Seismic Response of the Earth to a Gravitational Wave in the 1-Hz Band, Dyson estimated that the Earth absorbs about $10^{-21}$ of the energy of a 1-Hz gravitational wave passing through it, and that this ratio varies as the inverse square of the frequency. LIGO detects gravitational waves with frequencies of order 100 Hz, so for them the absorption ratio would be of order $10^{-25}$.

Thus the Earth is indeed essentially transparent. A detector on the far side of the Earth (i.e., away from the source of the gravitational wave) is just as likely to detect the wave as a detector on the near side.

The energy flux for a gravitational wave is


where $h$ is the dimensionless wave amplitude and $f$ the frequency. LIGO waves have $h$ of order $10^{-21}$ and $f$ of order $100$ Hz, and thus energy fluxes of about 3 milliwatts per square meter (around twice the energy flux a full moon). The Earth's radius is about $6\times 10^6$ meters, and its cross sectional area about $5\times 10^{14}$ square meters. Thus the gravitational wave power hitting the Earth is on the order of a terawatt, for maybe a tenth of a second. The gravitational wave energy that Earth absorbs is on the order of $10^{-14}\,\text{J}$... miniscule!

  • $\begingroup$ Wait - so it's about the same (perhaps "twice") the power of the full moon???? But only for 1/10th of a second. Is that correct? :O $\endgroup$
    – Fattie
    Commented Mar 4, 2021 at 20:02
  • $\begingroup$ @Fattie Yes. There is an enormous amount of energy in the wave, but it has spread out over a huge sphere... perhaps a billion light years in radius. That doesn’t leave much energy per square meter. And it is a short-lived wave because the merging black holes only radiate copiously in their final moments. It was there earlier, but too weak to detect. $\endgroup$
    – G. Smith
    Commented Mar 4, 2021 at 21:38
  • $\begingroup$ (thanks for the reply!) Wait, I may misunderstand. I thought you meant, (i) when it arrives on Earth (ii) considering only "what hits the Earth" (iii) in fact it is about the same flux of energy as a full moon. To me that seems incredibly high?? A full moon is (obviously) easy to see, it can light up a sheet of paper. Is my misunderstanding that the energy is the same but it's incredibly hard to detect the gravitational wave?? (But trivial to detect "that much" light??) Thanks! $\endgroup$
    – Fattie
    Commented Mar 4, 2021 at 22:18
  • $\begingroup$ sorry if i have a drastic misunderstanding, @G.Smith ! $\endgroup$
    – Fattie
    Commented Mar 4, 2021 at 22:19
  • $\begingroup$ @Fattie Yes, gravitational waves interact with matter much more weakly than electromagnetic waves. They mostly just pass through without interacting. $\endgroup$
    – G. Smith
    Commented Mar 4, 2021 at 22:30

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