A bit of both.  <a href="https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Gravitational_wave#Wave_amplitudes_from_the_Earth.E2.80.93Sun_system" rel="nofollow">https:&#x2F;&#x2F;en.wikipedia.org&#x2F;wiki&#x2F;Gravitational_wave#Wave_amplit...</a> runs through the numbers for the gravitational waves the Earth produces by orbiting the sun; it claims that at 1 light-year the amplitude would be 1 part in 1e26.<p>For comparison, the wave that was detected is claimed to be &quot;four one-thousandths of the diameter of a proton&quot;.  That&#x27;s about 7e-18 meters, on a baseline arm length of 4 km, so about one part in 6e20 -- about 175,000 times stronger than the waves Earth&#x27;s orbit produces.  And that was about 40x as strong as minimal sensitivity on LIGO, according to the article (&quot;can detect changes in the length of one of those arms as small as one ten-thousandth the diameter of a proton&quot;).<p>Obviously if we were closer to the black hole collision we&#x27;d see much stronger waves.  But you really do need very massive bodies accelerating very much (or equivalently orbiting very fast) to produce something that&#x27;s detectable by LIGO over interstellar distances at all.  The key part from this article is that the orbital period was about 1&#x2F;250 of a second at the end; compare to Earth&#x27;s orbital period.  Going back to the formula given in the above Wikipedia entry, the frequency dependence is hiding in the &quot;1&#x2F;r&quot; factor for the amplitude.  1&#x2F;r is proportional to w^{2&#x2F;3} (though it&#x27;s not clear to me whether that&#x27;s still true in a general-relativistic treatment; it&#x27;s true enough for the Earth&#x27;s orbit), which tells you how the wave amplitude scales with frequency...