When we shrink Earth radius, from 6371000 m to a value of 0.00886 m, increasing its material density, we created a black hole with a gravity value of 5.07071e18 ms-2.
A decrease in radius of,
\(-\Delta x\) = 6371000 m (ignoring 0.00886)
results in an increase in force per unit mass of,
\(\Delta \cfrac{F}{m}\) = 5.07071e18 ms-2 (ignoring 9.80665)
Consider the region just above and below the surface of such a planet, where space density is equal. The space density there decreases in both direction above and below the surface. Gravity also decreases in both direction above and below the surface. Gravity is maximum in the region. It is reasonable to assume that increasing space density requires a force. This is the same force that hold the material of the planet together; it is gravity. An opposite and equal force pushes space around the planet surface back. If space behave like a spring, we might have,
\(\cfrac{-k}{m}= \cfrac{\Delta F}{m \Delta x}\)
\(\cfrac{k}{m}\) = 5.07071e18/6371000 = 7.9590e11 Nm-1kg-1
And therefore resonance,
\(f_{res}=\cfrac{1}{2\pi}\sqrt{\cfrac{k}{m}}\)
\(f_{res}\) = sqrt(5.07071e18/6371000)/2/pi
\(f_{res}\) = 1.4199e5 Hz = 141.99 kHz
If we compare this with the frequency of the gravitational wave, \(f_{BH}\), from a black hole,
\(f_{BH}=\cfrac{1}{2\pi}\cfrac{c}{r_{eo}}\) = 1/(2pi)*299792458/0.00886 = 5.3853e9 Hz
They are not the same. Waves at resonance frequency travels further with the least attenuation. The wave from the black hole decrease exponentially from the source. However if we have a planet of the size,
\(\cfrac{1}{2\pi}\cfrac{c}{r_{res}}=f_{res}\)
\(r_{res}=\cfrac{1}{2\pi}\cfrac{c}{f_{res}}\)
\(r_{res}\) = 1/(2pi)*299792458/1.4199e5 = 336.033 m
At this radius, it will generate resonance gravity waves that travels over long distances. A gravity wave antenna will have such a dimension; radius of 336.033 m
