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Congrats to the LIGO team on the announcement of their discovery of gravitational waves!
The articles I've read say that the distortion we see here is much smaller than a proton. What about at the source? Would these waves have been strong enough to see macroscopic effects near the binary black holes themselves? Could you orbit the system at a "safe" distance, observe the merger, and "feel" the waves?
The strain (ratio of displacement from equilibrium to equilibrium separation) of gravitational waves decreases as $1/r$ for a distance $r$ from the source. Since the strain in this case peaked at $10^{-21}$ at a distance of $1.3\times10^9\ \mathrm{ly} = 1.3\times10^{25}\ \mathrm{m}$, you would expect strains on the order of $1\%$ at a distance of $1300\ \mathrm{km}$. For reference, the observed signal is from black holes that were about $100\ \mathrm{km}$ in radius initially.
Much closer than this, and the linearized theory of GR breaks down. Gravitational waves are only well-defined in the small-amplitude limit. Close-in, we have the near-field regime where nonlinear effects that can't really be described as simple waves dominate. The only simple statement that really can be made here is that distortions are even stronger closer in.
According to the press announcement, the merger of the black holes released an amount of energy equivalent to three solar masses (an energy given by $E=mc^{2}$ with $m$ being three times the mass of the sun), which is an absolutely mind-boggling amount of energy. If you were too close, the rapidly fluctuating tidal forces on your body would cause alternating compressive and tensile stresses on your body, tearing your body to bits, which would certainly count as a "macroscopic effect".
Sure, at an appropriate distance, you'd still be able to feel the gravitational wave without being killed. However, with such an enormous amount of energy being released from two objects that are quite small and close together by astronomical standards, for the experience to be survivable you'd have to be so far away that you wouldn't really be able to "observe the merger" with the naked eye. I.e., you wouldn't be able to see two black blobs in the sky that merge into one.
Knowing that the emitted energy is of about 3 solar masses, and if we use the famous Einstein energy formula $E=mc^2$ we have then :
$\text{A.N.:}\ \ E=1,989\cdot10^{30} \times 3 \times c^2\approx2\cdot10^{43} \ \text{J}$
which is a huge amount of energy. This is a bigger emission of energy than a supernova (!) and similar to gamma-ray bursts (biggest emission of light after the Big-Bang)...
Making "popular science", this is the caloric intake (energy) that $5\cdot10^{34}$ Nutella pots of 400kg would give you.
If that amount of energy would be inside the gravitational waves passing threw us, we would be looking like spaghettis. Our body and Earth would be visibly distorted or elongated. Maybe the word "feel" is not appropriate because it would be more "die". Orbiting the system at a safe distance is plausible, knowing that we have to take care the combined ISCO (innermost stable circular orbit or last stable orbit) of both black holes before they merge. We can determine it by :
$R_{ISCO} = \dfrac{6GM}{c^2}$
where $M$ is the mass of the black hole in $kg$ and $G$ the universal gravitational constant in $N\cdot m^2\cdot kg^{-2}$ and $c$ the velocity of light in $m\cdot s^{-1}$. We then have, for the most massive black hole (here $36$ solar masses) :
$R_{ISCO} = \dfrac{6\times6,67384\cdot10^{-11}\times36\times1,989\cdot10^{30}}{299 792 458^2}=319,0\cdot10^3m=319,0km$
For the ring-down black hole of $62$ solar masses, we have ${{R}_{rd}}_{ISCO}=5,494\cdot10^5m=549,4km$
