Groups in Seattle, Colorado, and perhaps others managed to measure and verify Newton's inverse-square law at submillimeter distances comparable to 0.1 millimeters, see e.g.
Sub-millimeter tests of the gravitational inverse-square law: A
search for "large" extra dimensions
Motivated by higher-dimensional theories that predict new effects, we
tested the gravitational $\frac{1}{r^{2}}$ law at separations ranging down to 218
micrometers using a 10-fold symmetric torsion pendulum and a rotating
10-fold symmetric attractor. We improved previous short-range
constraints by up to a factor of 1000 and find no deviations from
Newtonian physics.
This is a 14 years old paper (with 600+ citations) and I think that these experiments were very hot at that time because the warped- and large-dimensions models in particle physics that may predict violations of Newton's law had been proposed in the preceding two years.
But I believe that there's been some extra progress in the field. At that time, the very fine measurement up to 200 microns etc. allowed them to deduce something about the law of gravity up to 10 microns. These are extremely clever, fine mechanical experiments with torsion pendulums, rotating attractors, and resonances. The force they are able to see is really tiny.
To see the gravitational force of a single atom is obviously too much to ask (so far?) – the objects whose gravity is seen in the existing experiments contain billions or trillions of atoms. Note that the (attractive) gravitational force between two electrons is about $10^{45}$ times weaker than the (repulsive) electrostatic one!
Most of the research in quantum gravity has nothing whatever to do with proposals to modify Newton's laws at these distance scales. Indeed, gravity is the weakest force and it's so weak that for all routinely observable phenomena involving atoms, it can be safely neglected. The research in quantum gravity is dealing with much more extreme phenomena – like the evaporation of tiny black holes – that can't be seen in the lab.
Plots and links to new papers available over here (thanks, alemi)
Does it have something to do with the curvature of the Earth which is
assumed to be spherical
You'll probably groan when you read this answer since it isn't nearly as complicated as you might think.
Essentially, there is factor of $\pi$ since the angular frequency $\omega = 2\pi f = \frac{2\pi}{T}$
A well know result from the linearized pendulum problem is that, for small angular displacements, the angular frequency is
$$\omega = \sqrt{\frac{g}{L}}$$
which follows from the differential equation in the angular displacement $\theta$:
$$\ddot \theta + \frac{g}{L}\sin \theta = 0 \approx \ddot \theta + \frac{g}{L}\theta\;,\quad \theta \ll 1$$
Thus,
$$\left(\frac{2\pi}{T}\right)^2 = \frac{g}{L} \Rightarrow g = \frac{4 \pi^2 L}{T^2}$$
Best Answer
It’s because people are squashy.
Specifically in this case, it’s the squashy discs between the vertebrae which compress while a person is vertical, but can expand while a person is horizontal. The same expansion happens in zero gravity: astronauts return to Earth taller than they left. The nighttime difference is half-centimeter/quarter-inch for most people.
But at some level, all matter is compressible. The amount of vertical compression in Earth’s gravity is a question of the material’s strength and its rebound time constant.