What are tidal forces of Earth gravity at Earth surface?

Question

8 tiny particles are arranged in a perfect cube 1x1x1 m³
formation originally at rest. 4 edes of the cube are
aligned vertically, and 8 edges are aligned horizontally.

All 8 particles are simutaneouslty released and
allowed to free fall for 1 second. By how much
the will distances between pairs of particles,
which share a same edge change?

Answer 1

Use a local coordinate system Oxyz, with Ox and Oy horizontal, and Oz vertical.

Then the local relative (tidal) gravitational accelerations experienced at a point with coordinates (x, y, z) w.r.t. O are as follows:

T (- x, - y, + 2z),

where T is the "tidal acceleration coefficient" G Me / Re^3. Here Me is the mass of the Earth, and Re the radial distance from the centre of the Earth, respectively.

Because of these relative accelerations, the distance BETWEEN the edge-sharing pairs of particles will change as follows in 1 second:

Horizontally, they will SHORTEN by 1/2 T*1m*(1s)^2;

Vertically, they will LENGTHEN by T*1m*(1s)^2.

(This SQEEZING by a factor of 1 and STRETCHING by a factor of 2 times some basic quantity is a characteristic property of local differential forces in an inverse square law force field.)

Working out T from the usual SI units, its magnitude is

T = 1.5405... x 10^(-6) SI units.

So,

Horizontally, lengths will SHORTEN by 7.7023... x 10^(-7) m;

Vertically, they will LENGTHEN by 1.5405... x 10^(-6) m.

Live long and prosper.

Answer 2

Let's assume the particles exert no forces on each other, and are in a vacuum chamber. You didn't specify the mass of the particles, so I cannot calculate a tidal force. However, I can calculate a tidal acceleration:

According to Wikipedia, the Earth's average radius is 6,372,797 m, and its mass is 5.9736*10^24 kg. The gravitational constant is 6.67428*10^-11 m^3/(kg*s^2). Above the surface of the Earth, the gravitational strength is

g = G*M/r^2
where G is the gravitational constant, M is the Earth's mass, and r is the distance from the center.

If we give the particles some room to fall without going underground, and place the bottom row 10 meters off the ground (average Earth radius), while the upper particles are 11 meters off the ground, we have respective radii of
upper = 6,372,797 + 11 = 6,372,808
lower = 6,372,797 + 10 = 6,372,807

Calculating the gravitational strength, we get
upper = 9.81700229 m/s^2
lower = 9.81700537 m/s^2

The difference between these is the tidal acceleration per meter:
t = 3.08*10^-6 m/s^2 per meter
t = 3.08*10^-6 s^-2

So, the tensile tidal force on the object would be approximated by
TF = m*h*t
where m is the object's mass, h is its height, and t is the tidal acceleration per meter.