Being stuck together in an elevator brings particles closer together

Assuming two particles of masses m_1 and m_2 in the gravitational field of the earth.The earth has a mass of 5,972 \times 10^{24} kg according to Wikipedia. What is the tidal force on the two particles as a function of their distance to the center of the earth?

For point masses, the Newtonian gravity states that they attract each other with a force \vec{F_G}=\frac{GMm}{r^2}\hat{r}. So if we assume two particles with a distance d between them where each particle has a distance R from the center of mass (COM) of the earth, then we can calculate the force on the particles in the direction of each other. we can do so by neglecting their mutual attraction through gravity or we can account for it. This depends on their masses and on the distances involved.

If we do the math (simple euclidean geometry in the plane) we get a very acute angle between the direct lines connecting the particles with the earth and the line which is perpendicular to the line connecting the particles. The angle is \phi = \arcsin(0.5 d/R). Now we can calculate the force on a particle along the line connecting it with the COM of the earth, as given above, and then divide this force into a component in the direction of the direct line connecting the particles and its perpendicular. We get \vec{F}_{attract} = |F_G|\sin(\phi) \cdot \hat{d}.

When one calculates this for two people (70 kg) standing 1m apart on the surface of the earth (remember: all point masses), it turns out that the tidal force (5.39456507 \cdot 10^{-5} kg\cdot m / s^2) is seven orders of magnitude smaller than the the force attracting the people towards earth (-6.87375482\cdot 10^{2} kg\cdot m / s^2). The mutual gravitation (-3.27029920 \cdot 10^{-7} kg\cdot m / s^2) on the other hand is still two orders of magnitude smaller than the tidal force.

Lifting the people into space only makes the distance between the people more insignificant compared to the distance to earth and thus the relative tidal force gets smaller and smaller, making the relative impact of the mutual gravitation in the free-falling elevator more important. Once we have to humans on a spaceship (say 100km above ground) the tidal force is completely irrelevant vs. the mutual gravitation.

Here’s my Python code for that (and btw. astropy is cool):

import matplotlib.pyplot as plt
import numpy as np
import astropy.units as u
from astropy import constants as c

M = 5.972e24 * u.kg
m = 70.0 * u.kg
earth = {'mass': M, 'loc': np.array([0.0, 0.0]) * u.m}

def gravity(one, two):
m1 = one['mass']
p1 = one['loc']
m2 = two['mass']
p2 = two['loc']

distance_squared = (p1 - p2).dot(p1 - p2)
return c.G*m1*m2/distance_squared*(p1-p2)/np.sqrt(distance_squared)

# These are the numbers
A = [6371000.0*0.1, 6371000.0, 6371000.0*10.0, 6371000.0*100.0] # earth radius
d = [1.0] # distance between particles

for i in range(len(A)):
for j in range(len(d)):
# The Center of Mass (COM) of the earth is supposed to sit at the origin
# Since our problem has rotational symmetry, along the perpendicular towards the COM
# we consider it 2-dimensional
p = [{'mass': m, 'loc': np.array([-d[j]/2.0, A[i]])*u.m},
{'mass': m, 'loc':np.array([d[j]/2.0, A[i]])*u.m}]

print("Gravitational force with earth: {0}".format(gravity(earth, p[0])))
print("Gravitational force with eachother: {0}".format(gravity(p[0], p[1])))

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