The earth is an infinite slab with a finite gravitational pull.
If we actually consider an infinite homogeneous slab with density
ρ and total thickness
H, then, due to symmetry, the graviational field must be directed along the normal of the slab and towards it. Similarly, we deduce that the intensity of the graviational field must be equal at all points lying in a plane parallel to the sides of the slab. This allows us to use Gauss' theorem to calculate the intensity of the graviational field
g at any plane. The plane passing through the middle of the slab and being parallel to the sides is a plane of symmetry. We choose it as the yz plane of a coordinate system, as depicted in the following figure:
Then, the vector graviational field
g has a direction along
x when
x>0 and along -
x when
x<0. Furthermore, the intensity of this field
g depends only on
x and
g(-
x) =
g(
x), which is in accordance to the above mentioned symmetry.
If we consider a prism with a basis parallel to the sides of the slab and height 2
x, then, the flux of the gravitaitonal field through this Gaussian surface is:
Φ = -2*g*ΔA,
because the flux from the sides is zero, since the field is parallel to them, and, along the bases, the field intensity is constant and the field is perpendicular to them.
The total mass contained within this prism is:
M = ρ*ΔA*2|x|, |x| <= H/2,
M = ρ*ΔA*H, |x| > H/2.
According to Gauss' Law for the graviational field:
Φ = -4*π*G*M,
where
G is the Universal Gravitational Constant, we can get:
g(x) = 4*π*G*ρ*|x|, |x|<= H/2
g(x) = 2*π*G*ρ*H, |x| > H/2
We see that the field intensity is zero in the middle of the slab and it rises linearly with height as we approach the surface, where it reaches the maximum value. After that, the field is constant. The maximum value is:
g0 = 2*π*G*ρ*H.
The graviational potential in this field
V(
x) is given by:
-g(x) = -dV/dx, x > 0,
g(x) = -dV/dx, x < 0.
We see that the derivative of the potential is an odd function of
x (because
g(
x) was an even one and the extra minus sign coming from the change in direction). Therefore, the potential is an even function of
x and it is sufficient to consider only the region
x > 0. Furthermore, we choose a reference level
V(0) = 0. Doing the integration and making sure the potential is continuous at
x =
H/2, we get:
V(x) = 2*π*G*ρ*x2, 0 < x < H/2,
V(x) = 2*π*G*ρ*H*x, H/2 < x.
Next, we want to find the total potential energy of the slab in its own gravitational field. But, since the slab is infinite, this is an infinite quantity. The potential energy per unit area, however, is a well defined quantity, because, the mass per unit area is finite. The mass in an infintesimal slab between (
x,
x +
dx) is:
dm/A = ρ*dx
The potential energy of this element is simply
V(
x)*
dm/
A. Integrating with respect to
x from -
H/2 to
H/2. We get:
U/A = π*G*ρ2*H3/6
Other things equal, the potential energy is positive and proportional to the cube of the thickness of the slab. We consider the following possibility: What if some material in the center of the slab liquifies and drips off the edges (the assumption of infinite slab is not a contradiction, since infinite means we can neglect edge effects, but a boundary, albeit moved to infinity is still present). Energy is required for this process, since it is a phase transition. But, by the liquid dripping off, the thickness decreases and the potential energy decreases. If the decrease in the potential energy is greater than the heat of melting, then this process is quite possible. If the thickness decreases by an infinitesimal amount
dH, then the mass (per unit area) that is melted is:
dM/A = ρ*dH,
so the heat of melting (per unit area) is:
dQ/A = L*ρ*dH,
where
L is the latent heat of melting (energy per unit mass). The decrease in potential energy is the differential of U/A with respect to H:
|dU|/A = π*G*ρ2*H2*dH/2.
The process we are describing is impossible if
dQ/
A >= |
dU|/
A, since there would not be enough energy for the phase transition to occur. This gives an upper bound of the slab's thickness:
H <= [2*L/(π*G*ρ)]1/2.
Other things equal, the gravitational field at the surface of the slab is proportional to its thickness. This means, there is a maximal value for the value of the gravitational field, since there is a maximal thickness. This is:
g0 <= 2*[2*π*G*ρ*L]1/2.
Why did we do this? Well, we can estimate the order of magnitude of
ρ and
L for known materials, and, knowning the exact value of
g0 = 9.8 m/s
2, we can get an estimate of the order of magnitude for the Universal gravitational constant. The density of the Earth's crust is not less than the density of water so
ρ ~ 10
3 kg/m
3. The latent heat of melting for most materials is of the order of magnitude 10
4 - 10
5 J/kg. This gives an order of magnitude estimate:
g02 ≲ 8*π*G*ρ*L,
G ≳ g02/(8*π*ρ*L),
G ≳ 10-8 - 10-7 m3/(kg*s2).
This is several orders of magnitude greater than the experimental value of the Universal Gravitational constant G = 6.67 x 10
-11 m
3/(kg*s
2)! This means that, assuming Newtonian Gravitation Law is a correct approximation to a high degree of accuracy, it is highly unlikely that the shape of the Earth is an infinite slab with finite thickness, strictly by considering issues of stability of such a system under perturbations of the above mentioned kind.