Page 351 - Physical Principles of Sedimentary Basin Analysis

10.9 Gravity from a prism with rectangular cross-section 333

x x 0 x 1 x 2
x x
z 1
z dz
dx
z 0 z 2

z z
(a) (b)
Figure 10.11. (a) The rectangular prism extending from the origin to (x 0 , z 0 ) is made of a (large)
number of (small) line loads with mass per length dx dz. (b) The gravity from the rectangular
cross-section (x 1 , z 1 ) to (x 2 , z 2 ) is found by superposition.

The gravity from any rectangular cross-section is obtained by superposition of cross-
sections as the one found (with the origin as one corner). The gravitational accelera-
tion (10.69) is now written
g z = f (x 0 , z 0 ) (10.70)

and the gravity from a rectangular cross-section from (x 1 , z 1 ) to (x 2 , z 2 ) is then

g z = f (x 2 , z 2 ) − f (x 1 , z 2 ) − f (x 2 , z 1 ) + f (x 1 , z 1 ), (10.71)

see Figure 10.11b. The gravity from several rectangular cross-sections are the superposition
of the gravity from each rectangular cross-section.
The gravity from a rectangular cross-section that extends laterally to infinity is g z =
2
G πz 0 (using that ln(1 + (z 0 /x 0 ) ) → 0 and that tan −1 (x 0 /z 0 ) → π/2 when x 0 →∞ in
equation (10.69)). Bouguer’s formula is then obtained by adding a factor 2, which accounts
for the cross-section along the negative x-axis.
Note 10.3 The integration over z in equation (10.68)gives

du 2 2 2
x 0 z 0 x 0

g z = 2G dx = G ln(x + z 0 ) − ln(x ) dx (10.72)
0 0 2u 0
2
2
using the substitution u = x + z , and the integration over x is done using the integrals
x

2
2
2
2
ln(x + a ) dx = x ln(x + a ) + 2a tan −1 − 2x (10.73)
a

2 2
ln(x ) dx = x ln(x ) − 2x. (10.74)
2
The term x ln(x ) = 2x lnx is evaluated at x = 0 using l’Hopital’s rule, where lnx/(1/x)
is an ∞/∞ expression that approaches 0 when x → 0.