Sometimes geophysicists must develop mathematical formulas for new models they d

Question

Sometimes geophysicists must develop mathematical formulas for new models they devise, to see if the new models fit their data better than existing models. Consider a model consisting of a long thin horizontal semi-infinite rod of mass per unit length µ. The rod is at a depth z below the ground surface. The rod starts at x = 0 and extends in the positive x direction to infinity. A gravity survey is done in the x direction along the rod, with the survey line being directly above the rod, and on the surface. (a) Derive a formula for the gravity effect of the rod (i.e., the z-component of the gravitational pull of the rod, gz) as a function of the observer position x. (b) Plot a graph of gz(x) vs. x. If desired, you can just sketch a graph using curve-sketching techniques learned in calculus, e.g., examine the function gz(x) for x ±, and |x| greater than z, etc., and sketch it based on those results. Examine the features of the curve to see if they make sense. Are the values of gz at x ± what you might expect? Explain. (c) Devise a method to determine values for z and µ using the maximum value of the gz curve and the “three-quarter” width x3/4. This is the value of x corresponding to the value of gz at three-quarters of the maximum of gz. Explain how you would use the method to compute z and µ. (d) Devise another method to determine values for z and µ using the maximum slope of the curve and the value of gz(x) at x = 0. The slope is dgz/dx. Explain how you would use the method to compute z and µ. (e) Practically, such a thin rod might not even be detectable because its gravitational pull would be too weak. For example, what would µ have to be if z = 20 m and the maximum value of gz is 0.1 g.u. (near the limits of measurability for typical instruments)? Does this seem like a reasonable value of µ?

Explanation / Answer

Gravity effect of a horizontal rod

The effect at P(x,y,0) of a segment of length dl of a horizontal rod with u as mass per unit length

dgr= gamma *u dl.r^2

=gamma*u(r1 dphi/cos^2phi=ydphi/r1

Component along r1

dg1=dgr cosphi

=gamma u cosphi dphi ri^2

Vertical component

dg=dgi*cosphi=dz1(z/r1)

=gamma*u*z*cosphi dphi/ ri^2

Integrating from tan-1(gamma-L)/r1 to tan-1(gamma+L)/r2

=(gamma*u*z) r1^2 ((gamma+L/ (gamma + L)^2+r1^2))^1/2- (( gamma- L)/( gamma-L)^2+r1^2)

=gamma u/ z( x^2+y^2+z^2)(1/(1+x^2+z^2)/(gamma+L)^2)^1/2 -1/(1+x^2+y^2/gamma-L)^2)^1/2

Since the rod is semi infinite in length limits of integration would be +- pi/2

L>10 z

the depth to the center of rod would be z

z=x1/2

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