Chapter 13, Problem 13.62AP

(a) Show that the rate of change of the free-fall acceleration with vertical position near the Earth’s surface is

$\frac{dg}{dr}=\frac{2G{M}_E}{R_E^3}$

This rate of change with position is called a gradient.

(b) Assuming h is small in comparison to the radius of the Earth, show that the difference in free-fall acceleration between two points separated by vertical distance h is

$\left|\Delta g\right|=\frac{2G{M}_{E}h}{R_E^3}$

(c) Evaluate this difference for h = 6.00 m, a typical height for a two-story building.