Serway & Jewett — Universal Gravitation: Problem 35

(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 \[ |\Delta g| = \frac{2G M_E h}{R_E^3} \] (c) Evaluate this difference for \( h = 6.00 \text{ m} \), a typical height for a two-story building.