Solution:
The acceleration due to gravity at depth \(h\) is \(g_d = gleft(1 - frac{h}{R}right)\. The change in gravity is \(Delta g = g - g_d = gfrac{h}{R}\). Thus, the fractional change \(frac{Delta g}{g}\) is \(frac{h}{R}\).
The acceleration due to gravity at depth \(h\) is \(g_d = gleft(1 - frac{h}{R}right)\. The change in gravity is \(Delta g = g - g_d = gfrac{h}{R}\). Thus, the fractional change \(frac{Delta g}{g}\) is \(frac{h}{R}\).
Leave a Reply