Solution:
The acceleration due to gravity at depth \(h\) is \(g_d = g\left(1 - \frac{h}{R}\right)\). The change in gravity is \(\Delta g = g - g_d = g\frac{h}{R}\). Thus, the fractional change \(\frac{\Delta g}{g}\) is \(\frac{h}{R}\).
The fractional change in acceleration due to gravity on surface of earth (g) when the body is taken to a depth h from the surface of earth, is (where R = radius of Earth)
Leave a Reply