Solution:
The change in gravity over height \(h\) is \(dg = \frac{2g}{R} h\). Using \(g = \frac{4}{3} \pi G \rho R\), we get \(dg = \frac{8}{3} \pi G \rho h\). Thus, the difference in weight (error) is \(m \cdot dg = \frac{8}{3} \pi G \rho m h\).
Two equal masses \(m\) and \(m\) are hung from a balance whose scale pans differ in vertical height by \(h\). The error in weighing in terms of density of the earth \(\rho\) is :
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