Discussion
At a point on the Earth at height 1000 m, the observed value of gravity is 979700 mGal. The value at sea level is 980000 mGal. Calculate the Bouguer anomalies :
Answer: A
The Bouguer anomaly is about -1046 gu (gravity units; 1 mGal = 10 gu).
Apply the free-air correction (0.3086 mGal/m) and the Bouguer correction \(0.0419\,\rho\) mGal/m, the stem leaving the density to the standard value. With \(\rho = 2.7\) g/cm\(^{3}\), the Bouguer factor is 0.1132 mGal/m:
\[ \begin{aligned} FA &= 0.3086 \times 1000 = 308.6 \\ BC &= 0.1132 \times 1000 = 113.2 \\ \Delta g_B &= 979700 + 308.6 - 113.2 - 980000 \\ &= -104.6\ \text{mGal} = -1046\ \text{gu} \end{aligned} \]
The density 2.67 g/cm\(^{3}\) gives -1033 gu, still nearest to -1046. Forgetting the free-air correction gives about -4130 gu, and the remaining options do not follow from any standard reduction.
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