Discussion
At a point on the Earth at height 1000 m, the observed value of gravity is 979700 mGal. At sea level it is 980000 mGal. Calculate the free-air anomaly.
Answer: D
The free-air anomaly is 86 gu (8.6 mGal).
The free-air correction adds back the decrease of gravity with height, 0.3086 mGal per metre:
\[ \begin{aligned} \Delta g_{FA} &= 0.3086 \times 1000 \\ &= 308.6\ \text{mGal} \end{aligned} \]
\[ \begin{aligned} g_{FA} &= g_{obs} + \Delta g_{FA} - g_0 \\ &= 979700 + 308.6 - 980000 \\ &= 8.6\ \text{mGal} \end{aligned} \]
One mGal is 10 gravity units (gu, micrometres per second squared), so the anomaly is 86 gu. Forgetting the correction gives a large negative anomaly of -300 mGal, and none of the other printed values comes from the standard correction.
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