SME
Geophysics
Gravity and magnetics
Use this for “A station 250 m above sea level reads 40 mGal below normal gravity. With 2.67 g/cm³ rock, find the free-air and Bouguer anomalies.”
Chart
Free-air anomaly: ΔgFA = 37.15 mGal
This is an example. Type your own values to start a new problem.
h
Station elevation
example · m
ρ
Slab density
example · g/cm³
Δg
Observed minus normal
example · mGal
FA
Free-air correction
calculated · mGal
B
Bouguer correction
calculated · mGal
ΔgFA
Free-air anomaly
calculated · mGal
ΔgB
Bouguer anomaly
calculated · mGal
Formulas
FA
=
0.3086
×
h
77.15 = 0.3086 × 250
B
=
0.04193
×
ρ
×
h
27.9883 = 0.04193 × 2.67 × 250
ΔgFA = Δg + FA
37.15 = −40 + 77.15
ΔgB
=
ΔgFA
−
B
9.1617 = 37.15 − 27.9883
Assumptions
•
Gravity falls 0.3086 mGal for each meter up, the free-air gradient; 1 mGal = 10⁻⁵ m/s².
•
The rock between the station and sea level is a flat slab: it pulls 2πGρh = 0.04193ρh mGal, with ρ in g/cm³ and G = 6.674 × 10⁻¹¹ N·m²/kg².
•
Δg is the observed gravity minus normal gravity at the station’s latitude; terrain corrections are left out.
Step-by-step
WE KNOW
h = 250 m, ρ = 2.67 g/cm³, Δg = −40 mGal
FIND
free-air correction (FA), Bouguer correction (B), free-air anomaly (ΔgFA), Bouguer anomaly (ΔgB)
Step 1 · Find free-air correction (FA)
Use FA = 0.3086 × h
The station sits higher, farther from Earth’s centre: add back 0.3086 mGal per meter.
FA = 0.3086 × 250
FA = 77.15 mGal
Step 2 · Find Bouguer correction (B)
Use B = 0.04193 × ρ × h
The slab of rock under the station adds its pull: take it away.
B
=
0.04193
×
2.67
×
250
B
=
0.11195
×
250
B
=
27.9883
mGal
Step 3 · Find free-air anomaly (ΔgFA)
Use ΔgFA = Δg + FA
The free-air anomaly is the reading corrected to sea level.
ΔgFA = −40 + 77.15
ΔgFA = 37.15 mGal
Step 4 · Find Bouguer anomaly (ΔgB)
Use ΔgB = ΔgFA − B
What is left after the slab is the pull of denser or lighter rock below.
ΔgB = 37.15 − 27.9883
ΔgB = 9.1617 mGal
Check
Put the numbers back in. Both sides should match.
77.15 = 0.3086 × 250 ✓
27.9883 = 0.04193 × 2.67 × 250 ✓
37.15 = −40 + 77.15 ✓
9.1617 = 37.15 − 27.9883 ✓
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The anomaly over a buried sphere
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