SME
Geophysics · Problem type
The anomaly over a buried sphere
Use this for “An ore body 100 m in radius is 500 kg/m³ denser than its host and centred 200 m down. Find its excess mass, peak anomaly and half-width.”
Diagram
Excess
mass
=
(
4
3
)πR³Δρ
=
(
4
3
)
×
π
×
100
3
×
500
=
2.094 ×
10
9
kg.
Δgₘₐₓ
=
G
×
mass
z
2
=
6.674 ×
10
-11
×
2.094 ×
10
9
200
2
×
10
5
=
0.3494
mGal.
It falls to half at x½ = 0.766z = 153.2 m either side: a deeper body gives a lower, wider anomaly.
This is an example. Type your own values to start a new problem.
R
Radius
example · m
Δρ
Density contrast
example · kg/m³
z
Depth to centre
example · m
M
Excess mass
calculated · kg
Δgₘₐₓ
Peak anomaly
calculated · mGal
x½
Half-width
calculated · m
Formulas
M
=
4
3
×
π
×
R
3
×
Δρ
2.0944 ×
10
9
=
4
3
×
π
×
100
3
×
500
Δgₘₐₓ
=
6.674 ×
10
-11
×
M
z
2
×
10
5
0.3494
=
6.674 ×
10
-11
×
2.0944 ×
10
9
200
2
×
10
5
x
1
2
=
0.766
×
z
153.2 = 0.766 × 200
Assumptions
•
The body is a sphere, so its excess mass pulls as if it were all at its centre.
•
G = 6.674 × 10⁻¹¹ N·m²/kg², and 1 mGal = 10⁻⁵ m/s².
•
The anomaly falls to half its peak 0.766z either side, so the half-width gives the depth.
•
A gravimeter reads to about 0.01 mGal, so a smaller peak is lost in the noise.
Step-by-step
WE KNOW
R = 100 m, Δρ = 500 kg/m³, z = 200 m
FIND
excess mass (M), peak anomaly (Δgₘₐₓ), half-width (x½)
Step 1 · Find excess mass (M)
Use M = 4 ÷ 3 × π × R³ × Δρ
The sphere’s volume times how much denser it is than the rock around it.
M
=
4
3
×
π
×
100
3
×
500
M
=
4
3
×
π
×
1,000,000
×
500
M
=
1.3333
×
π
×
1,000,000
×
500
M
=
4.1887
×
1,000,000
×
500
M
=
4,188,700
×
500
M
=
2.094395 ×
10
9
kg
Step 2 · Find peak anomaly (Δgₘₐₓ)
Use Δgₘₐₓ = 6.674 × 10⁻¹¹ × M ÷ z² × 10⁵
Newton’s pull of the excess mass on the station straight above it, in mGal.
Δgₘₐₓ
=
6.674 ×
10
-11
×
2.094395 ×
10
9
200
2
×
10
5
Δgₘₐₓ
=
(6.674
×
10
-11
)
×
2.094395 ×
10
9
40,000
×
100,000
Δgₘₐₓ
=
0.1398
40,000
×
100,000
Δgₘₐₓ
=
(3.495
×
10
-6
)
×
100,000
Δgₘₐₓ = 0.3494 mGal
Step 3 · Find half-width (x½)
Use x½ = 0.766 × z
The
anomaly
falls
to
half
its
peak
0.766
z
either
side
of
the
centre.
x
1
2
=
0.766
×
200
x
1
2
=
153.2
m
Check
Put the numbers back in. Both sides should match.
2.094395 ×
10
9
=
4
3
×
π
×
100
3
×
500
✓
0.3494
=
6.674 ×
10
-11
×
2.094395 ×
10
9
200
2
×
10
5
✓
153.2 = 0.766 × 200 ✓
Topic check
See what you know: 5 questions on this topic
New questions each time. Pass to mark it complete.
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