CE678A Physical Geodesy - Midsem Solutions
Q1. Gravimeter under a water tank (5 pts)
Tank volume = 100,000 L = 100 m³. Mass m = 100,000 kg.
Distance r = 20 m. Gravitational constant G = 6.6743×10■¹¹ m³kg■¹s■².
(1) Acceleration from tank:
(1.1) ∆g = G m / r² = (6.6743×10■¹¹ × 1.0×10■) / (20²)
(1.2) ∆g = 1.67×10■■ m/s² = 1.67 µGal.
Instrument precision = ±5 µGal, so effect is below detection.
Q2. Superposition principle (2 pts)
The gravitational field from multiple masses adds linearly:
(2.1) g = Σ g■, V = Σ V■.
Significance: allows separation of Earth's main field, tides, topography, etc. in modeling gravity.
Q3. Gradient of centrifugal acceleration (5 pts)
Centrifugal acceleration:
(3.1) a_cf = Ω × (Ω × r).
For Ω=(0,0,Ω), r=(x,y,z):
(3.2) a_cf = (-Ω²x, -Ω²y, 0).
Gradient tensor:
(3.3) ∇a_cf = diag(-Ω², -Ω², 0).
With Ω=7.2921×10■■ rad/s, Ω²=5.317×10■■ s■² ≈ -5.317 E.
Physical meaning: constant horizontal gradient, none vertically.
Q4. Why pure gravity cannot be measured (3 pts)
Observed acceleration includes gravity + inertial terms:
(4.1) g_app = g_Earth + Ω×(Ω×r) + a_tides + a_local.
By equivalence principle, instruments cannot distinguish gravity from inertial acceleration. Hence,
models & corrections are needed.
Q5. Error in pendulum timing (3 pts)
Period: T=2π√(L/g). Differentiate:
(5.1) dT/T = -½ (dg/g).
So |dT| = (|dg| / 2g) T.
For L=10 m, g≈9.8, T≈6.35 s, dg=0.05 m/s²:
(5.2) dT ≈ 0.016 s (16 ms).
Answer: timing must be precise within ±16 ms.
Q6. Drop vs Launch principle (10 pts)
Equations:
(6.1) Drop: g = 2s / t².
(6.2) Launch: g ∝ 1/t² (using round-trip time).
Sensitivity:
(6.3) ∆g/g = -2 ∆t/t.
Larger t reduces relative error.
Launch experiments allow longer flight times than short drops, hence smaller timing error impact.
Conclusion: Launch principle is preferable when timing precision limits accuracy.
