Question

The speed of a ship is given as \(V_1\) (km/h) and \(V_2\) (knots). The latitude of observation and the direction of the ship with respect to the North are \(\theta_1\) and \(\theta_2\), respectively. The CORRECT expression(s) for the Eötvös correction in mGal is/are:

• A. \(4.040\, V_1 \cos \theta_1 \sin \theta_2 + 0.001211\, V_1^{2}\)
• B. \(7.503\, V_2 \cos \theta_1 \sin \theta_2 + 0.004154\, V_2^{2}\)
• C. \(4.040\, V_2 \cos \theta_2 \sin \theta_1 + 0.001211\, V_2^{2}\)
• D. \(7.503\, V_1 \cos \theta_1 \sin \theta_2 + 0.004154\, V_1^{2}\)

Hint:

Remember the ready-to-use constants for the Eötvös correction (mGal):
\(\;\bullet\;V\) in km/h: \(4.040\,V\cos\phi\sin\alpha + 0.001211\,V^{2}\).
\(\;\bullet\;V\) in knots: \(7.503\,V\cos\phi\sin\alpha + 0.004154\,V^{2}\).
Here, \(\phi\) is latitude and \(\alpha\) is heading from North (eastward component via \(\sin\alpha\)).

Answer 1

The Correct Option is A, B

Step 1: Formula and geometry.
The Eötvös correction accounts for the effect of ship motion on measured gravity: \[ \Delta g_E = 2\,\omega\,V \cos\phi\,\sin\alpha \;+\; \frac{V^{2}}{R}\quad \text{(in m/s$^{2}$)}, \] where \(\omega\) is Earth's angular speed, \(R\) is Earth's mean radius, \(\phi\) is latitude, and \(\alpha\) is the heading measured clockwise from North.
Here, \(\phi=\theta_1\) and \(\alpha=\theta_2\). The term with \(\sin\alpha\) gives the eastward component of velocity; the latitude factor is \(\cos\phi\). Step 2: Convert to mGal and express in given speed units.
Use \(\omega = 7.292115\times10^{-5}\,\text{s}^{-1}\), \(R \approx 6.371\times10^{6}\,\text{m}\), and \(1\,\text{mGal}=10^{-5}\,\text{m/s}^2\). \underline{If \(V\) is in km/h}: \(1\,\text{km/h}=0.27778\,\text{m/s}\).
Coefficient of the linear term: \[ \frac{2\omega(0.27778)\times 10^{5}}{1} \approx 4.040. \] Coefficient of the quadratic term: \[ \frac{(0.27778)^{2}}{R}\times 10^{5} \approx 0.001211. \] Hence, \[ \Delta g_E(\text{mGal}) = 4.040\,V_1\cos\theta_1\sin\theta_2 + 0.001211\,V_1^{2}. \] This matches option (A). \underline{If \(V\) is in knots}: \(1\,\text{knot}=0.51444\,\text{m/s}\).
Linear coefficient: \[ 2\omega(0.51444)\times10^{5}\approx 7.503. \] Quadratic coefficient: \[ \frac{(0.51444)^{2}}{R}\times10^{5}\approx 0.004154. \] Hence, \[ \Delta g_E(\text{mGal}) = 7.503\,V_2\cos\theta_1\sin\theta_2 + 0.004154\,V_2^{2}, \] which is option (B). Step 3: Eliminate the incorrect options.
(C) mixes units (uses \(V_2\) with km/h coefficients) and swaps \(\cos\theta_1\) and \(\sin\theta_2\).
(D) mixes units (uses \(V_1\) with knot coefficients).
Therefore, only (A) and (B) are correct. \[ \boxed{\text{Correct options: (A) and (B)}} \]