GATE-2018 (24) :Wyllie – Time average equation

GATE-2007

62. The difference in gravity measurement aboard two ships sailing towards each other in opposite directions (E-W) with a constant speed of 10 knots is 130 mgal at the crossing point of both the ships. At what latitude are the ships sailings? (thanks manoj AU)

A) 15°                B)30°                   C)45°                        D)60°

Solution:

Using Eotvos correction for moving objects in gravity

ΔE= ((7.505 *V*cosφ*sinα )+ (0.0041* V^{^2} )) mgal ----------------------(1)

Velocity of the ship (V) in knots

The difference in gravity measurements (ΔE)=ΔE₁-ΔE₂

ΔE=130 mgal

ΔE = 7.505*V * cosφ (sinα₁ –sinα₂) -------------------------------(2)

The deviation of the ship, is the angle from the north side

(Because we want to find the latitude, so it will become North side or South side)

One ship moving from North –East, the angle is 90˚

Another ship moving from North- West, the angle is = 270˚

(if you take angle form south side also means South-East and South-West, will give the same result)

Substitute values in eq(2)

130 = 7.505 * 10* cosφ (sin90˚-sin270˚)

sin90˚=1

sin270˚=-1

130 = 7.505* 10 * cosφ (1+1)

130 = 7.505* 10*2*cosφ

cosφ=130/150

φ = cos^-1(0.866)

φ= 30˚

Therefore ships sailing at latitude (φ) =30˚