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

Two survey vessels with ship borne gravimeters are cruising towards each other at a speed of 6 knots each along east – west course. The difference in gravity readings of the two gravimeters is 63.5 mgals at the point of which the survey vessels cross each other. The latitude along at which the survey vessels are cruising is --------⁰N.

Solution:

 

Eotvos correction for moving objects in gravity is 

 

$\triangle E = 7.505 V \cos\phi\sin\alpha+ 0.0041V^{2}$

Where V is Velocity of ship in Knots

$\phi$ is the lattitutde

$ \alpha$ is the heading of the ship from north.

the Eotvos correction for the two ships

$\triangle E_{1} = 7.505 V \cos\phi\sin \alpha_{1}+ 0.0041V^{2}$

$\triangle E_{2} = 7.505 V \cos\phi\sin \alpha_{2}+ 0.0041V^{2}$

The difference between the two Eotvos corrections is 

 

$\triangle E =\triangle E_{1} -\triangle E_{2} $ =63.5 mgals(given)

$\triangle E=(7.505 V \cos\phi\sin \alpha_{1}+ 0.0041V^{2}) – (7.505 V \cos\phi\sin \alpha_{2}+ 0.0041V^{2})$

$\triangle E =7.505 V \cos\phi(\sin \alpha_{1}-\sin \alpha_{2})$---(1)

$\alpha$ is the deviation of the ship , the angle from the north side

One ship is moving from N- E at angle 90⁰

Another ship is moving N-W at angle 270⁰

By substituting these values in equation (1) 

 

 $\triangle E =7.505 V \cos\phi(\sin \alpha_{1}-\sin \alpha_{2})$

$63.5 = 7.505*6*\cos\phi(\sin 90-\sin 270)$

$63.5 = 45.03 *\cos\phi(1+1)$

$63.5 = 90.06\cos\phi$

$\cos\phi=63.5/90.06$

$\cos\phi = 0.705$

 $\phi = 45^0$