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

CSIR NET – 2016 JUNE

89) A is a gravity station at 45⁰ N latitude and B is another station 20 km north at A. The measured gravity value at B is 16.2 m Gal larger than that at A, then the elevation at B is

A)Same as that of A 

B) 80m higher than that of A

C) 40m lower than that of A 

D) 80 m lower than that of A 

(Thanks to Chandrasekhar,ANU) 

Solution:

The average value of Gravity at a given latitude is 

 

$g_{t}=g_{e}(1+0.005278 sin^{2}\phi+0.00002346sin^{4}\phi )$

Where 

 

$g_{t}$ is the theoretical gravity for the latitude of the observation point (m gal)

$ g_{e}$ is the theoretical gravity of equator(978,031.85 m gals)

$\phi$ is the latitude of the observation point (degrees)

This equation takes into account the fact that the earth is an imperfect sphere, bulging out at equator and rotating about an axis through the poles (figure).

For such an oblate spheroid it estimates that the gravitational acceleration at the equator ($\phi$= 0) would be 978,031.85 m gals and gradually increasing with latitude to 983,217.72 m gals at the poles ($\phi$= 90⁰ ), so the latitude correction is 0.8140 sin 2$\phi$ m gals/km of N-S displacement.

Therefore the latitude correction at 45⁰ N is 

 

$\triangledown g_{lat} = 0.8140 \sin(2\times45) mgal/km$

                                              = 0.814 sin 90 mgals/km

$\triangledown g_{lat}= 0.814 mgal/km$ (since sin 90 = 1)

For 1 km = 0.8140 m gals 

 

20 km = ?

Therefore X = 20 * 0.814

X= 16.28 mgals

If station B is moving 20 km from A in north direction it’s latitude correction is 16.28 m gals.

Whether presented gravity anomaly is due to latitude variations not due to topographic.

Hence station B is presented at same height as that of A.

Reference: Whole Earth Geophysics by Robert J.Lillie