Introduction to gpsurya blog

$Dear$ $Friends,$ In this blog you will find around 320 solved Geophysics GATE and CSIR-NET and other competitive exams solutions with a better explanation. Please follow and share if you like it. $Thanks,$ $gpsurya$ and $group$

CSIR-NET 2014 (June)-116

CSIR-NET 2014

116.  Δg(x,y) are the gravity anomalies recorded on a horizontal plane underlain by a distribution of mass M which a density d surrounded by a material of constant density d0. G is the Newton's gravitational constant. If

A= Δg(x,y)ds , then M=
     s

 A) A/4πG

 B) A/4πG

 C) A/4πG *(d/d0)

 D) A/2πG *(d/d-d0)

Answer: D



Explanation:


In gravity total excess mass is


Mexcess = 1 / 2πG (volume of the anomaly)






The volume is the volume ‘under’ the anomaly, which needs to be

 considered in 3D. In practice, the anomaly is divided into little 

columns whose volumes are added (shown in above figure).


Each column should have an area small enough that there is little 

variations of g across it, so that it’s value equals it’s area times it’s 

height.


To calculate it’s total mass perhaps the mass of an ore the

 densities of the both body and it’s surroundings are needed. 

Assuming the body and its surrounding each have a uniform

 density.


Mbody = Mexcessbody / ρbody - ρsurroundings]


Mbody = A /2πG ( d / d – d0)


d = ρbody = density of the body


d0 = ρsurroundings = density of the surrounding rock


G = universal gravitational constant


(6.67408 * 10-11 m3 / kg s2 )


A = volume of the anomaly.





Related Posts

Post a Comment

Subscribe Our Newsletter