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

74. A gravity reading is taken in a stationary helicopter hovering 1 km above mean-sea level at a particular location. The difference in the value of g measured in the helicopter and at mean sea level vertically beneath the helicopter is ____________ mGals.

Solution:

Using free-air correction,

The gravity value reduced if distance from the Earth’s center increases.

Free-air correction = 0.3086 mGal/m

Distance =1000 m

1 m = 0.3086 mGal

1000 m = ?

The difference in the value of g measured in the helicopter and at mean sea level vertically beneath the helicopter is 

   = 0.3086* 1000 mGal

    = 308.6 mGal