find the value of g at poles and at equator.

Asked by thakursonali2000 | 7th Oct, 2015, 05:11: PM

Expert Answer:

The expression for variation of g with latitude is given as,
g'=g-Reω2cos2λ     ... (1)
where,
g' is the acceleration due to gravity in presence of Earth's rotation
λ is latitude
g is the acceleration due to gravity
Re is the radius of the Earth
ω is the angular velocity
 
At the equator:
λ=0, cos λ=cos 0=1
Thus from eqn (1), we get,
g'=g-Reω2
Substituting Re=6.37×106 m and begin mathsize 14px style straight omega equals fraction numerator 2 straight pi over denominator straight T end fraction equals fraction numerator 2 cross times 3.14 over denominator 24 cross times 60 cross times 60 end fraction equals 7.27 cross times 10 to the power of negative 5 end exponent space straight s to the power of negative 1 end exponent end style, we get,
g-g' = Reω= (6.37×106 m) (7.27×10-5 s-1)= 0.034 m s-2.
We see that g', the acceleration due to gravity at the equator of the rotating Earth, is less than g, the expected value if the Earth were not rotating by only 0.034/9.8 or 0.35%. 
The effect diminishes as we go to higher latitudes at the poles.
 
At the poles:
λ=90°, cos λ=cos 90°=0
Thus, from eqn(1),
g'=g
 
Hence, we conclude that as we go from the equator towards the poles, the value of g goes on increasing.

Answered by Faiza Lambe | 7th Oct, 2015, 05:50: PM