# 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-R_{e}ω^{2}cos^{2}λ ... (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
R_{e} 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-R_{e}ω^{2}
Substituting R_{e}=6.37×10^{6} m and , we get,
g-g' = R_{e}ω^{2 }= (6.37×10^{6} m) (7.27×10^{-5} s^{-1})^{2 }= 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.

_{e}ω

^{2}cos

^{2}λ ... (1)

_{e}is the radius of the Earth

**At the equator**:

_{e}ω

^{2}

_{e}=6.37×10

^{6}m and , we get,

_{e}ω

^{2 }= (6.37×10

^{6}m) (7.27×10

^{-5}s

^{-1})

^{2 }= 0.034 m s

^{-2}.

**At the poles**:

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

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