Question

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why the value of "g" is more at pole than that of the equator ¿?
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Answer 1

$\red{HEY\:MISS!!}$

HERE'S THE ANSWER

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♠️ Firstly , we know

✔️ $\boxed{ g\:= \:( G M ) \:/ \:r^2}$

▶️ Where

⏺️g is Acceleration due to gravity

⏺️G is universal gravitational constant

⏺️M is mass of earth

⏺️ r is Radius of Earth

♠️ $\underline{Earth\: is \: elliptical\: in \:shape}$ , So distance between center and pole is more than distance between centre and equator .

⏺️ From above relation .

✔️ $\boxed{ g\:\alpha \:1 \: /\:r^2}$

▶️ So g is inversely proportional to r

♠️ That's why value of Acceleration due to gravity is more at pole

HOPE HELPED...

$\red{JAI HIND..}$

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Answer 2

Thanks For The Question.

May-be You earlier know that [g] at at the pole is about 9.83N/Kg, However On the Equator It is about 9.78N/Kg.

There are Innumerable reasons for it:

You are aware of that Rotation Of Earth Desire a small Centripetal force and this scale down the observed strength of gravity by about 0.03 N/Kg.

You are cognizant about that Our Equatorial Diameter is terrific than the polar diameter thus equator from the centre of the Earth than the pole and gravitational attraction slightly less by about 0.02N/Kg.

At the Mid-point of Earth there would be no gravitational force due to Earth So, Of course as the mass of Earth Assigned Fairly and Symmetrically all around it.

So, Acceleration Due to gravity [g] = GM/R^2

We can See that Value Of 'g' is inversely proportional to the Radius R and the value of R at the Equator is greater than at the pole(already Mentioned).
Hence, 'g' at the pole is greater than 'g' at the Equator.


Regards
Vickysky
@Brainly User (INDIA)