__Variation of Gravitational acceleration ‘g’__

**At poles or longitude:**

It is ‘constant’ and ‘maximum’ i.e (9.8 m/s.s)

**At equator or latitude:**
It ‘decreases’ with increasing distance from center of Earth.

**At some altitude (height):**
It decreases exponentially , and is inversely proportional to the square of height H above the surface of earth. The formula for g at the surface of earth is given by:

g = GM/R

^{2}
Here ‘g’ is the acceleration due to gravity, ‘G’ is the gravitational constant, ‘M’ is the mass of Earth and ‘R’ is the mean radius of Earth.

Let ‘g

_{o}’ be the value of acceleration due to gravity at radius ‘R’ from center of Earth. ‘g_{H}’ be the value of acceleration due to gravity at a height ‘H’ from Earth’s surface.
So,

g

_{o}= GM/R^{2}
And

g

_{H }= GM / [(R+H)]^{2}
Dividing the equations we get :

g

_{H }= g_{o}(R / (R+H))^{2}**In Depth:**

It decreases linearly, and is inversely proportional to the depth in Earth. Let ‘g

_{o}' be the value of gravitational acceleration on Earth’s surface from center of Earth and the radius be ‘R’ and 'g

_{x}' be the value of gravitational acceleration at a depth of ‘(R – x)’ , So, a formula can be derived which gives

g

_{x}= g_{o}(1- x/R )
When x = R i.e. center of Earth g

_{x}= 0. So at center of Earth acceleration due to gravity is zero.
Its value at sun is

**274 m/s**. At moon it is^{2}**1.6 m/s**^{2}**Some important concepts:-**

Just imagine the Earth starts rotating very fast. What will happen to the person standing on it? Obviously, he will fly off tangentially to the surface of the Earth. It means the gravitational force on that person has decreased. In this case the value of g will be ‘MINIMUM ‘.

The variation of gravitational acceleration with rotation of earth is given by

g

^{' }= g - R_{e}ω^2 cos^2θ
When ‘ω’ increases g’ becomes more and more less.

Now what happens if the Earth stops suddenly?

The value of g will become maximum or infinity in that case.

The value of g will become maximum or infinity in that case.

In this case ‘ω’ becomes zero so does the factor R

_{e}ω^2 cos^2θ so g^{' }= g, which is maximum value.
As we know gravitational force of attraction is the force that attracts a body towards the center of Earth. Now if a body is placed at the center of the Earth what will be the gravitational force of attraction in this case. It will be zero as the body is already placed at the center of the Earth. (already discussed in variation of ‘g’ with depth.)

Now, if a body starts moving from equators to poles value of g will increase continuously increases till it reaches its maximum value at poles. And if a body starts moving from poles to equators, value of g will decrease till it reaches its minimum value at equators.

This can also be explained on the basis of this formula

g

^{' }= g - R_{e}ω^2 cos^2θ
At poles the value of θ is 90 which make the factor R

_{e}ω^2 cos^2θ equal to zero so gravitation acceleration is maximum i.e g’ = g.
At equator the value of θ is 0 which gives the factor R

_{e}ω^2 cos^2θ its maximum value so g^{' }< g.
Now, as we move from equator to poles the value of θ increases and cosine value gets lowers making g' greater and vice verse.

This effect can also be simply under by Newton’s law of gravitation which gives

g= GM/R^2

At poles the radius is smaller so gravitational acceleration is large and at equator radius is large so value of acceleration due to gravity gets smaller.

__Post by : Tuba Bari__
thanks

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