Question

discuss variation of 'g' with the depth derive an expression for it what value of g at center of the earth​

Answer 1

Hey mate here is ur ans

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variation of g

the acceleration due to gravity at a place is inversely proportional to the square of the distance of the point from the centre of the earth. 

Hence acceleration due to gravity is maximum at the poles and minimum at the equator.

derivation

As per Newton's law, Force = mass × acceleration = m×a .....................(1)

 

where m is mass and a is acceleration

 

Gravitational force acting on a mass m on earth's surface = ..........................(2)

where G is Gravitational force constant, M is mass of earth and R is radius of earth

 

By comparing (1) and (2), acceleration due to gravity = a = (G×M)/R2

value of g at the center of earth is 0

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Answer 2

Answer:

9.8m/s^2

Explanation:

value of 'g' on the surface of earth. Due to gravitational pull of the earth , the force acting on an object is

F = G x Me x m / R^2

where , the value of Me = Mass of the earth

m = Mass of the object

R = Radius of earth

and if acceleration is g due to force F

then F = m x g

mg = G x Me x m / R^2

g = G x Me / R^2

If G = 6.67 x 10^11 Nm^2/Kg^2

Me = Mass of the earth = 6 x 10^24kg

R = Radius of the earth = 6.4 x 10^6

Then , g = 6.67x10^11 x 6 x 10^24 / 6.4 x 10^6 = 9.8m/s^2

So , the value of g ( Acceleration due to gravity) = 9.8m/s^2

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