Question

derive an experassion for the value of 'g' with depth

Answer 1

hey mate here is your answer

let the object be at depth d metre

so,,,,

g'=g(1-(d/R))

hope it will help you

mark me brainliest ✌✌✌✌

Answer 2

When we go deeper into earth, the attraction at that point is due to a sphere of earth of radius (R – d) where d is the depth below the surface of earth.

 
Let a body of mass m be kept on the surface of earth. The force exerted by earth on it,
 where M is the total mass of earth and R is the radius of earth.
But, 
Substituting for M,

or 
 ————– (1)
When the object is at a depth d below the surface of earth, it is attracted by the sphere of radius (R-d) only as the attraction due to the rest of earth cancel out.
(At the centre of earth, the object would be attracted equally in all directions and the net force experienced by the object and hence the value of g would be zero)
The force on the body, 

Substituting for M’ and simplifying as above, we get:
 ——————(2)

OR

This equation shows that the value of acceleration due to gravity decreases as we go deeper.
(Here we have made an assumption that the density of earth is uniform. But in actual case it is not uniform. However, this will not contradict the fact that g decreases with depth.)
The point raised in your question is true if we are on the surface of earth and there is a variation in the radius of earth. this is the reason why the value of g is maximum at poles and minimum at equator.