I will be thank full to u if u help me in solving the below problem

If the change in the value of g at a height h above the surface of the earth is same as depth x below the surface of the earth, then (h<<R)

If the two values of g' are equal, and we let g be the value of the acceleration of gravity at the Earth's surface,

g*R^2/(R+h)^2 = g*(R-x)/R

because g' is proportional to distance from the center of the earth, below the surface, if the earth has constant density

R^3 = (R-x)*(R+h)^2

1 = (1- x/R)(1 + h/R)^2
= 1 + (2h/R) +h^2/R^2 -(x/R)[1 + (2h/R) +h^2/R^2]
= 1 + 2h/R -x/R + higher order terms
2h/R = x/R + higher order terms (neglect them)
x = 2h

Solve for the values of U and V below?

To solve this problem, we need to use the formula for the change in the value of the acceleration due to gravity, denoted as Δg. Δg can be calculated using the formula:

Δg = -g (h/R)

Where:
Δg is the change in the value of g
g is the acceleration due to gravity at the surface of the Earth
h is the height above the surface of the Earth
R is the radius of the Earth

According to the problem statement, the change in the value of g at a height h above the surface of the Earth is the same as the depth x below the surface of the Earth. Mathematically, this can be expressed as:

Δg = -g (h/R) = -g (x/R)

Now, we can equate the two expressions for Δg:

-g (h/R) = -g (x/R)

We can cancel out the negative sign and the factor of g/R, giving us:

h = x

Since h and x are equal, we can conclude that if the change in the value of g at a height h above the surface of the Earth is the same as the depth x below the surface of the Earth, then h is approximately equal to x. Additionally, the problem statement mentions that h is much smaller than R (the radius of the Earth).