The decrease in the value of g at height h from earth's surface is

The correct answer is 2hg/R.

To find the decrease in the value of the acceleration due to gravity (g) at a height h above the Earth's surface, we can derive the expression step-by-step.

At the surface of the Earth, the acceleration due to gravity is given by:
$g=\frac{GM}{R^2}$
where G is the universal gravitational constant, M is the mass of the Earth, and R is the radius of the Earth.

At a height h above the Earth's surface, the distance from the center of the Earth is R + h. Thus, the acceleration due to gravity becomes:
$g^{\prime }=\frac{GM}{{(R+h)}^2}$

We can rewrite the denominator by factoring out R:
$g^{\prime }=\frac{GM}{R^2{(1+\frac{h}{R})}^2}$

Substituting the expression for g at the surface, we get:
$g^{\prime }=g{(1+\frac{h}{R})}^{-2}$

If we assume the height h is much smaller than the radius of the Earth (h << R), we can apply the binomial approximation:
${(1+x)}^n\approx 1+nx$
where x = h/R and n = -2.

Applying this approximation:
$g^{\prime }\approx g(1-\frac{2h}{R})$

Multiplying g through the terms:
$g^{\prime }\approx g-\frac{2hg}{R}$

The decrease in the value of gravity, Δg, is the difference between the acceleration due to gravity at the surface and that at height h:
$\Delta g=g-g^{\prime }$

Substituting our approximation for g':
$\Delta g=g-(g-\frac{2hg}{R})$

Simplifying the expression gives the decrease in the value of g:
$\Delta g=\frac{2hg}{R}$