Question

derive variation in small g above the earth​

Answer 1

Answer:

Explanation:

Consider a test mass (m) at a height (h) from the surface of the earth. Now, the force acting on the test mass due to gravity is;

F = GMm/(R+h)2

Where M is the mass of earth and R is the radius of the earth. The acceleration due to gravity at a certain height is ‘h’ then,

mgh= GMm/(R+h)2

⇒ gh= GM/[R2(1+ h/R)2 ] . . . . . . (2)

The acceleration due to gravity on the surface of the earth is given by;

g = GM/R2 . . . . . . . . . (3)

On dividing equation (3) and (2) we get,

gh = g (1+h/R)-2. . . . . . (4)

This is the acceleration due to gravity at a height above the surface of the earth. Observing the above formula carefully we can say that the value of g decreases with increase in height of an object and the value of g becomes zero at infinite distance from the earth.

Approximation Formula:

From Equation (4)

when h << R, the value of g at height ‘h’ is given by gh = g/(1 – 2h/R)