Calculate the distance below and above the surface of the earth, at which the value of acceleration due to gravity becomes ¼ th of that at earth’s surface?

Formula Setup:

- g at height h from the Earth's surface = g' = g (R/(R+h))^2
- g at depth h below the Earth's surface = g' = g(1 - h/R)

To find: h where g' = g/4

At height h above the Earth's surface:

g' = g(R/(R+h))^2

g/4 = g (R/(R+h))^2

1/4 = (R/(R+h))^2

1/2 = R/(R+h)

R + h = 2R

h = 2R - R

**So, h = R**

At depth h above the Earth's surface:

g' = g (1 - h/R)

or, g/4 = g (1 - h/R)

or, 1/4 = 1 - h/R

or, h /R = 1- 1/4

or, h/R = 3/4

**So, h = 3R/4**

Hence, at height h = R above the Earth's surface and at a depth h = 3R/4 below the Earth's surface, the acceleration due to gravity (g') is equal to one-fourth of that at the surface of the Earth.