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:
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.