Question

At what height above the Earth's surface is the acceleration due to gravity 1% less than its value at the surface. Radius of Earth is 6400 km. (Take (1 + x) = 1 - 2r when x << 1)​

Answer 1

Decrease by 1% implies that g’ = 99/100g

Substitute In 1,we get

99/100g = g (1 - 2h/R)

99/100 = 1 - 2h/R

2h/R = 1 - 99/100

2h/R = 1/100

h = R/200

h = 6400/200 (R = 6400KM)

H = 32KM From surface of earth