Question

At the surface of the earth the acceleration due to gravity is 9.8 m/s². How far above the surface of the earth would the acceleration due to gravity be 10% smaller? In other words, at what height is the acceleration due to gravity 8.82 m/s²?

Answer 1

Answer 2

SOLUTION :

Let that height be h  m from the earth’s surface at which altitude, g’ = 8.82 m/s^2

g at the earth’s surface is = 9.8 m/s^2.

As we go up, value of acceleration due to gravity decreases.

Value of g’ at height h is given by the following formula:

g’ = g (1 - 2h / R) 

Where,

g’ = acceleration due to gravity at height h = 8.82  m/s^2 (given in the problem)

g = acceleration due to gravity at earth’s surface = 9.8 m/s^2

h = height at which g’ = 8.82 m/s^2. (To be determined)

R = radius of the earth = 6371 km = 6.371*10^6 m

So, as per above formula :

8.82 = 9.8( 1 - 2 h / (6.371 * 10^6) )

=> h = (1 - 8.82/9.8) *  (6.371 * 10^6) / 2 

=> h = 318 550 m = 318.55 km

=> h = height at which g’ is 8.82 m/s^2 = 318.55 km   (ANSWER).