Question

acceleration due to gravity at earth surface is g .find effective value of acceleration due to gravity at a height of 32 km from sea level

Answer 1

Assume-
     h= 32km = 32000m = 3.2 x10^4
     R= 64km = 64000m = 6.4 x 10^4
     g= 9.8m/s^2
We know that-
           g'= g[R^2/(R+h)^2]
              = 9.8(40.96 x. 10^8/92.16 x 10^8)
              = 9.8 x 4.44
              = 4.355m/s^2 ~ 4.4m/s^2

Answer 2

The value of Acceleration due to gravity changes when we go to a higher altitude or go into a depth or go to a different latitude.

The change in acceleration due to gravity with altitude is given by,

g' = g ( R / R+ h)²

When h <<<< R,

Then g' = g ( 1 - h/R)

Radius of earth = 6400 km.

Since 32 <<< 6400,

g' = g ( 1 - h/R)

g' = g ( 1 - 32/6400)

g' = g ( 1 - 1/200)

g' = g ( 1 - 0.005)

g' = g ( 0.995)

g' = 0.995 g.

Therefore, Effective value of acceleration due to gravity at a height of 32 km from sea level is 0.995 g.