At what height above the earths surface is the acceleration due to gravity 4% less than its value on surface of earth RE is radius of Earth?

Text Solution

Solution : Acceleration due to gravity at a height, h=4% of g. <br> Radius of earth, R=6400 K.M=`6.4xx10^(6)` m. <br> `:.g_h=(4)/(100)g` But `g_h=(g*R^2)/(1+(h)/(R ))^2` <br> `:.(4)/(100)g=(g)/(1+h/R^2) implies (1+(h)/(R ))^2=(100)/(4)` <br> Take roots on both sides then, `1+(h)/(R )=(10)/(2)` <br> `=5 implies 1+(h)/(R )=5 implies (h)/(R )=5-1=4` <br> `:. h=4R=6400xx4=25,600` k.m.