Question

If the acceleration due to gravity at a height ‘h’ from the surface of the earth is 96% less than its value on the surface, then h is (where R is the radius of the earth).

Answer 1

the acceleration due to gravity at a height 'h' from the surface of earth is 96% less than its value on the surface .
Let g is the acceleration due to gravity on the earth's surface and g' is the acceleration due to gravity at a height h from the surface of earth.
then, g' = g - 96 % of g
g' = 0.04g....... (1)

we know, acceleration due to gravity at a height h from the surface of earth is given by
$g'=\frac{g}{\left(1+\frac{h}{R}\right)^2}$

so, g' = g/(1 + h/R)²
0.04g = g/(1 + h/R)² [ from equation (1), ]
(0.2)² = 1/(1 + h/R)²
0.2 = 1/(1 + h/R)²
(1 + h/R) = 1/0.2
1 + h/R = 5
h/R = 4
h = 4R

hence, h = 4R