The percentage decrease in the weight of a body when taken to a height of $48$ km above the surface of the earth is (Radius of the earth is $6400$ km)

The acceleration due to gravity at a height $h$ above the surface of the earth, where $h \ll R$, is given by

$g' = g \left(1 - \dfrac{2h}{R}\right)$

The fractional decrease in weight is

$\dfrac{\Delta W}{W} = \dfrac{g - g'}{g} = \dfrac{2h}{R}$

The percentage decrease in weight is

$\dfrac{\Delta W}{W} \times 100 = \dfrac{2h}{R} \times 100$

Substituting $h = 48$ km and $R = 6400$ km:

Percentage decrease $= \dfrac{2 \times 48}{6400} \times 100$

Percentage decrease $= \dfrac{96}{64} = 1.5\%$

Answer: $1.5\%$