To solve this problem, we need to determine the gravitational force on a body at a height equal to half the radius of the Earth.$\\$Given:$\\$• Weight of the body on the surface of the Earth $= 72$ N$\\$• Height above the Earth's surface $= \frac{R}{2}$, where $R$ is the radius of the Earth.$\\$The gravitational force at a distance $r$ from the center of the Earth is given by:$\\ F = \frac{G M m}{r^2} \\$On the surface of the Earth, the gravitational force (weight) is:$\\ F_{surface} = \frac{G M m}{R^2} = 72$ N$\\$At a height $\frac{R}{2}$, the distance from the center of the Earth is:$\\ r = R + \frac{R}{2} = \frac{3R}{2} \\$The gravitational force at this height is:$\\ F_{height} = \frac{G M m}{\left(\frac{3R}{2}\right)^2} \\$Simplify the expression:$\\ F_{height} = \frac{G M m}{\frac{9R^2}{4}} = \frac{4 G M m}{9 R^2} \\$We know $F_{surface} = \frac{G M m}{R^2} = 72$ N$\\$Therefore, $F_{height} = \frac{4}{9} \times F_{surface} \\ F_{height} = \frac{4}{9} \times 72 \\ F_{height} = \frac{288}{9} \\ F_{height} = 32$ N$\\$Thus, the gravitational force on the body at a height equal to half the radius of the Earth is $32$ N.$\\$This corresponds to Option 1.