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A body of weight 72 N moves from the surface of earth to a height half of the radius of the earth, then gravitational force exerted on it will be :
A
36 N
B
32 N
C
144 N
D
50 N
Detailed Solution
On the surface: $F = mg = 72$ N
At height h: $g' = g\left(\dfrac{R_e}{R_e + h}\right)^2 = g\left(\dfrac{R_e}{R_e + R_e/2}\right)^2 = g\left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}g$
$F' = mg' = mg \times \dfrac{4}{9} = 72 \times \dfrac{4}{9}$
$F' = 32$ N
At height h: $g' = g\left(\dfrac{R_e}{R_e + h}\right)^2 = g\left(\dfrac{R_e}{R_e + R_e/2}\right)^2 = g\left(\dfrac{2}{3}\right)^2 = \dfrac{4}{9}g$
$F' = mg' = mg \times \dfrac{4}{9} = 72 \times \dfrac{4}{9}$
$F' = 32$ N