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A body weighs 72 N on the surface of the earth. What is the gravitational force on it, at a height equal to half the radius of the earth?

1

48 N

2

32 N

3

30 N

4

24 N

Step-by-Step Solution

The weight at height h is given by W=mg0(1+hR)2W = \frac{mg_0}{(1 + \frac{h}{R})^2}. Substituting h=R/2h = R/2, we get W=72(1+R/2R)2=72(3/2)2=729/4=72×49=32 NW = \frac{72}{(1 + \frac{R/2}{R})^2} = \frac{72}{(3/2)^2} = \frac{72}{9/4} = \frac{72 \times 4}{9} = 32 \text{ N}.

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