At surface of earth weight of a person is 72 N then his weight at height R/2 from surface of earth is (R = radius of earth)

Question

At surface of earth weight of a person is 72 N then his weight at height R/2 from surface of earth is (R = radius of earth) (A) 28 N (B) 16 N (C) 32 N (D) 72 N

Tardigrade Admin · Accepted Answer

Weight of the body at height R, W' = W((R/R+h))2 = W((R/R+(R)2))2 = W((2/3))2 = (4/9) W = (4/9)×72 = 32 N

Q. At surface of earth weight of a person is $72 N$ then his weight at height $R /2$ from surface of earth is ( $R =$ radius of earth)

$28 N$

$16 N$

$32 N$

$72 N$

Weight of the body at height $R$ ,

$W^{'} = W (\frac{R}{R + h})^{2}$

$= W (\frac{R}{R + \frac{R}{2}})^{2} = W (\frac{2}{3})^{2}$

$= \frac{4}{9} W = \frac{4}{9} \times 72 = 32 N$

Q. At surface of earth weight of a person is 72N then his weight at height R/2 from surface of earth is (R= radius of earth)

28N

16N

32N

72N