1.

The altitude at which the weight of a body is only 64% of its weight on the surface of earth is

Answer»

ANSWER:

h = 1.6×10^6 m

Explanation:

LET h be the height at which weight of the BODY is 64 % of its weight on earth. W' = 64W/100.

g' = g [R/(R+h)]²  

[(R+h)/R]^2 = g/g'  

(h/R + 1)^2 = mg / mg'  

(h/R + 1)^2 = W / W'  

(h/R + 1)^2 = W / (64W/100)  

(h/R + 1)^2 = 100/64  

h/R + 1 = 10/8  

h/R = 1.25 - 1  

h/R = 0.25  

h = 0.25 R  

h = 0.25 × 6.4×10^6  

h = 1.6×10^6 m

Hence, altitude is 1.6×10^6 m.

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