Question

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

Answer 1

Answer:

Explanation:

gh=get(r/r+h)2

64=100(r/r+h)2

8/10=r/(r+h)

8r+8h=10r

h=r/4

Answer 2

Dear Student,

◆ 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.

Thanks dear. Hope this helps you...