At what height from the surface of earth will the value of g be reduced by `36%` from the value on the surface? Take radius of earth `R = 6400 km`.
A
3.2×10^6 m
B
1.6×10^6 m
C
2.5×10^6 m
D
4.6×10^6 m
Text Solution
Verified by Experts
The correct Answer is:
B
Since the acceleration due to gravity reduces by `36%` the value of acceleration due to gravity there is `100 -36=64%`. It means `g'=64/100g.` If `h` is the height of location above the surface of earth then `g'=gR^(2)/((R+h)^2)` or `64/100g=gR^(2)/((R+h^(2))^(2))` `8/10=R/(R+h)` or `8R+8h=10R` `impliesh=(2R)/8=R/4=(6.4xx10^(6))/4=1.6xx10^(6)m`
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