How much above the surface of the earth does the acceleration due to gravity reduce by `36%` of its value on the surface of the earth.

Explanation

Since `g` reduces by `36%`, the value of `g` there is `100-36%=64 %`. It means `g'=64/100g`,
If `h` is the height of location above the surface of the earth, then,
`g'=g(R^(2))/((R+h)^(2))rArr 64/100 g=g(R^(2))/((R+h)^(2))`
`rArr 8/10=R/(R+h) rArr h=R/4=(6.4xx10^(6))/4=1.6xx10^(6) m`