At what height over the Earth's pole the free-fall acceleration decreases by one per cent, by half?

Explanation

Let h be the sought height in the first case. so
`(99)/(100)g=(gammaM)/((R+h)^2)`
`=(gammaM)/(R^2(1+h/R)^2)=(g)/((1+h/R)^2)`
or `(99)/(100)=(1+h/R)^-2`
From the statement of the problem, it is obvious that in this case `h lt lt R`
Thus `(99)/(100)=(1-(2h)/(R))` or `h=(R)/(200)=((6400)/(200))km=32km`
In the other case if `h^'` be the sought height, than
`g/2=g(1+h^'/R)^-2` or `1/2=(1+h^'/R)^-2`
From the language of the problem, in this case `h^'` is not very small in comparison with R.
Therefore in this case we cannot use the approximation adopted in the previous case.
Here, `(1+h^'/R)^2=2` So, `h^'/R=+-sqrt2-1`
As `-ve` sign is not acceptable
`h^'=(sqrt2-1)R=(sqrt2-1)6400km=2650km`