A place where g is half

Gravitational acceleration at height above sea level Gh is given by

Gh = g (r/r+h)^2

where g = 9.8

r is Earths mean radius

h is height above sea level

Calculating h = 0.414 R

Using Newton's law of gravitation

$F =\frac { G m_1 m_2}{r^2} = m_2g$

( $m_2$ is mass of the object)

Divides $m_2$ on both sizes, we get

$g = \frac {G m_1}{r^2}$

Since $G m_1$ is a constant (Gravitational constant x Mass of the Earth)

We can conclude that

$g$ $\alpha \frac {1}{r^2}$

So R must be $\sqrt{2}$

Thus, the answer is $\sqrt{2} - 1 = 0.414$

This is not a solution, this is how i calculated this first we know that as we reach close to the surface of earth the gravitational force decreases so the the value of 'g' half on the surface of the earth should be <1 meaning the only choice left is 0.414 R

Let the height be h,we know that g=GM/R^2 so g/2=GM/(R+h)^2 or GM/2R^2=GM/(R+h)^2 or (R+h)^2=2R^2 or (R+h)^2=(sqrt 2R)^2 or R+h=sqrt 2R or h=sqrt 2R-R or h=0.414 R