Effect of altitude on g:
Let us consider \( g \) is the acceleration due to gravity on the surface of the earth. The mass of the earth is \( M \) and the radius is \( R \).
Then the gravitational force on a body of mass \( m \), placed on earth’s surface, exerted by the earth is
\( mg=G\frac{Mm}{R^2} \\or,\ g=\frac{GM}{R^2} \tag{1} \)Where \( G \) is the gravitational constant.
Let \( g’ \) is the acceleration due to gravity at a point \( P \), which is at a height \( h \) from the earth’s surface.
Now, \( mg’=G\frac{Mm}{({R+h}^2)} \\or,\ g’=\frac{GM}{({R+h}^2)} \tag{2} \)
Comparing eqn. 1 & eqn. 2 we get,
\( \frac{g}{g’}={\left(\frac{R+h}{R}\right)}^2 = {\left(1+\frac{h}{R}\right)}^2\)or, \( g’=g {\left(1+\frac{h}{R}\right)}^{-2} \)
or, \( g’=g\left(1-\frac{2h}{R}\right) \)
neglecting higher power of \( \frac{h}{R} \), since \( h<<R \)