A body weight 63 N on the surface of the earth. what is the gravitation force on it due to the earth at a height equal to half the radius of the earth 3

Weight of body $W=63N$

Acceleration due to gravity at height 'h' from

the earth surface in given by the relation

$9^{\prime }=9/{[1+\frac{h}{Re}]}^2$

Where 9= Acceleration due to gravity

$R_e$= radius of earth

For $h=\frac{R_e}{2}$

$\therefore 9^{\prime }=\frac{9}{{(1+\frac{R_e}{2R_e})}^2}=\frac{9}{1+(\frac{1}{2}{)}^2}=\left(\frac{4}{9}\right)g$

weight of body man m at height h in given as

$w^{\prime }=mg$

$=m\times \left(\frac{4}{9}\right)9=\frac{4}{9}mg$

$=\frac{4}{9}w$

$=\frac{4}{9}\times 63=28N$

$\therefore w^{\prime }=28N$