The height above the Earth's surface at which the value of acceleration due to gravity reduces to half its value on the Earth's surface is h. Assume the Earth to be a sphere of radius 6400 Km. Find h.

The correct option is B. 2650 km

We know that $\frac{g_h}{g}$=${\left(\frac{R}{R+H}\right)}^2$;

But $g_h$ = $\frac{g}{2}$

$\therefore$$\frac{1}{2}$ = ${\left(\frac{R}{R+H}\right)}^2$ or $\frac{R}{R+H}=\frac{1}{\sqrt{2}}$

or $\frac{R}{R+H}$ = $\frac{1}{\sqrt{2}}$ or $\frac{H}{R}$=$\sqrt{2}$ - 1 = 0.414

h = 0.414$\times$R = 0.414 $\times$ 6400 km/h = 2649.6 km.

$\therefore$ At a height of 2649.6 km from the Earth's surface, the acceleration due to gravity will be half its value on the surface.