The value of acceleration due to gravity at Earth's surface is $9.8m{s}^{-2}$. The altitude above its surface at which the acceleration due to gravity decreases to $4.9m{s}^{-2}$, is close to (Radius of earth $=6.4\times {10}^{6}m$ )

The correct option is D $2.6\times {10}^{6}m$
Formula used:$g=\frac{GM}{R^2}$
The value of acceleration due to gravity at Earth's surface is $9.8m{s}^{(-2)}$.

$g=\frac{GM}{R^2}\dots \left(i\right)$

The acceleration due to gravity decreases at altitude to $4.9m{s}^{(-2)}$.

$\frac{g}{2}=\frac{GM}{(R+h{)}^2}\dots \left(ii\right)$

$\frac{1}{2}=\frac{R^2}{(R+h{)}^2}$

$R+h=\sqrt{2}R$

$h=0.41R$

$=0.41\times 6.4\times {10}^{6}m$

$=2.6\times {10}^{6}M$

Final Answer: (d)