Question

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 o f earth =$6.4\times {10}^{6}m)$

• A. $1.6\times {10}^{6}m$
• B. $6.4\times {10}^{6}m$
• C. $9.0\times {10}^{6}m$
• D. $2.6\times {10}^{6}m$

Answer 1

The correct option is D $2.6\times {10}^{6}m$

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

$R+h=\sqrt{2}R$
$h=(\sqrt{2}-1)R$
$\approx 2.6\times {10}^{6}m$