Question

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

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

Answer 1

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

Given:
Acceleration due to gravity at Earth's surface $=9.8{\text{ms}}^{-2}$,
Acceleration due to gravity at altitude $=4.9{\text{ms}}^{-2}$,
Radius of earth $=6.4\times {10}^6\text{m}$
We know that,
Acceleration due to gravity at a height $h$ from earth's surface is
$g_h=g{(1+\frac{h}{R_e})}^{-2}\Rightarrow 4.9=9.8{(1+\frac{h}{R_e})}^{-2}$

$\Rightarrow \sqrt{2}=(1+\frac{h}{R_e})$

$h=R_e(\sqrt{2}-1)$

$h=6400\times 0.414\text{km}$

$h=2.6\times {10}^6\text{m}$

Hence option $\left(A\right)$ is correct.