Question

The height of a point vertically above the earth's surface at which the acceleration due to gravity becomes $9\%$ of its value at the surface is (Given, R$=$ radius of earth)

• A. $2R$
• B. $\frac{7}{3}R$
• C. $3R$
• D. $\frac{2}{3}R$

Answer 1

Answer: (B)

$\frac{7}{3}R$

Acceleration due to gravity at height h is given as
$g_h={\left(\frac{R}{R+h}\right)}^{2}g$
$\Rightarrow \frac{g_h}{g}={\left(\frac{1}{1+\frac{h}{R}}\right)}^2$ ..........(i)
$\because g_h$ is $9\%$ of g
$\therefore g_h=\frac{9}{100}$g
$=0.09$g
From Eq. (i),
$\frac{0.09g}{g}={\left(\frac{1}{1+\frac{h}{r}}\right)}^2$
${\left(1+\frac{h}{R}\right)}^2=\frac{1}{0.09}$
$\Rightarrow 1+\frac{h}{r}=\frac{1}{0.3}$
$\Rightarrow \frac{1}{R}=\frac{10}{3}-1=\frac{7}{3}$
$h=\frac{7}{3}R$.