Question

The height at which the acceleration due to gravity becomes $g/9$ (where $g=the$ acceleration due to gravity on the surface of the earth) in terms of $R$, the radius of the earth, is

• A. $R/2$
• B. $\sqrt{2}R$
• C. $2R$
• D. $\frac{R}{\sqrt{2}}$

Answer 1

Answer: (C)

$2R$

Acceleration due to gravity at a height ${}^{\prime }{h}^{\prime }$ is given by:-

${}^{\prime }{g}^{\prime }=g{\left(\frac{R}{R+H}\right)}^2$

where $g\to$ on the surface of the earth.

$R\to$ Radius of earth.

As $g^{\prime }=g/9,$ we get :

$\Rightarrow \frac{g}{9}=g{\left(\frac{R}{R+H}\right)}^2$

$\Rightarrow \frac{R}{R+H}=\frac{1}{3}\Rightarrow h=2R$