Q84 / 90 · JEE Main 2009 · Physics

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

  • Correct answer
    A
    $2 R$
  • B
    $\frac{R}{\sqrt{2}}$
  • C
    $\frac{R}{2}$
  • D
    $\sqrt{2} \mathrm{R}$

Solution

$g^{\prime}=\frac{G M}{(R+h)^2}$, acceleration due to gravity at height $h$ $$ \begin{aligned} & \Rightarrow \frac{g}{9}=\frac{G M}{R^2} \cdot \frac{R^2}{(R+h)^2}=g\left(\frac{R}{R+h}\right)^2 \\ & \Rightarrow \frac{1}{9}=\left(\frac{R}{R+h}\right)^2 \Rightarrow \frac{R}{R+h}=\frac{1}{3} \\ & \Rightarrow 3 R=R+h \Rightarrow 2 R=h \end{aligned} $$

Hence, the correct answer is option: 'A'.

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