Q82 / 160 · JIPMER 2007 · Physics

Imagine a new planet having the same density as that of earth but it is $3$ times bigger than the earth in size. If the acceleration due to gravity on the surface of earth is $g$ and that on the surface of the new planet is $g^{\prime}$, then

  • Correct answer
    A
    $g^{\prime}=3 g$
  • B
    $g^{\prime}=\frac{g}{9}$
  • C
    $g^{\prime}=9 g$
  • D
    $g^{\prime}=27 g$

Solution

The acceleration due to gravity on the new planet can be found using the relation $g=\frac{G M}{R^2}...(i)$ but $M=\frac{4}{3} \pi R^3 \rho, \rho$ being density. Thus, Eq. (i) becomes $\therefore g=\frac{G \times \frac{4}{3} \pi R^3 \rho}{R^2}$ $=G \times \frac{4}{3} \pi R \rho$ $\begin{array}{ll}\Rightarrow & g \propto R \\ \therefore & \frac{g^{\prime}}{g}=\frac{R^{\prime}}{R} \\ \Rightarrow & \frac{g^{\prime}}{g}=\frac{3 R}{R}=3 \\ \Rightarrow & g^{\prime}=3 g\end{array}$

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

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