There is a planet which is 8 times massive and 27 times denser than the earth. If $g^{\prime}$ and $g$ are…
There is a planet which is 8 times massive and 27 times denser than the earth. If $g^{\prime}$ and $g$ are the accelerations due to gravity on the surfaces of the planet and the earth respectively then
$g^{\prime}=8 \mathrm{~g}$
$g^{\prime}=27 \mathrm{~g}$
$g^{\prime}=18 \mathrm{~g}$
$g^{\prime}=\frac{9}{4} g$
Solution
Let, $d=$ Density of planet,
$\begin{aligned} & M=\text { Mass of planet } \\ & R=\text { Radius of planet }\end{aligned}$
then density, $d=\frac{\text { Mass }}{\text { Volume }}$
$\begin{array}{ll}\Rightarrow & \quad d=\frac{M}{\left(\frac{4}{3} \pi R^3\right)} \\ \Rightarrow & R^3=\frac{3 M}{4 \pi d}\end{array}$
$\Rightarrow \quad R=\left(\frac{3 M}{4 \pi d}\right)^{\frac{1}{3}}$...(i)
Now, acceleration due to gravity $g$ is given by
$g=\frac{G M}{R^2}$
Substituting $R$ from Eq. (i) we get,
$\Rightarrow g=\frac{G M}{R^2}=\frac{G M}{\left(\frac{3}{4 \pi} \cdot \frac{M}{d}\right)^{\frac{2}{3}}}$
$\Rightarrow \quad g \propto M^{\frac{1}{3}} d^{\frac{2}{3}}$
Hence, ratio of acceleration due to gravity on surface of planet and that on surface of earth will be
$\frac{g^{\prime}}{g}=\frac{(M)^{\frac{1}{3}}}{M^{\frac{1}{3}}} \frac{(d)^{\frac{2}{3}}}{(d)^{\frac{2}{3}}}=\left(\frac{M^{\prime}}{M}\right)^{\frac{1}{3}} \cdot\left(\frac{d^{\prime}}{d}\right)^{\frac{2}{3}}$
Given, $\frac{M^{\prime}}{M}=8$ and $\frac{d^{\prime}}{d}=27$
$\begin{array}{ll}\text { So } & \frac{g^{\prime}}{g}=8^{\frac{1}{3}} \cdot 27^{\frac{2}{3}}=2 \times 9 \\ \text { or } & g^{\prime}=18 g\end{array}$