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The ratio of the weights of a body on the Earth's surface to that on the surface of a planet is $9 : 4$. The mass of the planet is $\frac{1}{9}^{th}$ of that of the Earth. If $'R'$ is the radius of the Earth, what is the radius of the planet ? (Take the planets to have the same mass density)
$\frac{R}{3}$
$\frac{R}{4}$
$\frac{R}{9}$
$\frac{R}{2}$
Solution
Since mass of the object remains same
$\therefore $ Weight of object will be proporational to $'g'$ $(acceleration\, due\, to\,gravity)$ Given ;
$\frac{{{W_{earth}}}}{{{W_{planet}}}} = \frac{9}{4} = \frac{{{g_{earth}}}}{{{g_{planet}}}}$
Also,
${g_{surface}} = \frac{{GM}}{{{R^2}}}$ $(M is mass planet,
G is unversal gravitational constant, R is radius of planet)$
$\therefore \frac{9}{4} = \frac{{G{M_{earth}}R_{planet}^2}}{{G{M_{planet}}R_{earth}^2}} = \frac{{{M_{earth}}}}{{{M_{planet}}}} \times \frac{{R_{planet}^2}}{{R_{earth}^2}}$
$ = 9\frac{{R_{planet}^2}}{{R_{earth}^2}}$
$\therefore {R_{planet}} = \frac{{{R_{earth}}}}{2} = \frac{R}{2}$
