Weight on a planet’s surface

The weight of a body on the surface of a spherical planet is $W = mg = m\frac{GM}{R^2}$ where $M$ is the mass of the planet and $R$ its radius.

Given ratio of weights

Let Earth’s mass and radius be $M_E$ and $R$.

Let the planet’s mass and radius be $M_P$ and $r$.

Then $\frac{W_E}{W_P} = \frac{\dfrac{GM_E}{R^2}}{\dfrac{GM_P}{r^2}} = \frac{M_E}{R^2}\cdot\frac{r^2}{M_P}$

Given: $\frac{W_E}{W_P} = \frac{9}{4}, \qquad M_P = \frac{1}{9}M_E$

Substituting: $\frac{9}{4} = \frac{M_E}{R^2}\cdot\frac{r^2}{(1/9)M_E} = 9\frac{r^2}{R^2}$

Solve for $r$

$\frac{9}{4} = 9\frac{r^2}{R^2}$ $\frac{1}{4} = \frac{r^2}{R^2}$ $\frac{r}{R} = \frac{1}{2}$

Hence, $r = \frac{R}{2}$

Check with same density condition

If the Earth and the planet have the same density, then mass is proportional to volume: $M \propto r^3$ So if $M_P = \frac{1}{9}M_E$ then $\left(\frac{r}{R}\right)^3 = \frac{1}{9} \implies \frac{r}{R} = \left(\frac{1}{9}\right)^{1/3}$ which is not equal to $\frac{1}{2}$.

Thus, the given data are internally inconsistent if we use the same-density condition.

Choosing the answer from options

Using the weight ratio and mass ratio directly, we get: $r = \frac{R}{2}$

So the correct option is: B. $\dfrac{R}{2}$