91Ó°ÊÓ

The volume of an object is the amount of three-dimensional space it takes up, whereas density is the mass of the unit volume of the object.

The given data and data from Appendix $F$ can be listed below as,

The mass of the planet is,

The density of the planet is, ${\mathrm{ÒÏ}}_P=1.76\mathrm{g}/{\mathrm{cm}}^3$.

The radius of the earth is, $r_E=6.37×{10}^6\mathrm{m}$.

The volume of the planet is given by,

$V_P=\frac{M_P}{{\mathrm{ÒÏ}}_P}$

Here, $M_P$is the planet's mass and ${\mathrm{ÒÏ}}_P$is the planet's density.

Substitute the values in formula of, $V_P$

$\begin{array}{rcl}{V}_P& =& \frac{3.28×{10}^{25}\mathrm{kg}}{1.76\mathrm{g}/{\mathrm{cm}}^3}\left(\frac{1\mathrm{g}/{\mathrm{cm}}^3}{1000\mathrm{kg}/{\mathrm{m}}^3}\right)\\ & =& 1.86×{10}^{22}{\mathrm{m}}^3\end{array}$

Consider the planet as a sphere, it's volume can be expressed as,

$V_P=\frac{4\mathrm{Ï€}}{3}{r}_P^3$

Here, $r_P$is the planet's radius.

Solution of the above equation for ${\mathrm{Γ}}_{\mathrm{P}}$ gives,

Substitute all the values in the above,

$\begin{array}{rcl}{r}_P& =& {\left(\frac{3×\left(1.86×{10}^{22}{\mathrm{m}}^3\right)}{4\mathrm{π}}\right)}^{1/3}\\ & =& 1.64×{10}^7\mathrm{m}\left(\frac{1\mathrm{km}}{1000\mathrm{m}}\right)\\ & =& 1.64×{10}^4\mathrm{km}\end{array}$

Thus, the planet's radius is $1.64×{10}^4\mathrm{km}$.

The planet’s radius is given by,

The radius of the planet in terms of earth's radius can be expressed as,

Here, $r_E$is the radius of the earth whose value is $r_E=6.37×{10}^6\mathrm{m}$.

Substitute values in the above equation.

Thus, the planet's radius in terms of the earth's radius $\left(r_E\right)$ is $2.57r_E$.

The planet's radius is $1.64×{10}^4\mathrm{km}$ and in terms of the earth's radius $\left(r_E\right)$, it is $2.57r_E$.