91Ó°ÊÓ

• Radius of Rhea is $R_R=764\text{ k³¾}×\frac{{10}^3\text{″¾}}{1\text{ k³¾}}=764×{10}^3\text{″¾}$.
• Acceleration due to gravity at the surface of Rhea is $g_R=0.265\text{″¾}/{\text{s}}^{\text{2}}$.

The acceleration due to gravity at the surface of Rhea can be expressed as,

$g_R=\frac{G{M}_R}{{R_R}^2}$

Here, is the mass of the Rhea, G is the gravitational constant, is the radius of Rhea.

Hence the mass of Rhea can be expressed as,

$M_R=\frac{g_R{R_R}^2}{G}$

Substitute $764×{10}^3\text{″¾}$for R_R, $0.265\text{″¾}/{\text{s}}^{\text{2}}$for g_R,$6.67×{10}^{−11}\text{ N}â‹\dots {\text{m}}^{\text{2}}â‹\dots {\text{kg}}^{−\text{2}}$ for G.

$\begin{array}{c}{M}_R={\frac{\left(0.265\text{″¾}/{\text{s}}^{\text{2}}\right){\left(764×{10}^3\text{″¾}\right)}^2}{6.67×{10}^{−11}\text{ N}â‹\dots {\text{m}}^{\text{2}}â‹\dots {\text{kg}}^{−\text{2}}}}^{}\left(\frac{\text{1}×\text{kg}}{{\text{kg}}^{\text{2}}\text{m}/{\text{Ns}}^{\text{2}}}\right)\\ =2.32×{10}^{21}\text{ k²µ}\end{array}$

Thus, the mass of Rhea is 2.32 x 10²¹Kg.

Consider the shape of Rhea is spherical. Then density of Rhea can be expressed as,

$\begin{array}{l}{\mathrm{ÒÏ}}_R=\frac{M_R}{V_R}\\ {\mathrm{ÒÏ}}_{{}_R}=\frac{M_R}{\frac{4}{3}\mathrm{Ï€}{R}_R^3}\end{array}$

Substitute $2.32×{10}^{21}\text{ k²µ}$for M_R, $764×{10}^3\text{″¾}$for R_R.

$\begin{array}{c}{\mathrm{ÒÏ}}_{{}_R}=\frac{2.32×{10}^{21}\text{ k²µ}}{\frac{4}{3}\mathrm{Ï€}{\left(764×{10}^3\text{″¾}\right)}^3}\\ =1242\text{ k²µ}/{\text{m}}^{\text{3}}\end{array}$

Thus, the density of Rhea is 1242 kg/m³.