Rhea, one of Saturn’s moons, has a radius of 764 km and an acceleration due to gravity of 0.265 m/s²at its surface. Calculate its mass and average density.

• Radius of Rhea is $R_R=764\text{\hspace{0.17em}km}\times \frac{{10}^3\text{\hspace{0.17em}m}}{1\text{\hspace{0.17em}km}}=764\times {10}^3\text{\hspace{0.17em}m}$.
• Acceleration due to gravity at the surface of Rhea is $g_R=0.265\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$.

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 $764\times {10}^3\text{\hspace{0.17em}m}$for R_R, $0.265\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}$for g_R,$6.67\times {10}^{-11}\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{-\text{2}}$ for G.

$\begin{array}{c}{M}_R={\frac{\left(0.265\text{\hspace{0.17em}m}/{\text{s}}^{\text{2}}\right){\left(764\times {10}^3\text{\hspace{0.17em}m}\right)}^2}{6.67\times {10}^{-11}\text{\hspace{0.17em}N}\cdot {\text{m}}^{\text{2}}\cdot {\text{kg}}^{-\text{2}}}}^{}\left(\frac{\text{1}\times \text{kg}}{{\text{kg}}^{\text{2}}\text{m}/{\text{Ns}}^{\text{2}}}\right)\\ =2.32\times {10}^{21}\text{\hspace{0.17em}kg}\end{array}$

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,

$\begin{array}{l}{\rho }_R=\frac{M_R}{V_R}\\ {\rho }_{{}_R}=\frac{M_R}{\frac{4}{3}\pi R_R^3}\end{array}$

Substitute $2.32\times {10}^{21}\text{\hspace{0.17em}kg}$for M_R, $764\times {10}^3\text{\hspace{0.17em}m}$for R_R.

$\begin{array}{c}{\rho }_{{}_R}=\frac{2.32\times {10}^{21}\text{\hspace{0.17em}kg}}{\frac{4}{3}\pi {\left(764\times {10}^3\text{\hspace{0.17em}m}\right)}^3}\\ =1242\text{\hspace{0.17em}kg}/{\text{m}}^{\text{3}}\end{array}$

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