91影视

• Radius of Rhea is $R_R=764\text{鈥塳尘}脳\frac{{10}^3\text{鈥尘}}{1\text{鈥塳尘}}=764脳{10}^3\text{鈥尘}$.
• Acceleration due to gravity at the surface of Rhea is $g_R=0.265\text{鈥尘}/{\text{s}}^{\text{2}}$.

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 $764脳{10}^3\text{鈥尘}$for R_R, $0.265\text{鈥尘}/{\text{s}}^{\text{2}}$for g_R,$6.67脳{10}^{鈭�11}\text{鈥塏}鈰�{\text{m}}^{\text{2}}鈰�{\text{kg}}^{鈭�\text{2}}$ for G.

$\begin{array}{c}{M}_R={\frac{\left(0.265\text{鈥尘}/{\text{s}}^{\text{2}}\right){\left(764脳{10}^3\text{鈥尘}\right)}^2}{6.67脳{10}^{鈭�11}\text{鈥塏}鈰�{\text{m}}^{\text{2}}鈰�{\text{kg}}^{鈭�\text{2}}}}^{}\left(\frac{\text{1}脳\text{kg}}{{\text{kg}}^{\text{2}}\text{m}/{\text{Ns}}^{\text{2}}}\right)\\ =2.32脳{10}^{21}\text{鈥塳驳}\end{array}$

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,

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

Substitute $2.32脳{10}^{21}\text{鈥塳驳}$for M_R, $764脳{10}^3\text{鈥尘}$for R_R.

$\begin{array}{c}{蚁}_{{}_R}=\frac{2.32脳{10}^{21}\text{鈥塳驳}}{\frac{4}{3}蟺{\left(764脳{10}^3\text{鈥尘}\right)}^3}\\ =1242\text{鈥塳驳}/{\text{m}}^{\text{3}}\end{array}$

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