## University Physics with Modern Physics (14th Edition)

(a) The acceleration due to gravity at the surface of Titania is $0.369~m/s^2$. (b) The average density of Titania is $1660~kg/m^3$.
(a) We can find the radius of Titania. $R = \frac{1}{8}~(6380~km)$ $R = 797.5~km$ We can find the mass of Titania. $M = \frac{1}{1700}~(5.98\times 10^{24}~kg)$ $M = 3.52\times 10^{21}~kg$ We can find the acceleration due to gravity $g'$ at the surface of Titania. $g' = \frac{G~M}{R^2}$ $g' = \frac{(6.67\times 10^{-11}~m^3/kg~s^2)(3.52\times 10^{21}~kg)}{(7.975\times 10^5~m)^2}$ $g' = 0.369~m/s^2$ The acceleration due to gravity at the surface of Titania is $0.369~m/s^2$. (b) We can find the volume of Titania. $V = \frac{4}{3}~\pi~R^3$ $V = \frac{4}{3}~\pi~(7.975\times 10^5~m)^3$ $V = 2.12\times 10^{18}~m^3$ We can find the average density of Titania. $\rho = \frac{Mass}{Volume}$ $\rho = \frac{3.52\times 10^{21}~kg}{2.12\times 10^{18}~m^3}$ $\rho = 1660~kg/m^3$ The average density of Titania is $1660~kg/m^3$.