Answer
(a) The acceleration due to gravity at the surface of Titania is $0.369~m/s^2$.
(b) The average density of Titania is $1660~kg/m^3$.
Work Step by Step
(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 $1660~kg/m^3$.
