Question

Titania, the largest moon of the planet Uranus, has 1/8 the radius of the earth and 1/1700 the mass of the earth.What is the acceleration due to gravity at the surface of Titania?What is the average density of Titania? (This is less than the density of rock, which is one piece of evidence that Titania is made primarily of ice.)

Answer 1

Answer:

(a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³

Explanation:

Given that,

Radius of Titania $R_{t}= \dfrac{1}{8}R_{e}$

Mass of Titania $M_{t}= \dfrac{1}{1700}M_{e}$

We need to calculate the acceleration due to gravity at the surface of Titania

Using formula of the acceleration due to gravity on earth

$g_{e}=\dfrac{GM_{e}}{R_{e}^2}$

The acceleration due to gravity on Titania

$g_{t}=\dfrac{GM_{t}}{R_{t}^2}$

Put the value into the formula

$g_{t}=\dfrac{G\times\dfrac{1}{1700}M_{e}}{(\dfrac{1}{8}R_{e})^2}$

$g_{t}=\dfrac{64}{1700}\times G\dfrac{M_{e}}{R_{e}^2}$

$g_{t}=0.004705 g_{e}$

$g_{t}=0.03764\times9.8$

$g_{t}=0.37\ m/s^2$

The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). We assume Titania is a sphere

The average density of the earth is 5500 kg/m³.

We need to calculate the average density of Titania

Using formula of density

$\rho=\dfrac{m}{V}$

$\rho_{t}=\dfrac{M_{t}}{\dfrac{4}{3}\pi\times R_{t}^2}$

$\rho_{t}=\dfrac{\dfrac{M_{e}}{1700}}{\dfrac{4}{3}\pi\times(\dfrac{R_{e}}{8})^2}$

$\rho_{t}=\dfrac{512}{1700}\times\rho_{e}$

$\rho_{t}=\dfrac{512}{1700}\times5500$

$\rho_{t}=1656.47\ kg/m^{3}$

The average density of Titania is 1656.47 kg/m³

Hence, (a). The acceleration due to gravity at the surface of Titania is 0.37 m/s².

(b). The average density of Titania is 1656.47 kg/m³