Chapter 7, Problem 6Q

To determine

The mass and average density of Mars, if the diameter of Mars is $6794\text{ km}$, the orbital period of Phobos is $0.31891\text{ days}$ and the altitude of Phobos is $5980\text{}\text{ km}$ above the planet’s surface.

Answer to Problem 6Q

Solution:

$6.425\times {10}^{23}\text{ kg}$ and $3.89\times {10}^{12}{\text{ kg/m}}^3$.

Explanation of Solution

Given data:

The diameter of Mars is $6794\text{ km}$.

Phobos’ orbital period is $0.31891\text{ days}$.

The altitude of Phobos is $5980\text{ km}$ from the surface of Mars.

Formula used:

The expression for Kepler’s Third Law, for the orbital period of a planet or a satellite, is given as:

$p^2=\frac{4{\pi }^2{a}^3}{GM}$

Here, $P$ is the orbital period of the star, $a$ is the semi-major axis, $r$ is the distance from the star, $M$ is the mass of the planet and $G$ is the constant of gravitation $\left(6.673\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{3}}{\text{/kg}}^2\right)$.

Write the expression for radius.

$r=\frac{D}{2}$

Here, $D$ is the diameter.

Write the formula for the volume of a sphere.

$V=\frac{4}{3}\pi r^3$

Here, $r$ is the radius.

Write the expression for density $\left(\rho \right)$.

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

Here, $m$ is the mass and $V$ is the volume.

Explanation:

Phobos’ orbital radius is:

$\begin{array}{c}a=\text{Average altitude}+\frac{1}{2}\left(\text{Mar's Diameter}\right)\\ =5980\text{ km}+\frac{1}{2}\left(6794\text{ km}\right)\\ =9377\text{ km}\left(\frac{{10}^3\text{ m}}{1\text{ km}}\right)\\ =9.377\times {10}^6\text{ m}\end{array}$

Recall the expression for Kepler’s Third Law for the orbital period of a satellite:

$p^2=\frac{4{\pi }^2{a}^3}{GM}$

Rearrange the expression in terms of the mass of planet.

$M=\frac{4{\pi }^2{a}^3}{G{p}^2}$

Substitute $0.31891\text{ days}$ for $p$, $9.377\times {10}^6\text{ m}$ for $a$ and $6.673\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{3}}{\text{/kg}}^2$ for $G$.

$\begin{array}{c}M=\frac{4{\pi }^2{\left(9.377\times {10}^6\text{ m}\right)}^3}{\left(6.673\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{3}}{\text{/kg}}^2\right){\left(0.31891\text{ days}\right)}^2}\\ =6.425\times {10}^{23}\text{ kg}\end{array}$

Recall the expression for radius.

$r=\frac{D}{2}$

Substitute $6794\text{ km}$ for $D$.

$\begin{array}{c}r=\frac{6794\text{ km}}{2}\\ =3397\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ =3.4\times {10}^3\text{ m}\end{array}$

Recall the formula for the volume of a sphere.

$V=\frac{4}{3}\pi r^3$

Substitute $3.4\times {10}^3\text{ m}$ for $r$.

$\begin{array}{c}V=\frac{4}{3}\pi {\left(3.4\times {10}^3\text{ m}\right)}^3\\ =\frac{4}{3}\pi {\left(3.4\times {10}^3\text{ m}\right)}^3\\ =1.65\times {10}^{11}{\text{ m}}^3\end{array}$

Recall the expression for density $\left(\rho \right)$.

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

Substitute $6.425\times {10}^{23}\text{ kg}$ for $m$ and $1.65\times {10}^{11}{\text{ m}}^3$ for $V$.

$\begin{array}{c}\rho =\frac{6.425\times {10}^{23}\text{ kg}}{1.65\times {10}^{11}{\text{ m}}^3}\\ =3.89\times {10}^{12}{\text{ kg/m}}^3\end{array}$

Conclusion:

The mass of Mars is $6.425\times {10}^{23}\text{ kg}$ and the average density of Mars is $3.89\times {10}^{12}{\text{ kg/m}}^3$.