Chapter 7, Problem 6Q

To determine

The mass and the average density of the planet mars are to be found.

Answer to Problem 6Q

The mass of the planet mars is $1.665\times {10}^{23}\text{kg}$ and the average density of planet mars is $1.014{\text{g/cm}}^{\text{3}}$.

Explanation of Solution

Given:

The orbital period of the satellite Phobosis $0.31891\text{day}$.

The average altitude of Phobosis $5890\text{km}$.

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

Concept used:

Mass is defined as the measure of sum total of all content present in the bulk and at the surface of the body.

The centripetal force is required for a body to be in a circular motion and that is provided by the gravitational force in case of celestial bodies.

Write the expression for the equality of the gravitational force on Phobos exerted by mars and the centripetal force to be in the circular motion.

$F_G=F_C$ ...... (1)

Write the expression for the gravitational force.

$F_G=\frac{GMm}{r^2}$

Write the expression for the centripetal force.

$F_C=\frac{m{v}^2}{r}$

Substitute $\frac{GMm}{r^2}$ for $F_G$ and $\frac{m{v}^2}{r}$ for $F_C$ in equation (1).

$\frac{GMm}{r^2}=\frac{m{v}^2}{r}$ ...... (2)

Here, $G$ is gravitational constant, $M$ is the mass of the mars, $m$ is the mass of the Phobos, $v$ is the orbital velocity of the Phobos and $r$ is the distance between the mars and the Phobos.

Write the expression for the orbital period of Phobos.

$T=\frac{2\pi r}{v}$

Rearrange the above expression for $v$.

$v=\frac{2\pi r}{T}$

Here, $T$ is the orbital period of the Phobos.

Substitute $\frac{2\pi r}{T}$ for $v$ in equation (2).

$\frac{GMm}{r^2}=\frac{m{\left(\frac{2\pi r}{T}\right)}^2}{r}$

Simplify the above expression for $M$.

$M=\frac{4{\pi }^2{r}^3}{G{T}^2}$ ...... (3)

Write the expression for the density of the mars.

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

Here, $\rho$ is the density of the mars and $V$ is the volume of the mars.

Write the expression for the volume of mars.

$V=\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$

Substitute $\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$ for $V$ in equation (4).

$\rho =\frac{M}{\left(\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3\right)}$ ...... (4)

Here, $D$ is the diameter of the mars.

Calculation:

Substitute $0.31891\text{day}$ for $T$, $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{-2}}$ for $G$ and $5890\text{km}$ for $r$ in equation (3).

$\begin{array}{c}M=\frac{4{\pi }^2{\left(5890\text{km}\right)}^3}{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{-2}}\right){\left(0.31891\text{day}\right)}^2}\\ =\frac{4{\pi }^2{\left(589\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^3}{\left(6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{-2}}\right){\left(0.31891\text{day}\left(\frac{86400\text{s}}{1\text{day}}\right)\right)}^2}\\ =1.665\times {10}^{23}\text{kg}\end{array}$

Substitute $1.665\times {10}^{23}\text{kg}$ for $M$ and $6794\text{km}$ for $D$ in equation (4).

$\begin{array}{c}\rho =\frac{1.665\times {10}^{23}\text{kg}}{\left(\frac{4}{3}\pi {\left(\frac{6794\text{km}}{2}\right)}^3\right)}\\ =\frac{1.665\times {10}^{23}\text{kg}}{\left(\frac{4}{3}\pi {\left(\frac{6794\text{km}}{2}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^3\right)}\\ =1014.5\text{kg}\left(\frac{1000\text{g}}{1\text{kg}}\right)\text{/}{\left(\text{m}\left(\frac{100\text{cm}}{1\text{m}}\right)\right)}^{\text{3}}\\ =1.014{\text{g/cm}}^{\text{3}}\end{array}$

Conclusion:

Thus,The mass of the planet mars is $1.665\times {10}^{23}\text{kg}$ and the average density is $1.014{\text{g/cm}}^{\text{3}}$