Chapter 9, Problem 38QAP

(a)

To determine

The mass of Electra.

Answer to Problem 38QAP

The mass of Electra is $8.17\times {10}^{18}\text{kg}$ .

Explanation of Solution

Write the expression for semi-major axis of moon orbit.

$a=r+d$        (I)

Here, $r$ is the radius of asteroid, $d$ is the distance between moon and asteroid and $a$ is the average distance between the objects.

Write the expression for orbital period using Kepler’s third law.

$T^2=\left(\frac{4{\pi }^2}{GM}\right)a^3$

Here, $G$ is gravitational constant, $M$ is the mass of Electra and $T$ is the orbital period.

Rearrange the above expression in term of $M$ .

$M=\left(\frac{4{\pi }^2}{G}\right)\frac{a^3}{T^2}$        (II)

Conclusion:

Calculate the radius of asteroid.

$\begin{array}{c}r=\frac{182\text{km}}{2}\\ =91\text{km}\end{array}$

Substitute $91\text{km}$ for $r$ and $1325\text{km}$ for $d$ in equation (I).

$\begin{array}{c}a=91\text{km}+1325\text{km}\\ =1416\text{km}\end{array}$

Substitute $1416\text{km}$ for $a$, $5.25\text{days}$ for $T$ and $6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^2$ for $G$ in equation (II).

$\begin{array}{c}M=\left(\left(\frac{4{\pi }^2}{6.67\times {10}^{-11}{\text{m}}^3/\text{kg}\cdot {\text{s}}^2}\right)\frac{{\left(\left(1416\text{km}\right)\left(\frac{{10}^3\text{m}}{1\text{km}}\right)\right)}^3}{{\left(\left(5.25\text{days}\right)\left(\frac{24\text{h}}{1\text{day}}\right)\left(\frac{3600\text{s}}{1\text{h}}\right)\right)}^2}\right)\\ =8.17\times {10}^{18}\text{kg}\end{array}$

Thus, the mass of Electra is $8.17\times {10}^{18}\text{kg}$ .

(b)

To determine

The density of Electra.

Answer to Problem 38QAP

The density of Electra is $2.6\times {10}^3\text{kg}/{\text{m}}^3$ .

Explanation of Solution

Write the expression for volume of asteroid.

$V=\frac{4}{3}\pi r^3$        (III)

Here, $V$ is the volume of sphere.

Write the expression for density of asteroid.

$\rho =\frac{M}{V}$        (IV)

Here, $\rho$ is the density of asteroid.

Conclusion:

Substitute $91\text{km}$ for $r$ in equation (III).

$\begin{array}{c}V=\frac{4}{3}\pi {\left(91\text{km}\right)}^3\\ =3.16\times {10}^6{\text{km}}^3\end{array}$

Substitute $3.16\times {10}^6{\text{km}}^3$ for $V$ and $8.17\times {10}^{18}\text{kg}$ for $M$ in equation (IV).

$\begin{array}{c}\rho =\frac{8.17\times {10}^{18}\text{kg}}{3.16\times {10}^6{\text{km}}^3{\left(\frac{{10}^3\text{m}}{1\text{km}}\right)}^3}\\ =2.6\times {10}^3\text{kg}/{\text{m}}^3\end{array}$

Thus, the density of Electra is $2.6\times {10}^3\text{kg}/{\text{m}}^3$ .