Chapter 12, Problem 38QP

(a)

To determine

The mass of the asteroid Electra.

Answer to Problem 38QP

The mass of the asteroid Electra is $\text{1}\text{.269}\times {\text{10}}^{19}\text{kg}$ .

Explanation of Solution

Write the expression for the Kepler’s law for the time period of a celestial body around the planet.

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

Rearrange the above expression for the mass of the planet.

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

Here, $M$ is the mass of the body around which another body revolves, $R$ is the distance between the two bodies, $G$ is the gravitational constant and $T$ is the time period of body.

Conclusion:

Substitute $1350\text{km}$ for $R$, $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$ and $3.92\text{days}$ for $T$ in equation (I).

  $\begin{array}{c}M=\frac{4{\pi }^2{\left(1350\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^3}{\left(6.67\times {10}^{-11}\right){\left(3.92\text{days}\left(\frac{24\text{hr}}{1\text{day}}\right)\left(\frac{60\min }{1\text{hr}}\right)\left(\frac{60\sec }{1\min }\right)\right)}^2}\\ =\frac{9.713\times {10}^{19}}{7.65}\text{kg}\\ \text{=1}\text{.269}\times {\text{10}}^{19}\text{kg}\end{array}$

Thus, the mass of the asteroid Electra is $\text{1}\text{.269}\times {\text{10}}^{19}\text{kg}$ .

(b)

To determine

The density of the asteroid Electra.

Answer to Problem 38QP

The density of the asteroid Electra is $4.021\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ .

Explanation of Solution

Write the expression for the density of the asteroid Electra.

  $\rho =\frac{M}{V}$        (II)

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

Write the expression for the volume of the asteroid.

  $V=\frac{4}{3}\pi r^3$        (III)

Here, $r$ is the radius of the asteroid.

The diameter of the asteroid is $182\text{km}$. Therefore, the radius of the asteroid is $91\text{km}$.

Conclusion:

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

  $\begin{array}{c}V=\frac{4}{3}\pi {\left(91\text{km}\left(\frac{1000\text{m}}{1\text{km}}\right)\right)}^3\\ =4.188\left(7.53\times {10}^{14}\right){\text{m}}^{\text{3}}\\ =3.156\times {10}^{15}{\text{m}}^{\text{3}}\end{array}$

Substitute $\text{1}\text{.269}\times {\text{10}}^{19}\text{kg}$ for $M$ and $3.156\times {10}^{15}{\text{m}}^{\text{3}}$ for $V$ in equation (II).

  $\begin{array}{c}\rho =\frac{\text{1}\text{.269}\times {\text{10}}^{19}\text{kg}}{3.156\times {10}^{15}{\text{m}}^{\text{3}}}\\ =4.021\times {10}^3\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Thus, the density of the asteroid Electra is $4.021\times {10}^3\text{kg}/{\text{m}}^{\text{3}}$ .