Chapter 15, Problem 58Q

(a)

To determine

The mass of Dactyl in kilogram if the density of Dactyl is $2500\text{kg}/{\text{m}}^3$ and the diameter of Dactyl is $1.4\text{km}$.

Answer to Problem 58Q

Solution:

$3.59\times {10}^{12}\text{kg}$.

Explanation of Solution

Given data:

The density of Dactyl is $2500\text{kg}/{\text{m}}^3$ and the diameter of Dactyl is $1.4\text{km}$.

Formula used:

Write the expression for the mass of Dactyl.

$m=\rho V$

Here, the mass of Dactyl is $m$, the density of Dactyl is $\rho$ and the volume of Dactyl is $V$.

Explanation:

Calculate the volume of Dactyl.

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

Here, the radius of Dactyl is $r$.

Substitute $\frac{1.4}{2}\text{km}$ for $r$.

$\begin{array}{c}V=\frac{4}{3}\pi {\left(\frac{1.4\text{km}}{2}\right)}^3\\ =1.436{\text{km}}^3\end{array}$

Write the expression for the mass of Dactyl.

$m=\rho V$

Substitute $1.436{\text{km}}^3$ for $V$ and $2500\text{kg}/{\text{m}}^3$ for $\rho$.

$\begin{array}{c}m=\left(2500\text{kg}/{\text{m}}^3\right)\left(1.436{\text{km}}^3\right)\\ =\left(2500\text{kg}/{\text{m}}^3\right)\left(1.436{\text{km}}^3\right)\left(\frac{{10}^9{\text{m}}^3}{1{\text{km}}^3}\right)\\ =3.59\times {10}^{12}\text{kg}\end{array}$

Conclusion:

Hence, the mass of Dactyl is $3.59\times {10}^{12}\text{kg}$.

(b)

To determine

The escape velocity from the surface of Dactyl if the density of Dactyl is $2500\text{kg}/{\text{m}}^3$ and the diameter of Dactyl is $1.4\text{km}$.

Answer to Problem 58Q

Solution:

$0.82\text{m}/\text{s}$.

Explanation of Solution

Given data:

The density of Dactyl is $2500\text{kg}/{\text{m}}^3$ and the diameter of Dactyl is $1.4\text{km}$.

Formula used:

Write the expression for the escape velocity at the surface of Dactyl.

$v=\sqrt{\frac{2Gm}{r}}$

Here, the escape velocity is $v$ and the gravitational constant is $G$.

Explanation:

From part (a), the mass of Dactyl is $3.59\times {10}^{12}\text{kg}$ and radius is $0.7\text{ km}$.

Write the expression for the escape velocity at the surface of Dactyl.

$v=\sqrt{\frac{2Gm}{r}}$

Substitute $6.67\times {10}^{-11}{\text{m}}^{\text{3}}\cdot {\text{kg}}^{-1}{\text{s}}^{-1}$ for $G$, $3.59\times {10}^{12}\text{kg}$ for $m$ and $0.7\text{km}$ for $r$.

$\begin{array}{c}v=\sqrt{\frac{2\left(6.67\times {10}^{-11}{\text{m}}^{\text{3}}\cdot {\text{kg}}^{-1}{\text{s}}^{-1}\right)\left(3.59\times {10}^{12}\text{kg}\right)}{\left(0.7\text{km}\right)}}\\ =\sqrt{\frac{2\left(6.67\times {10}^{-11}{\text{m}}^{\text{3}}\cdot {\text{kg}}^{-1}{\text{s}}^{-1}\right)\left(3.59\times {10}^{12}\text{kg}\right)}{\left(0.7\text{km}\right)\left(\frac{100\text{m}}{1\text{km}}\right)}}\\ =0.82\text{m}/\text{s}\end{array}$

Conclusion:

Hence, the escape velocity at the surface of Dactyl is $0.82\text{m}/\text{s}$.