Chapter 20, Problem 30Q

(a)

To determine

The average density of a $1{\text{M}}_{\Theta }$ dwarf star whose diameter is same as that of

Earth.

Answer to Problem 30Q

$1.83\times {10}^9\text{kg}\cdot {\text{m}}^{-3}$

Explanation of Solution

Given data:

The mass of the star is $1{\text{M}}_{\text{Θ}}$ and the radius of the star is $6378\text{km}$.

Formula used:

The average density of the star can be calculated by using the following relation:

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

Here, $\overline{\rho }$ is the average density of the star, $M$ is the mass of the star and $V$ is the volume of the star.

Write the expression for the volume of a sphere.

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

Here, $V$ and $R$ are the volume and radius of a spherical object, such as a star, respectively.

Explanation:

Since the mass of the star is given in solar mass, convert it into standard the mass unit, that is, kilograms.

$1{\text{M}}_{\odot }=1.989\times {10}^{30}\text{kg}$

Recall the expression for the volume of a sphere.

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

Substitute $6.378\times {10}^6\text{ m}$ for R.

$\begin{array}{c}V=\frac{4}{3}\pi {\left(6.378\times {10}^6\text{ m}\right)}^3\\ =1.059\times {10}^{21}{\text{m}}^{\text{3}}\end{array}$

Recall the expression for the average density of a star.

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

Substitute $1.98\times {10}^{30}\text{ kg}$ for M and $1.059\times {10}^{21}{\text{ m}}^3$ for V.

$\begin{array}{c}\overline{\rho }=\frac{1.98\times {10}^{30}\text{ kg}}{1.059\times {10}^{21}{\text{ m}}^3}\\ =1.83\times {10}^9\text{kg}\cdot {\text{m}}^{-3}\end{array}$

Conclusion:

The average density of the dwarf star, $\overline{\rho }=1.83\times {10}^9\text{kg}\cdot {\text{m}}^{-3}$.

(b)

To determine

Whether or not the star’s density is the same as the density of about an elephant per teaspoon, along with proper calculation.

Answer to Problem 30Q

No, the star’s density is approximately twice that of about an elephant per teaspoon.

Explanation of Solution

Given data:

Let us consider a general elephant with mass 4000 kg and an ordinary teaspoon with volume $V_{ts}=5\times {10}^{-6}{\text{m}}^{\text{3}}$.

Formula used:

Write the expression of the density as -

$\rho =\frac{m}{V_{ts}}$

Here, $\rho$ is the density of about an elephant per teaspoon, $m$ is the mass of an elephant and $V_{ts}$ is the volume of a teaspoon.

Explanation:

The density will be calculated as:

$\rho =\frac{m}{V_{ts}}$

Substitute $4\times {10}^3\text{kg}$ for $m$ and $5\times {10}^{-6}{\text{m}}^{\text{3}}$ for $V_{ts}$ in the expression of density.

$\begin{array}{c}\rho =\frac{4\times {10}^3\text{kg}}{5\times {10}^{-6}{\text{m}}^{\text{3}}}\\ =8.1\times {10}^8\text{kg}/{\text{m}}^3\end{array}$

Compare $\overline{\rho }$ (calculated in a-part) and $\rho$(calculated above).

$\overline{\rho }=2\rho$

Conclusion:

The density of the dwarf star is approximately twice that of about an elephant per teaspoon. So, the provided statement is incorrect.

(c)

To determine

The escape speed required for a gas to eject from the star’s surface.

Answer to Problem 30Q

$6450\text{km}\text{/s}$

Explanation of Solution

Given data:

The mass of the star is $1{\text{M}}_{\text{Θ}}$ and the radius of the star is $6378\text{km}$.

Formula used:

Write the expression for escape speed.

$v_e=\sqrt{\frac{2GM}{R}}$

Here, $v_e$ is the escape speed of the gas, $G$ is the gravitational constant and $R$ is the radius of the star.

Explanation:

The escape speed is the minimum speed required to send an object out of the gravitational field of the body it is being sent from.

Recall the expression for escape speed.

$v_e=\sqrt{\frac{2GM}{R}}$

Substitute $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}$ for $G$, $1.98\times {10}^{30}\text{ kg}$ for M and $6.378\times {10}^6\text{ m}$ for R in the expression of escape velocity.

$\begin{array}{c}{v}_e={\left(\frac{2\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^{\text{2}}/{\text{kg}}^{\text{2}}\right)1.98\times {10}^{30}\text{ kg}}{6.378\times {10}^6\text{ m}}\right)}^{\frac{1}{2}}\\ =6.45\times {10}^6\text{m/s}\left(\frac{1\text{ km/s}}{1000\text{ m/s}}\right)\\ =6450\text{km}\text{/s}\end{array}$

Conclusion:

The escape speed for the gas is $6450\text{km}\text{/s}$.