Chapter 14, Problem 3P

To determine

The density of white dwarfs of masses $1.0M_0$ and $1.4M_0$.

Answer to Problem 3P

The density of white dwarfs of masses $1.0M_0$ and $1.4M_0$ are $1.39\times {10}^9\text{kg}/{\text{m}}^{\text{3}}$ and $1.95\times {10}^9\text{kg}/{\text{m}}^{\text{3}}$ respectively.

Explanation of Solution

Necessary data is obtained from problem 1.

Write the empirical relation between mass and radius of star.

    $\frac{R_1}{R_2}={\left(\frac{M_2}{M_1}\right)}^{\frac{1}{3}}$        (I)

Here, $R_1,R_2$ are the radii of stars and $M_2,M_1$ are the masses.

Write the relation between mass and density.

    $\rho =\frac{m}{V}$

Here, $\rho$ is the density, $m$ is the mass, and $V$ is the volume of H1 cloud.

Write the equation to find $V$.

    $V=\frac{4}{3}\pi R^3$

Here, $R$ is the radius.

Rewrite the equation for $\rho$ by substituting the above relation for $V$.

    $\begin{array}{c}\rho =\frac{m}{\frac{4}{3}\pi R^3}\\ =\frac{3m}{4\pi R^3}\end{array}$        (II)

Rewrite the above equation by substituting $1.0M_0$ for $m$ and replace $R$ by $R_1$.

    $\rho =\frac{3\left(1.0M_0\right)}{4\pi R_1^3}$        (III)

Here, $M_0$ is the solar mass.

Rewrite equation (I) by substituting $1.4M_0$ for $m$ and replace $R$ by $R_2$.

    $\rho =\frac{3\left(1.4M_0\right)}{4\pi R_2^3}$        (IV)

Conclusion:

Case 1: Radius corresponds to $1.0M_0$.

Substitute $7000\text{km}$ for $R_1$, $0.75M_0$ for $M_1$ and $1.0M_0$ for $M_2$ to find $R_2$.

    $\begin{array}{c}\frac{7000\text{km}}{R_2}={\left(\frac{1.0M_0}{0.75M_0}\right)}^{\frac{1}{3}}\\ =1.10\end{array}$

Rewrite the expression for $R_2$.

    $\begin{array}{c}{R}_2=\frac{\text{7000}\text{km}}{\text{1}\text{.10}}\\ =6363\text{km}\end{array}$

Substitute $2\times {10}^{30}\text{kg}$ for $M_0$ and $6363\text{km}$ for $R_1$ in equation (III) to find $\rho$.

    $\begin{array}{c}\rho =\frac{3\left(1.0\left(2\times {10}^{30}\text{kg}\right)\right)}{4\pi {\left(6363\text{km}\left(\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1}\text{km}}\right)\right)}^3}\\ =\frac{6\times {10}^{30}\text{kg}}{3.23\times {10}^{21}{\text{m}}^{\text{3}}}\\ =1.85\times {10}^9\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Case 2: Radius corresponds to $1.4M_0$.

Substitute $7000\text{km}$ for $R_1$, $0.75M_0$ for $M_1$ and $1.4M_0$ for $M_2$ to find $R_2$.

    $\begin{array}{c}\frac{7000\text{km}}{R_2}={\left(\frac{1.4M_0}{0.75M_0}\right)}^{\frac{1}{3}}\\ =1.23\end{array}$

Rewrite the expression for $R_2$.

    $\begin{array}{c}{R}_2=\frac{\text{7000}\text{km}}{\text{1}\text{.23}}\\ =5691\text{km}\end{array}$

Substitute $2\times {10}^{30}\text{kg}$ for $M_0$ and $5691\text{km}$ for $R_2$ in equation (IV) to find $\rho$.

    $\begin{array}{c}\rho =\frac{3\left(1.4\left(2\times {10}^{30}\text{kg}\right)\right)}{4\pi {\left(5691\text{km}\left(\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1}\text{km}}\right)\right)}^3}\\ =\frac{8.4\times {10}^{30}\text{kg}}{2.31\times {10}^{21}{\text{m}}^{\text{3}}}\\ =3.6\times {10}^9\text{kg}/{\text{m}}^{\text{3}}\end{array}$

Therefore, the density of white dwarfs of masses $1.0M_0$ and $1.4M_0$ are $1.85\times {10}^9\text{kg}/{\text{m}}^{\text{3}}$ and $3.6\times {10}^9\text{kg}/{\text{m}}^{\text{3}}$ respectively.