Chapter 20, Problem 30Q

To determine

(a)

The average density of the $1-M_{\odot }$ white dwarf having the same diameter as Earth.

Answer to Problem 30Q

The density of the $1-M_{\odot }$ white dwarf having the same diameter as Earth is $1.84\times {10}^9\text{kg}{\text{m}}^{\text{-3}}$.

Explanation of Solution

Given:

$\text{The diameter of the}1-M_{\odot }\text{white dwarf}=\text{Earth's diameter}=12,756\text{km}$

$\text{Mass of the white dwarf}=2\times {10}^{30}\text{kg}$

Formula used:

Density of a material can be determined by the formula,

$\text{Density=}\frac{\text{mass}}{\text{volume}}$

Volume of a sphere is given by the formula,

$\text{Volume of a sphere}=\frac{4}{3}\pi {\left(\frac{D}{2}\right)}^3$

Where; $D=\text{diameter of the white dwarf in meters}$

Calculation:

$\begin{array}{c}\text{Density}=\frac{{\text{2×10}}^{\text{30}}\text{kg}}{\frac{\text{4}}{\text{3}}\text{×3}\text{.14×}{\left(\frac{\text{12,756}\text{km}}{\text{2}}\right)}^{\text{3}}}\\ =1.84\times {10}^9\text{kg}{\text{m}}^{\text{-3}}\end{array}$

Conclusion:

The density of the white dwarf having the same radius as Earth is $1.84\times {10}^9\text{kg}{\text{m}}^{\text{-3}}$.

To determine

(b)

To check: Whether the density of one teaspoon full of $1-M_{\odot }$ white dwarf matter is same as the density of an elephant.

Answer to Problem 30Q

The mass of one teaspoon of the white dwarf matter is about $9,000kg$; nearly as the mass of an African male elephant. Therefore, the white dwarf has the density of about one elephant per teaspoon.

Explanation of Solution

Given:

$\text{Volume}\text{of a tea spoon}=4.\text{92}\text{ml=4}\text{.92}{\text{cm}}^{\text{3}}$

$\text{density of the white dwarf}\text{=}1.84\times {10}^9\text{kg}{\text{m}}^{\text{-3}}$

$\text{Mass of an average elephant}\text{(African male)}\cong 7,\text{000}\text{kg}$

Formula used:

Mass of a material can be determined by the formula,

$\text{mass}\text{=}\text{Density}\times \text{volume}$

Calculation:

$\begin{array}{c}\text{mass}=\text{1}{\text{.84×10}}^{\text{9}}\text{kg}{\text{m}}^{\text{-3}}\text{×4}\text{.92}{\text{×10}}^{\text{-6}}{\text{m}}^{\text{3}}\\ \text{=9,052}\text{.8}\text{kg}\end{array}$

Conclusion:

The mass of one teaspoon of the white dwarf matter is about $9,000\text{kg}$; nearly as the mass of an African male elephant. Therefore, the white dwarf has the density of about one elephant per teaspoon.

To determine

(c)

Speed required for a gas to escape the surface of the white dwarf.

Answer to Problem 30Q

The speed required to eject a gas from the surface of the white dwarf is $6,469km{s}^{-1}$.

Explanation of Solution

Given data:

$\text{G }\left(\text{gravitational constant}\right)=\text{6}{\text{.673×10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}$

$\text{Mass of the white dwarf}(M_{wd})={\text{2×10}}^{\text{30}}\text{kg}$

$\text{Radius}\text{of}\text{the}\text{white dwarf}\text{=}\text{6,378}\times {\text{10}}^{\text{3}}\text{m}$

Let $V_{esc}$ be the escape speed from the white dwarf,

Formula used:

Escape speed is given by the formula,

$V_{esc}=\sqrt{\frac{2G{M}_{wd}}{R_{wd}}}$

Where; $\begin{array}{c}{V}_{esc}=\text{escape velocity}\\ G=\text{gravitational constant}\\ M_{wd}=\text{mass of the white dwarf}\\ R_{wd}=\text{radius of the white dwarf}\end{array}$

Calculation:

$\begin{array}{c}{V}_{esc1}=\sqrt{\frac{\text{2×6}{\text{.673×10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}{\text{×2×10}}^{\text{30}}\text{kg}}{\text{6,378,000}\text{m}}}\\ =6,469,16\text{6}\text{.08}\text{m}{\text{s}}^{\text{-1}}\\ \cong 6,469\text{km}{\text{s}}^{\text{-1}}\end{array}$

Conclusion:

The speed required to eject a gas from the surface of the white dwarf is $6,469\text{km}{\text{s}}^{\text{-1}}$.