Chapter 33, Problem 13P

To determine

The density of the white dwarf. Also, the density of the white dwarf that is larger than Earth’s density.

Answer to Problem 13P

The density of the white dwarf is $1.83\times {10}^9{\text{kg/m}}^3$ .

The density of the white dwarf is about $3.33\times {10}^5$ times larger than Earth’s density.

Explanation of Solution

Given:

Assume the mass of the Earth is $M_{\text{E}}=5.98\times {10}^{24}\text{kg}$

Assume the radius of the Earth is $R_{\text{E}}=6.38\times {10}^6\text{m}$ .

Assume the mass of the white dwarf is equal to the mass of the Sun and is equal to $M_{\text{wd}}=1.99\times {10}^{30}\text{kg}$ .

Assume the radius of the white dwarf is equal to the radius of the Earth and is equal to $R_{\text{wd}}=6.38\times {10}^6\text{m}$ .

Formula used:

Assume the planet is spherical in shape.

The expression for the volume ( V ) of the Planet (spherical shape) is,

  $V=\frac{4}{3}\pi R^3$

Here, R is the radius of the sphere.

The expression for the density ( D ) of the planet is,

  $D=\frac{M}{V}$

Here, M is the mass of the planet.

Calculation:

The volume of the white dwarf is,

  $\begin{array}{l}{V}_{\text{wd}}=\frac{4}{3}\pi R_{wd}^3\\ V_{\text{wd}}=\frac{4}{3}\pi {\left(6.38\times {10}^6\right)}^3\\ V_{\text{wd}}=1.087\times {10}^{21}{\text{m}}^3\end{array}$

The density of the white dwarf is,

  $\begin{array}{c}{D}_{\text{wd}}=\frac{M_{\text{wd}}}{V_{\text{wd}}}\\ =\left(\frac{1.99\times {10}^{30}\text{kg}}{1.087\times {10}^{21}{\text{m}}^3}\right)\\ =1.83\times {10}^9{\text{kg/m}}^3\end{array}$ …… (1)

The volume of the Earth is,

  $\begin{array}{l}{V}_{\text{E}}=\frac{4}{3}\pi R^3\\ V_{\text{E}}=\frac{4}{3}\pi {\left(6.38\times {10}^6\right)}^3\\ V_{\text{E}}=1.087\times {10}^{21}{\text{m}}^3\end{array}$

The density of the Earth is,

  $\begin{array}{c}{D}_{\text{E}}=\frac{M_{\text{E}}}{V_{\text{E}}}\\ =\left(\frac{5.98\times {10}^{24}\text{kg}}{1.087\times {10}^{21}{\text{m}}^3}\right)\\ =5.50\times {10}^3{\text{kg/m}}^3\end{array}$ …… (2)

Divide equation (1) by equation (2).

  $\begin{array}{l}\frac{D_{\text{wd}}}{D_{\text{E}}}=\frac{1.83\times {10}^9{\text{kg/m}}^3}{5.50\times {10}^3{\text{kg/m}}^3}\\ \frac{D_{\text{wd}}}{D_{\text{E}}}=\frac{1.83\times {10}^9{\text{kg/m}}^3}{5.50\times {10}^3{\text{kg/m}}^3}\\ D_{\text{wd}}=3.33\times {10}^5{D}_{\text{E}}\end{array}$

Conclusion:

Thus, the density of the white dwarf is $1.83\times {10}^9{\text{kg/m}}^3$ .

Thus, the density of the white dwarf is about $3.33\times {10}^5$ times larger than Earth’s density.