Q49P - University Physics with Modern Physics

Question

Recall that density is mass divided by volume, and consult Appendix B as needed. (a) Calculate the average density of the earth in $g/c{m}^3$, assuming our planet is a perfect sphere. (b) In about 5 billion years, at the end of its lifetime, our sun will end up as a white dwarf that has about the same mass as it does now but is reduced to about 15,000 km in diameter. What will be its density at that stage? (c) A neutron star is the remnant of certain supernovae (explosions of giant stars). Typically, neutron stars are about 20 km in diameter and have about the same mass as our sun. What is a typical neutron star density in $g/c{m}^3$?

Step-by-Step Solution

Answer

1. Earth's average density is $5.52 \mathrm{g}/{\mathrm{cm}}^3$.
2.  The white dwarf's density will be $1.12\times {10}^6 \mathrm{g}/{\mathrm{cm}}^3$.
3. The neutron star's density is $4.74\times {10}^{14} \mathrm{g}/{\mathrm{cm}}^3$.

Concept of mass, volume, and density

All the physical quantities are defined for matter, in general, have mass, volume, and density. These three are an object's most fundamental properties. Density is calculated by dividing mass by volume.

Identification of given data

The given data and data from Appendix B can be listed below,

The earth's mass is, $M_E=5.97\times {10}^{24} \mathrm{kg}$.

The earth's radius is, $r_E=6.37\times {10}^6 \mathrm{m}$.

 The radius of the sun is,

$r_s=7500 \mathrm{km}\left(\frac{1000 \mathrm{m}}{1 \mathrm{km}}\right).$

$=7.5\times {10}^6 \mathrm{m}$

The white dwarf's radius is,

$$\begin{gathered} r_W=\frac{15000 \mathrm{km}}{2}\left(\frac{1000 \mathrm{m}}{1 \mathrm{km}}\right). \\ =7.5\times {10}^6 \mathrm{m} \end{gathered}$$

The white dwarf's mass is,

$$\begin{gathered} M_W=M_S \\ =1.99\times {10}^{30} \mathrm{kg}. \end{gathered}$$

The neutron star's radius is,

$$\begin{gathered} r_N=\frac{20 \mathrm{km}}{2}\left(\frac{1000 \mathrm{m}}{1 \mathrm{km}}\right). \\ =1.0\times {10}^4 \mathrm{m} \end{gathered}$$

The neutron star's mass is

$$\begin{gathered} M_N=M_S \\ =1.99\times {10}^{30} \mathrm{kg} \end{gathered}$$

(a) Determination of the average density of the earth

The earth is taken as a sphere, so its volume is given by,

$V_E=\frac{4\pi }{3}{r}_E^3$

Here, $r_E$ is the radius of the earth.

Substitute values in the above expression, we have

$\begin{array}{rcl}{V}_E& =& \frac{4\pi }{3}{\left(6.37\times {10}^6 \mathrm{m}\right)}^3\\ & =& 1.08\times {10}^{21}{ \mathrm{m}}^3\end{array}$

The earth's density can be expressed as,

${\rho }_E=\frac{M_E}{V_E}$

Here, $M_E$ is the mass of the earth.

Substitute the values in the above equation.

$\begin{array}{rcl}{\rho }_E& =& \frac{5.97\times {10}^{24} \mathrm{kg}}{1.08\times {10}^{21}{ \mathrm{m}}^3}\\ & =& \left(5.52\times {10}^3 \mathrm{kg}/{\mathrm{m}}^3\right)\left(\frac{1000 \mathrm{g}}{1 \mathrm{kg}}\right){\left(\frac{1 \mathrm{m}}{100 \mathrm{cm}}\right)}^3\\ & =& 5.52 \mathrm{g}/{\mathrm{cm}}^3\end{array}$

Thus, the earth's average density is $5.52 \mathrm{g}/{\mathrm{cm}}^3$.

(b) Determination of the density of white dwarf

The white dwarf is taken as a sphere, so its volume is given by,

$V_w=\frac{4\pi }{3}{r}_w^3$

Here,$r_w$ is the white dwarf's radius.

Substitute value in the above expression.

$\begin{array}{rcl}{V}_w& =& \frac{4\pi }{3}{\left(7.5\times {10}^6 \mathrm{m}\right)}^3\\ & =& 1.77\times {10}^{21}{ \mathrm{m}}^3\end{array}$

The white dwarf"s density is expressed as,

${\rho }_W=\frac{M_W}{V_W}$

Here, $M_W$ is the white dwarf's mass.

Substitute the values in the above equation.

$\begin{array}{rcl}{\rho }_W& =& \frac{1.99\times {10}^{30} \mathrm{kg}}{1.77\times {10}^{21}{ \mathrm{m}}^3}\\ & =& 1.12\times {10}^9 \mathrm{kg}/{\mathrm{m}}^3\left(\frac{1 \mathrm{g}/{\mathrm{cm}}^3}{1000 \mathrm{kg}/{\mathrm{m}}^3}\right)\\ & =& 1.12\times {10}^6 \mathrm{g}/{\mathrm{cm}}^3\end{array}$

Thus, the white dwarf's density is $1.12\times {10}^6 \mathrm{g}/{\mathrm{cm}}^3$.

(c) Determination of the density of neutron star

The volume of the neutron star, which is taken as a sphere is given by,

$V_N=\frac{4\pi }{3}{r}_N^3$

Here, $r_N$ is the neutron star's radius.

Substitute value in the above,

$$\begin{gathered} V_N=\frac{4\pi }{3}{\left(1.0\times {10}^4 \mathrm{m}\right)}^3 \\ =4.19\times {10}^{12}{ \mathrm{m}}^3 \end{gathered}$$

The white dwarf's density is expressed as,

${\rho }_N=\frac{M_N}{V_N}$

Here, $M_N$ is the mass of the neutron star.

Substitute the values in the above equation.

$\begin{array}{rcl}{\rho }_N& =& \frac{1.99\times {10}^{30} \mathrm{kg}}{4.19\times {10}^{12}{ \mathrm{m}}^3}\\ & =& 4.74\times {10}^{17} \mathrm{kg}/{\mathrm{m}}^3\left(\frac{1 \mathrm{g}/{\mathrm{cm}}^3}{1000 \mathrm{kg}/{\mathrm{m}}^3}\right)\\ & =& 4.74\times {10}^{14} \mathrm{g}/{\mathrm{cm}}^3\end{array}$

Thus, the neutron star's density is $4.74\times {10}^{14} \mathrm{g}/{\mathrm{cm}}^3$.

Final Solution

The average density of the earth, the white dwarf and the neutron star is found to be $5.52 \mathrm{g}/{\mathrm{cm}}^3, 1.12 \times {10}^{6 }\mathrm{g}/{\mathrm{cm}}^3$ and $4.74 \times {10}^{14 }\mathrm{g}/{\mathrm{cm}}^3$respectively.