Chapter 5.3, Problem 15P

To determine

To Prove: That the average density of Earth is about four times that of the sun.

Explanation of Solution

Given information:

Mass of Sun $=2.0\times {10}^{30}$ kg

Mass of Earth $=6.0\times {10}^{24}$ kg

Radii of Sun $=7.0\times {10}^8$

Radii of Earth $=6.4\times {10}^6$

Shape of Earth and Sun is Sphere.

Therefore,

Volume $=V=\frac{4}{3}\pi r^3$

Radii of Sun $=r_s=7.0\times {10}^8$

Radii of Earth $=r_e=6.4\times {10}^6$

On substituting values

We get

  $\begin{array}{l}{V}_s=\frac{4}{3}\pi r_s{}^3\\ \Rightarrow V_s=\frac{4}{3}\pi {\left(7\times {10}^8\right)}^3\\ \Rightarrow V_s=\frac{4}{3}\pi \left(343\times {10}^{8\times 3}\right)\\ \Rightarrow V_s=1436.76\times {10}^{24}\\ \Rightarrow V_s=1436.76\times {10}^{24}\times \frac{1000}{1000}\\ \Rightarrow V_s=\frac{1436.76}{1000}\times {10}^{24}\times 1000\\ \Rightarrow V_s=1.437\times {10}^{24}\times {10}^3\\ \Rightarrow V_s=1.437\times {10}^{24+3}\\ \Rightarrow V_s=1.437\times {10}^{27}\end{array}$

Therefore,

Volume of Sun $=V_s=1.437\times {10}^{27}$

Mass of Sun $=m_s=2.0\times {10}^{30}$ kg

Average Density of Sun $=\frac{m_s}{V_s}$

  $\Rightarrow$ Average Density of Sun $=\frac{2.0\times {10}^{30}}{1.437\times {10}^{27}}$

  $\Rightarrow$ Average Density of Sun $=1.4\times {10}^{30-27}$

  $\Rightarrow$ Average Density of Sun $=1.4\times {10}^3$

Radii of Earth $=r_e=6.4\times {10}^6$

On substituting values

We get

  $\begin{array}{l}{V}_e=\frac{4}{3}\pi r_e{}^3\\ \Rightarrow V_e=\frac{4}{3}\pi {\left(6.4\times {10}^6\right)}^3\\ \Rightarrow V_e=\frac{4}{3}\pi {\left(6.4\right)}^3\times {10}^{6\times 3}\\ \Rightarrow V_e=1098.07\times {10}^{18}\\ \Rightarrow V_e=1098.07\times {10}^{18}\times \frac{1000}{1000}\\ \Rightarrow V_e=\frac{1098.07}{1000}\times {10}^{18}\times 1000\\ \Rightarrow V_e=1.098\times {10}^{18}\times {10}^3\\ \Rightarrow V_e=1.1\times {10}^{18+3}\\ \Rightarrow V_e=1.1\times {10}^{21}\end{array}$

Therefore,

Volume of Earth $=V_e=1.1\times {10}^{21}$

Mass of Earth $=m_e=6.0\times {10}^{24}$ kg

Average Density of Earth $=\frac{m_e}{V_e}$

  $\Rightarrow$ Average Density of Earth $=\frac{6.0\times {10}^{24}}{1.1\times {10}^{21}}$

  $\Rightarrow$ Average Density of Earth $=5.5\times {10}^{24-21}$

  $\Rightarrow$ Average Density of Earth $=5.5\times {10}^3$

  $\Rightarrow$ Average Density of Earth $\approx 4\times 1.4\times {10}^3$

  $\Rightarrow$ Average Density of Earth $\approx 4\times$ Average Density of Sun.

Therefore,

Average Density of Earth $=$ Average Density of Sun.

Hence Proved.