Chapter 33, Problem 49GP

To determine

Size of the Sun if its density equals the critical density of the universe.

To compare: The obtained result with the Earth-Sun distance and the size of the Milky-way Galaxy.

Answer to Problem 49GP

The size of the Sun is $400\text{ ly}$

The new size of the Sun is $2\times {10}^7$ times the distance between the Earth-Sun and $8\times {10}^{-3}$ times the size of the Milky-way Galaxy.

Explanation of Solution

Given:

The critical density of the Universe is ${\rho }_{\text{c}}\approx {10}^{-26}{\text{ kg/m}}^3$

Formula used:

Density is calculated as $\rho =\frac{M}{V}$

Calculation:

The density of the Sun is

  $\begin{array}{c}\rho =\frac{M}{\frac{4}{3}\pi r_{Sun}^3}\\ r_{Sun}={\left(\frac{3M}{4\pi \rho }\right)}^{1/3}\\ r_{Sun}={\left(\frac{3(1.99\times {10}^{30}\text{ kg})}{4\pi ({10}^{-26}{\text{ kg/m}}^3)}\right)}^{1/3}\\ =3.62\times {10}^{18}\text{ m}\left(\frac{1\text{ ly}}{9.46\times {10}^{15}\text{ m}}\right)\\ =382\text{ ly}\\ \approx \text{400 ly}\end{array}$

The new size of the Sun will vary as

  $\begin{array}{l}\frac{r_{Sun}}{d_{Earth-Sun}}=\frac{3.62\times {10}^{18}\text{ m}}{1.50\times {10}^{11}\text{ m}}\approx 2\times {10}^7\\ \frac{r_{Sun}}{r_{Galaxy}}=\frac{382\text{ ly}}{50000\text{ ly}}\approx 8\times {10}^{-3}\end{array}$

Conclusion:

Thus, the size of the Sun is $400\text{ ly}$

The new size of the Sun is $2\times {10}^7$ times the distance between the Earth-Sun and $8\times {10}^{-3}$ times the size of the Milky-way Galaxy.