Chapter 30, Problem 60GP

(a)

To determine

To Find: Mass number of the neutron star.

Answer to Problem 60GP

Mass number of the neutron star is $\mathrm{A}=7.23\times {10}^{55}$ .

Explanation of Solution

Given:

Diameter of the neutron star, $\mathrm{d}=10\text{ km}=10\times {10}^3\text{m}$

Formula used:

Radius of a nucleus is given by:

  $\mathrm{r}={\mathrm{r}}_0{\mathrm{A}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}$

Here,

  ${\mathrm{r}}_0=1.2\text{ fm}=1.2\times {10}^{-15}\text{m}$

  $\mathrm{A}$ is the mass number

Calculation:

Radius of the neutron star is given by:

  $\mathrm{r}=\frac{\mathrm{d}}{2}=5\times {10}^3\text{m}$

According to the statement, neutron star consists of neutrons at approximately nuclear density. So, the mass number can be calculated as follows:

  $\begin{array}{l}\mathrm{r}=\left(1.2\times {10}^{-15}\text{m}\right){\mathrm{A}}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\\ \mathrm{A}={\left(\frac{\mathrm{r}}{1.2\times {10}^{-15}\text{m}}\right)}^3\\ \mathrm{A}={\left(\frac{5\times {10}^3\text{m}}{1.2\times {10}^{-15}\text{m}}\right)}^3\\ \mathrm{A}=7.23\times {10}^{55}\end{array}$

So, the mass number of the neutron star is $\mathrm{A}=7.23\times {10}^{55}$ .

Conclusion:

Mass number of the neutron star is $\mathrm{A}=7.23\times {10}^{55}$

(b)

To determine

To Find: Mass of the neutron star.

Answer to Problem 60GP

The mass of the neutron star is $\mathrm{m}=1.2\times {10}^{29}\text{kg}$ .

Explanation of Solution

Given:

Diameter of the neutron star, $\mathrm{d}=10\text{ km}=10\times {10}^3\text{m}$

Formula used:

Mass can be given as: $\mathrm{m}=\mathrm{A}\mathrm{u}$

Here,

  $\mathrm{A}$ is the mass number.

  $\mathrm{u}$ is the atomic mass unit.

Calculation:

Mass of the neutron star is given by:

  $\begin{array}{l}\mathrm{m}=\mathrm{A}.\mathrm{u}\\ \mathrm{m}=\left(7.23\times {10}^{55}\right)\left(1.67\times {10}^{-27}\text{kg}\right)\\ \mathrm{m}=1.2\times {10}^{29}\text{kg}\end{array}$

Conclusion:

Hence, the mass of the neutron star is $\mathrm{m}=1.2\times {10}^{29}\text{kg}$ .

(c)

To determine

To Find: The acceleration of gravity at the surface of neutron star.

Answer to Problem 60GP

  $\mathrm{g}=3.2\times {10}^{11}{\text{m/s}}^2$

Explanation of Solution

Given:

Diameter of the neutron star, $\mathrm{d}=10\text{ km}=10\times {10}^3\text{m}$

Formula used:

Acceleration due to gravity is related to universal gravitational constant as follows:

  $\mathrm{g}=\frac{\mathrm{G}\mathrm{m}}{{\mathrm{r}}^2}$

Here,

  $\mathrm{g}$ is the acceleration of gravity

  $\mathrm{G}$ is universal gravitational constant equals to $6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2$

  $\mathrm{m}$ is the mass

  $\mathrm{r}$ is the radius

Calculation:

Acceleration of gravity at the surface of the neutron star can be calculated as follows:

  $\begin{array}{l}\mathrm{g}=\frac{\mathrm{G}\mathrm{m}}{{\mathrm{r}}^2}\\ \mathrm{g}=\frac{\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(1.2\times {10}^{29}\text{kg}\right)}{{\left(5\times {10}^3\text{m}\right)}^2}\\ \mathrm{g}=3.2\times {10}^{11}{\text{m/s}}^2\end{array}$

Conclusion:

The acceleration of gravity is $\mathrm{g}=3.2\times {10}^{11}{\text{m/s}}^2$ .