Chapter 14, Problem 6P

To determine

The escape velocity at the surface of Neutron stars of masses $1.4M_0$ and $3.0M_0$.

Answer to Problem 6P

The escape velocity at the surface of Neutron stars of masses $1.4M_0$ and $3.0M_0$ are $1.9\times {10}^8\text{m}/\text{s}$ and $3.2\times {10}^8\text{m}/\text{s}$ respectively.

Explanation of Solution

Necessary data is obtained from problem 2. It is found that from problem 2, radius of Neutron star having mass $3.0M_0$ is $7.8\text{km}$.

Write the equation to find the escape velocity.

    $v_{es}=\sqrt{\frac{2GM}{R}}$        (I)

Here, $v_{es}$ is the escape velocity, $G$ is the gravitational constant, $M$ is the mass of star, and $R$ is the radius of star.

Rewrite equation by substituting $1.4M_0$ for $M$

    $v_{es}=\sqrt{\frac{2G\left(1.4M_0\right)}{R}}$        (II)

Here, $M_0$ is the mass of sun.

Rewrite equation by substituting $3.0M_0$ for $M$

    $v_{es}=\sqrt{\frac{2G\left(3.0M_0\right)}{R}}$        (III)

Conclusion:

Case 1:Escape velocity star with mass $1.4M_0$.

Substitute $6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M_0$, and $10\text{km}$ for $R$ in equation (II) to find $v_c$.

    $\begin{array}{c}{v}_{es}=\sqrt{\frac{2\left(6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(1.4\left(1.99\times {10}^{30}\text{kg}\right)\right)}{10\text{km}\left(\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1}\text{km}}\right)}}\\ =\sqrt{3.71\times {10}^{16}}\text{m}/\text{s}\\ =1.9\times {10}^8\text{m}/\text{s}\end{array}$

Case 1: Escape velocity star with mass $3.0M_0$.

Substitute $6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}$ for $G$, $1.99\times {10}^{30}\text{kg}$ for $M_0$, and $7.8\text{km}$ for $R$ in equation (III) to find $v_{es}$.

    $\begin{array}{c}{v}_{es}=\sqrt{\frac{2\left(6.67\times {10}^{-11}{\text{m}}^{\text{3}}/\text{kg}\cdot {\text{s}}^{\text{2}}\right)\left(3.0\left(1.99\times {10}^{30}\text{kg}\right)\right)}{7.8\text{km}\left(\frac{{\text{10}}^{\text{3}}\text{m}}{\text{1}\text{km}}\right)}}\\ =\sqrt{10.2\times {10}^{16}}\text{m}/\text{s}\\ =3.2\times {10}^8\text{m}/\text{s}\end{array}$

Therefore, the escape velocity at the surface of Neutron stars of masses $1.4M_0$ and $3.0M_0$ are $1.9\times {10}^8\text{m}/\text{s}$ and $3.2\times {10}^8\text{m}/\text{s}$ respectively.