Chapter 20, Problem 45Q

To determine

(a)

Themass of the core.

Answer to Problem 45Q

The mass of the core in kilograms is $1.676\times {10}^{30}\text{kg}$, and in solar mass, the mass is $0.84{\text{M}}_{\odot }$.

Explanation of Solution

Given:

The diameter of the core is, $d=20\text{km}$.

The density of the core is, $\rho =4\times {10}^{17}\text{kg}/{\text{m}}^3$.

Formula used:

The expression for the volume of the core is given by,

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

Here, $r$ is the radius of the core.

The expression for the density of the core is given by,

$\rho =\frac{M}{V}$

Calculation:

The radius of the coreis calculated as,

$\begin{array}{c}r=\frac{d}{2}\\ =\frac{20\text{km}}{2}\\ =10\text{km}\end{array}$

The volume of the core is calculated as,

$\begin{array}{c}V=\frac{4}{3}\pi r^3\\ =\frac{4}{3}\left(\pi \right){\left(10\text{km}\right)}^3\\ =\frac{4}{3}\left(\pi \right){\left(10\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)}^3\\ =4.19\times {10}^{12}{\text{m}}^3\end{array}$

The mass of the core is calculated as,

$\begin{array}{c}M=\rho V\\ =\left(4\times {10}^{17}\text{kg}/{\text{m}}^3\right)\left(4.19\times {10}^{12}{\text{m}}^3\right)\\ =\left(1.676\times {10}^{30}\text{kg}\times \frac{1{\text{M}}_{\odot }}{1.989\times {10}^{30}\text{kg}}\right)\\ =0.84{\text{M}}_{\odot }\end{array}$

Conclusion:

The mass of the core in kilograms is $1.676\times {10}^{30}\text{kg}$; and in solar mass, the mass is $0.84{\text{M}}_{\odot }$.

To determine

(b)

Theforce of gravity on a $1\text{kg}$ object at the surface of the core and to compare it with the force if the object was kept on Earth.

Answer to Problem 45Q

The force of gravity on the surface of the core is $1.12\times {10}^{12}\text{N}$ and is approximately ${10}^{11}$ times the force on the surface of the Earth.

Explanation of Solution

Given:

The radius of the core is, $r=10\text{km}$.

The mass of the object is, $m=1\text{kg}$.

Formula used:

The gravitational force between the two bodies is given by,

$F=G\left(\frac{mM}{r^2}\right)$

Here, $M$ is the mass of the core.

Calculation:

The gravitational force between the two bodies is calculated as,

$\begin{array}{c}F=G\left(\frac{mM}{r^2}\right)\\ =\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left[\frac{\left(1\text{kg}\right)\left(1.676\times {10}^{30}\text{kg}\right)}{{\left(10\text{km}\right)}^2}\right]\\ =\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left[\frac{\left(1.676\times {10}^{30}{\text{kg}}^2\right)}{{\left(10\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)}^2}\right]\\ =1.12\times {10}^{12}\text{N}\end{array}$

The ratio of the gravitational force on the surface of the core of the supernova and on the surface of the Earth is calculated as,

$\begin{array}{c}\frac{F_{\text{c}}}{F_{\text{e}}}=\frac{1.12\times {10}^{12}\text{N}}{10\text{N}}\\ =1.12\times {10}^{11}\text{N}\end{array}$

Conclusion:

The force of gravity on the surface of the core is $1.12\times {10}^{12}\text{N}$ and is approximately ${10}^{11}$ times the force on the surface of the Earth.

To determine

(c)

The escape speed from the surface of the star’s core.

Answer to Problem 45Q

The escape speed from the surface of the star’s core is $1.5\times {10}^8\text{m}/\text{s}$ or $0.5c$. Therefore, the escape speed is very large and has a value equal to half the speed of light. Thus, in order for the particles to escape the surface of the core, they have to attain a speed greater than half the speed of the light. This indicates that the explosion should be very large if the particles have to attain such a velocity.

Explanation of Solution

Given:

The radius of the core is, $r=10\text{km}$.

The mass of the object is, $m=1\text{kg}$.

Formula used:

The escape velocity is given by,

$v_{\text{c}}=\sqrt{\frac{2GM}{r}}$

Calculation:

The escape velocity is calculated as,

$\begin{array}{c}{v}_{\text{c}}=\sqrt{\frac{2GM}{r}}\\ =\sqrt{\frac{2\left(6.67\times {10}^{-11}\text{N}\cdot {\text{m}}^2/{\text{kg}}^2\right)\left(1.676\times {10}^{30}\text{kg}\right)}{\left(10\text{km}\right)}}\\ =\sqrt{\frac{2.24\times {10}^{20}\text{N}\cdot {\text{m}}^2/\text{kg}}{\left(10\text{km}\times \frac{1000\text{m}}{1\text{km}}\right)}}\\ =1.5\times {10}^8\text{m}/\text{s}\end{array}$

Solve further,

$\begin{array}{c}{v}_{\text{c}}=\left(1.5\times {10}^8\text{m}/\text{s}\times \frac{1c}{3\times {10}^8\text{m}/\text{s}}\right)\\ =0.5c\end{array}$

Conclusion:

The escape speed from the surface of the star’s core is $1.5\times {10}^8\text{m}/\text{s}$ or $0.5c$ .Therefore, the escape speed is very large and has a value equal to half the speed of light. Thus, in order for the particles to escape the surface of the core, they have to attain a speed greater than half the speed of the light. This indicates that the explosion should be very large if the particles have to attain such a velocity.