Chapter 20, Problem 45Q

(a)

To determine

The mass of the core in kg and solar masses, if the core of diameter is $20\text{ km}$ and its density is $4\times {10}^{17}{\text{ kg/m}}^3$.

Answer to Problem 45Q

Solution:

$1.68\times {10}^{30}\text{ kg}$ and $0.85{\text{ M}}_{\odot }$

Explanation of Solution

Given data:

The core of diameter is $20\text{ km}$ and its density is $4\times {10}^{17}{\text{ kg/m}}^3$.

Formula used:

Write the expression for radius.

$r=\frac{D}{2}$

Here, $D$ is the diameter.

Write the formula for volume of sphere.

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

Here, $r$ is the radius.

Write the expression for density $\left(\rho \right)$.

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

Here, $m$ is the mass and $V$ is the volume.

Explanation:

Recall the expression for radius.

$r=\frac{D}{2}$

Substitute $20\text{ km}$ for $D$.

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

Recall the formula for volume of sphere.

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

Substitute $10\text{ km}$ for $r$.

$\begin{array}{c}V=\frac{4}{3}\pi {\left(10\text{ km}\right)}^3\\ =\frac{4}{3}\pi {\left(10\text{ km}\right)}^3{\left(\frac{1000\text{ m}}{1\text{ km}}\right)}^3\\ =4.2\times {10}^{12}{\text{ m}}^3\end{array}$

Recall the expression for density $\left(\rho \right)$.

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

Substitute $4\times {10}^{17}{\text{ kg/m}}^3$ for $\rho$ and $4.2\times {10}^{12}{\text{ m}}^3$ for $V$.

$\begin{array}{c}4\times {10}^{17}{\text{ kg/m}}^3=\frac{m}{4.2\times {10}^{12}{\text{ m}}^3}\\ m=\left(4\times {10}^{17}{\text{ kg/m}}^3\right)\left(4.2\times {10}^{12}{\text{ m}}^3\right)\\ m=1.68\times {10}^{30}\text{ kg}\end{array}$

Convert mass kg to solar mass.

$\begin{array}{c}m=1.68\times {10}^{30}\text{ kg}\left(\frac{1{\text{ M}}_{\odot }}{1.989\times {10}^{30}\text{ kg}}\right)\\ =0.85{\text{ M}}_{\odot }\end{array}$

Conclusion:

The mass of the core in kg is $1.68\times {10}^{30}\text{ kg}$ and in solar masses is $0.85{\text{ M}}_{\odot }$.

(b)

To determine

The force of gravity on a 1 kg object at the surface of the core, if the core of diameter is $20\text{ km}$ and density is $4\times {10}^{17}{\text{ kg/m}}^3$. Also, comapre the gravitational force on the on such object on the surface of earth and explain about 10 Newtons in comparison to this gravitaional force.

Answer to Problem 45Q

Solution:

$1.12\times {10}^{12}\text{ N}$

Explanation of Solution

Given data:

The core of diameter is $20\text{ km}$ and density is $4\times {10}^{17}{\text{ kg/m}}^3$.

The gravity of the mass is $1\text{ kg}$.

Formula used:

Write the expression for the radius.

$r=\frac{D}{2}$

Newton’s law of gravitation for the forces between two massive body can be stated by the following equation:

$F=G\left(\text{}\frac{m_1{m}_2}{r^2}\right)$

Here, $m_1$ is the mass of the first object, $m_2$ is the mass of the second object, r is the distance between the objects, and G is the universal constant of gravitation having a value of $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$.

Explanation:

Recall the result of part (a):

The mass of the core in kg is $1.68\times {10}^{30}\text{ kg}$.

Recall the expression for radius.

$r=\frac{D}{2}$

Substitute $20\text{ km}$ for $D$.

$\begin{array}{c}r=\frac{20\text{ km}}{2}\\ =10\text{ km}\left(\frac{1000\text{ m}}{1\text{ km}}\right)\\ ={10}^4\text{ m}\end{array}$

Newton’s law of universal gravitation can be stated by an equation as:

$F=G\left(\frac{m_1{m}_2}{r^2}\right)$

Substitute $1.68\times {10}^{30}\text{ kg}$ for $m_1$, $1\text{ kg}$ for $m_2$, ${10}^4\text{ m}$ for $r$, and $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}$ for $G$.

$\begin{array}{c}F=\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^{\text{2}}{\text{/kg}}^{\text{2}}\right)\left(\frac{\left(1.68\times {10}^{30}\text{ kg}\right)\left(1\text{ kg}\right)}{{\left({10}^4\text{ m}\right)}^2}\right)\\ =1.12\times {10}^{12}\text{ N}\end{array}$

The gravitational force due to the gravity of the Earth is provided as $10\text{ N}$. So, the gravitational force due to the core of the high mass star is approximately ${10}^{11}$ times than that of the gravitational force due to gravity of the earth.

Conclusion:

The gravitation force due to the core of the earth is $1.12\times {10}^{12}\text{ N}$. It is ${10}^{11}$ times higher than that of the gravitational force on the earth.

(c)

To determine

The escape speed from the surface of the core of the star, if the diameter of the core of the star is $20\text{ km}$ and its density is $4\times {10}^{17}{\text{ kg/m}}^3$.

Answer to Problem 45Q

Solution:

$1.5\times {10}^8\text{m/s}$, $\text{0}\text{.5}c$, and the explosion must be much higher.

Explanation of Solution

Given data:

The core of diameter is $20\text{ km}$ and density is $4\times {10}^{17}{\text{ kg/m}}^3$.

The gravity of the mass is $1\text{ kg}$.

Formula used:

Escape speed from Earth ($v_e$) is calculated by using the following expression,

$v_e=\sqrt{\frac{2Gm}{r}}$

Here, G is the universal gravitational constant, m is the mass of Earth and r is the radius of Earth.

Explanation:

Recall the result of part (a):

The mass of the core in kg is $1.68\times {10}^{30}\text{ kg}$.

The radius is ${10}^4\text{ m}$.

Escape speed from Earth ($v_e$) is given by the following expression,

$v_e=\sqrt{\frac{2Gm}{r}}$

Substitute $6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2{\text{/kg}}^{\text{2}}$ for G, $1.68\times {10}^{30}\text{ kg}$ for m and ${10}^4\text{ m}$ for r.

$\begin{array}{c}{v}_e=\sqrt{\frac{2\left(6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2{\text{/kg}}^{\text{2}}\right)\left(1.68\times {10}^{30}\text{ kg}\right)}{{10}^4\text{ m}}}\\ =1.5\times {10}^8\text{m/s}\end{array}$

Further solve,

$\begin{array}{c}{v}_e=1.5\times {10}^8\text{m/s}\left(\frac{c}{3\times {10}^8\text{ m/s}}\right)\\ =1.5\times {10}^8\text{m/s}\left(\frac{c}{3\times {10}^8\text{ m/s}}\right)\\ =\text{0}\text{.5}c\end{array}$

The escape velocity is large, as it is half of the speed of the light. So, it can be concluded that the gases must have velocity higher than half of the velocity of the light to escape from the core surface of the star. The explosion must be much higher to attained this velocity.

Conclusion:

The speed required by a gas to escape from the surface of the star’s core in m/s is $1.5\times {10}^8\text{m/s}$ and in terms of the speed of light is $\text{0}\text{.5}c$. The explosion must be significantly powerful, so that the material can blow from the core of supernova.