Chapter 12, Problem 40QAP

(a)

To determine

The escape velocity when the Sun becomes a red giant.

Answer to Problem 40QAP

The escape velocity when the Sun becomes a red giant is $83\text{km}/\text{s}$ .

Explanation of Solution

Write the expression for escape velocity.

$v_{\text{esc}}=\sqrt{\frac{2G{M}_{\text{Sun}}}{R_{\text{Sun}}}}$

Here, $v_{\text{esc}}$ is the escape velocity on the Sun, $M_{\text{Sun}}$ is the mass of the Sun, $R_{\text{Sun}}$ is the radius of the Sun and $G$ is the gravitational constant.

Compare the velocities for different masses and radius of the Sun and simplify.

$\frac{{\left(v_{\text{esc}}\right)}_2}{{\left(v_{\text{esc}}\right)}_1}=\sqrt{\frac{{\left(M_{\text{Sun}}\right)}_2{\left(R_{\text{Sun}}\right)}_1}{{\left(R_{\text{Sun}}\right)}_2{\left(M_{\text{Sun}}\right)}_1}}$

Here, subscript 1 is used for mass, radius and velocity of initial phase and subscript 2 is used for mass, radius and velocity of final phase.

Rearrange the above expression in term of ${\left(v_{\text{esc}}\right)}_2$ .

${\left(v_{\text{esc}}\right)}_2={\left(v_{\text{esc}}\right)}_1\sqrt{\frac{{\left(M_{\text{Sun}}\right)}_2{\left(R_{\text{Sun}}\right)}_1}{{\left(R_{\text{Sun}}\right)}_2{\left(M_{\text{Sun}}\right)}_1}}$        (I)

Conclusion:

Substitute $618\text{km}/\text{s}$ for ${\left(v_{\text{esc}}\right)}_1$, $0.9{\left(M_{\text{Sun}}\right)}_1$ for ${\left(M_{\text{Sun}}\right)}_2$ and $50{\left(R_{\text{Sun}}\right)}_1$ for ${\left(R_{\text{Sun}}\right)}_2$ in equation (I).

$\begin{array}{c}{\left(v_{\text{esc}}\right)}_2=\left(618\text{km}/\text{s}\right)\sqrt{\frac{0.9{\left(M_{\text{Sun}}\right)}_1{\left(R_{\text{Sun}}\right)}_1}{50{\left(R_{\text{Sun}}\right)}_1{\left(M_{\text{Sun}}\right)}_1}}\\ =83\text{km}/\text{s}\end{array}$

Thus, the escape velocity when the Sun becomes a red giant is $83\text{km}/\text{s}$ .

(b)

To determine

The escape velocity when the Sun becomes AGB star.

Answer to Problem 40QAP

The escape velocity when the Sun becomes AGB star is $36.6\text{km}/\text{s}$ .

Explanation of Solution

Conclusion:

Substitute $618\text{km}/\text{s}$ for ${\left(v_{\text{esc}}\right)}_1$, $0.7{\left(M_{\text{Sun}}\right)}_1$ for ${\left(M_{\text{Sun}}\right)}_2$ and $200{\left(R_{\text{Sun}}\right)}_1$ for ${\left(R_{\text{Sun}}\right)}_2$ in equation (I).

$\begin{array}{c}{\left(v_{\text{esc}}\right)}_2=\left(618\text{km}/\text{s}\right)\sqrt{\frac{0.7{\left(M_{\text{Sun}}\right)}_1{\left(R_{\text{Sun}}\right)}_1}{200{\left(R_{\text{Sun}}\right)}_1{\left(M_{\text{Sun}}\right)}_1}}\\ =36.6\text{km}/\text{s}\end{array}$

Thus, the escape velocity when the Sun becomes a red giant is $36.6\text{km}/\text{s}$ .

(c)

To determine

The effect on mass loss with the changes in escape velocity.

Answer to Problem 40QAP

Mass loss increases with the decrease in escape velocity and vice versa.

Explanation of Solution

The star loses it mass when it leaves the main-sequence phase. The mass loss depends on the nuclear burning and gravity of the star. The escape velocity is the velocity of an object with which the object can escape the atmosphere of the star.

The escape velocity depends upon the mass and radius of the Sun; it is directly proportional to the mass of Sun and inversely proportional to the radius of Sun. During evolution of star, its mass reduces and radius increases for red giant and AGB star.

The escape velocity reduces with the increase in size and decrease in mass; hence object can easily escape the surface of star. Therefore, mass loss from the surface of the Sun increases.

Conclusion:

Thus, the mass loss increases with the decrease in escape velocity and vice versa.