Chapter 19, Problem 46Q

(a)

To determine

The escape speed from the surface of the present-day Sun.

Answer to Problem 46Q

Solution: $6.18\times {10}^5\text{m/s}\text{.}$

Explanation of Solution

Given data:

For present-day Sun, $M=1.99\times {10}^{30}\text{kg},R=696,000\text{ km}=6.96\times {10}^8\text{m}.$

Formula used:

Write the expression for the escape speed.

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Here,

M is the mass of the Sun

R is the radius of the Sun

G is a constant having value $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Explanation:

Recall the equation for the escape speed.

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Substitute $1.99\times {10}^{30}\text{kg for }M\text{, 6}\text{.96}\times {\text{10}}^8\text{m for }R\text{ and }6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{ for G}\text{.}$

$\begin{array}{c}{v}_{esc}=\sqrt{\frac{2\times 6.67\times {10}^{-11}\text{ N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 1.99\times {10}^{30}\text{ kg}}{6.96\times {10}^8\text{m}}}\\ =6.29\times {10}^5\text{ m/s}\text{.}\end{array}$

Conclusion:

Therefore, the escape speed for the present-day Sun is $6.29\times {10}^5\text{ m/s}\text{.}$

(b)

To determine

The escape speed from the surface of the Sun when it becomes a red giant with the mass as of the present-day but having a radius 100 times larger than of present-day.

Answer to Problem 46Q

Solution: $6.18\times {10}^4\text{m/s}\text{.}$

Explanation of Solution

Given data:

For a red giant Sun, $M=\text{1}{\text{.99×10}}^{\text{30}}\text{kg, }R=69,600,000\text{ km}=6.96\times {10}^{10}\text{m}\text{.}$

Formula used:

Write the expression for the escape speed.

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Here,

M is the mass of the red giant Sun

R is the radius of the red giant Sun

G is a constant having value $6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}$

Explanation:

Recall the equation for escape speed.

$v_{esc}=\sqrt{\frac{2GM}{R}}$

Substitute $1.99\times {10}^{30}\text{kg for }M\text{, 6}\text{.96}\times {\text{10}}^8\text{m for }R\text{ and }6.67\times {10}^{-11}{\text{Nm}}^{\text{2}}{\text{/kg}}^{\text{2}}\text{ for G}\text{.}$

$\begin{array}{c}{v}_{esc}=\sqrt{\frac{2\times 6.67\times {10}^{-11}\text{ N}{\text{.m}}^{\text{2}}{\text{/kg}}^{\text{2}}\times 1.99\times {10}^{30}\text{ kg}}{6.96\times {10}^{10}\text{m}}}\\ =6.29\times {10}^4\text{ m/s}\text{.}\end{array}$

Conclusion:

Therefore, the escape speed when the Sun becomes a red giant with a radius 100 times that of the present-day Sun is $6.29\times {10}^4\text{ m/s}\text{.}$

(c)

To determine

The reason that a red giant can lose mass more easily than a main-sequence star based on the results obtained in (a) and (b).

Answer to Problem 46Q

Solution: In the case of the red giant, the matter can escape more easily because, in the red giant, the matter has to travel at one-tenth of the speed required to escape the present-day Sun. Hence, the red giant will lose mass more easily.

Explanation of Solution

Introduction:

The escape velocity is the minimum speed required to escape the gravitational attraction of a planet or a star.

Explanation:

The escape speed for present-day Sun is $6.29\times {10}^5\text{ m/s}$ and that of the red giant phase of the Sun is $6.29\times {10}^4\text{ m/s}\text{.}$ As compared to the present-day, the escape velocity for the red giant phase is one-tenth. This indicates that its easy for the matter present in the red giant to escape from the surface of the Sun. Hence, the red giant will lose mass more easily compared to the present-day Sun.

Conclusion:

A red giant will lose mass more easily as compared to the present-day Sun.