Chapter 19, Problem 46Q

To determine

(a)

Escape speed from the present-day Sun.

Answer to Problem 46Q

The escape speed from the present-day surface of the Sun is approximately $619\text{km}{\text{s}}^{\text{-1}}$.

Explanation of Solution

Given:

$\text{G }\left(\text{Gravitational Constant}\right)=6.673\times {\text{10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}$

$\text{Mass of the Sun}(M_{Sun})=2\times {10}^{30}\text{kg}$

$\text{Radius}\text{of}\text{the}\text{Sun}\text{=}\text{696,000}\times {\text{10}}^3\text{m}$

Let $V_{esc1}$ be the escape speed from the present-day surface of the Sun.

Formula used:

Escape speed is given by the formula

$V_{esc}=\sqrt{\frac{2G{M}_{Sun}}{R}}$

Where, V is the escape velocity, G is the gravitational constant, M is the sun and R is the radius.

Calculation:

$\begin{array}{c}{V}_{esc1}=\sqrt{\frac{\text{2×6}{\text{.673×10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}{\text{×2×10}}^{\text{30}}\text{kg}}{{\text{696,000×10}}^{\text{3}}\text{m}}}\\ =619,278.40\text{m}{\text{s}}^{\text{-1}}\\ \cong 619\text{km}{\text{s}}^{\text{-1}}\end{array}$

Conclusion:

The escape speed from the present-day surface of the Sun is approximately $619\text{km}{\text{s}}^{\text{-1}}$.

To determine

(b)

Escape speed from the Sun after it became a red giant.

Answer to Problem 46Q

The escape speed from the Sun after it became a red giant is approximately $61.9km{s}^{-1}$.

Explanation of Solution

Given:

$\text{G }\left(\text{Gravitational Constant}\right)=6.673\times {\text{10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}$

$\text{Mass of the Sun}(M_{Sun})=2\times {10}^{30}\text{kg}$

$\text{Radius}\text{of}\text{the}\text{Sun}\text{=}\text{696,000}\times {\text{10}}^3\text{m}$

Let $V_{esc2}$ be the escape speed from the surface of the Sun after it became a red giant.

Formula used:

Escape speed is given by the formula,

$V_{esc}=\sqrt{\frac{2G{M}_{Sun}}{R}}$

Calculation:

Substitute the values:

$\begin{array}{c}{V}_{esc2}=\sqrt{\frac{\text{2×6}{\text{.673×10}}^{\text{-11}}{\text{m}}^{\text{3}}{\text{kg}}^{\text{-1}}{\text{s}}^{\text{-2}}{\text{×2×10}}^{\text{30}}\text{kg}}{{\text{696,000×10}}^{\text{5}}\text{m}}}\\ =61,927.84\text{m}{\text{s}}^{-1}\\ \cong 61.9\text{km}{\text{s}}^{\text{-1}}\end{array}$

Conclusion:

The escape speed from the present-day surface of the Sun is approximately $61.9\text{km}{\text{s}}^{\text{-1}}$.

To determine

(c)

The reason for mass of a red giant star loses easily rather than a main-sequence star.

Answer to Problem 46Q

The escape speed from a red giant star is much lower than the escape speed from a main-sequence star.

Explanation of Solution

Introduction:

According to the calculations, the escape speed from the Sun at present-day is approximately $619\text{km}{\text{s}}^{\text{-1}}$ and the escape speed from the Sun after it became a red giant is approximately $61.\text{9}\text{km}{\text{s}}^{\text{-1}}$. Escape speed determines the lowest speed that a mass required in order to permanently escape the gravitational field of a certain planet or a celestial body.

The results imply that after the Sun became a red giant, the speed that a mass requires to escape into the outer space is approximately10 times lower than the speed required to escape the present day Sun, which is still a main-sequence star. Hence, more mass from a red giant star would escape from it to the outer space than a main-sequence star.