Chapter 14, Problem 16A

a. How fast would a particle have to be ejected from the sun to leave the solar system?

b. What speed would be needed if an ejected particle started at a distance from the sun equal to Earth’s distance from the sun?

To determine

The speed of a particle for the given condition.

Answer to Problem 16A

  $620\text{ km/s}$

Explanation of Solution

Given:

A particle is ejected from the sun to leave the solar system.

Formula used:

According to the energy conservation principle, the total energy of an isolated system remains constant, neither it can be created nor can be destroyed.

The total energy is equal to the sum of potential energy ${\text{U}}_{\text{g}}=\text{mgh}$ and kinetic energy $\text{K}=\frac{1}{2}{\text{mv}}^2$ of an object.

Calculation:

The escape velocity $v_e$ is the minimum speed attained by an object, to exert an equal amount of force to balance the effects of force of gravity and get away from its hold.

Using the energy conservation principle, total energy initial and finally must be constant. Final velocity after escaping the gravity is zero since all the thrust is applied to work against gravity. And, the gravitational potential energy after escaping the star’s gravity is zero at infinite distance ${\text{U}}_{\text{gf}}=0$ .

Thus,

  ${(\text{K}+{\text{U}}_{\text{g}})}_{\text{i}}={(\text{K}+{\text{U}}_{\text{g}})}_{\text{f}}$

  $\frac{1}{2}{\text{mv}}_e{}^2+\frac{-\text{GMm}}{{\text{r}}^2}=0+0$

  $v_e=\sqrt{\frac{2\text{GM}}{\text{r}}}$

Escape velocity after substituting the values of constant of gravitation, mass of Sun, and radius of Sun is

  $v_e=\sqrt{\frac{2\times (6.67\times {10}^{-11}\text{ N}·{\text{m}}^2{\text{kg}}^{-2})(333000\times 5.98\times {10}^{24}\text{ kg})}{(109\times 6.38\times {10}^6\text{ m})}}=620\text{ km/s}$

Conclusion:

Thus, speed needed to escape the solar system from the Sun’s gravity is $620\text{ km/s}$ .

To determine

The speed of a particle for the given condition.

Answer to Problem 16A

  $42.2\text{ km/s}$

Explanation of Solution

Given:

A particle is ejected at a distance from the Sun which is equal to the distance of Earth from Sun.

Formula used:

Using the energy conservation principle, total energy initial and finally must be constant. Final velocity after escaping the gravity is zero since all the thrust is applied to work against gravity. And, the gravitational potential energy after escaping the star’s gravity is zero at infinite distance ${\text{U}}_{\text{gf}}=0$ .

Thus,

  ${(\text{K}+{\text{U}}_{\text{g}})}_{\text{i}}={(\text{K}+{\text{U}}_{\text{g}})}_{\text{f}}$

  $\frac{1}{2}{\text{mv}}_e{}^2+\frac{-\text{GMm}}{{\text{r}}^2}=0+0$

  $v_e=\sqrt{\frac{2\text{GM}}{\text{r}}}$

Calculation:

Escape velocity after inserting the values of constant of gravitation, mass of Earth and distance from the Sun equal to Earth’s distance from the sun is

  $v_e=\sqrt{\frac{2\times (6.67\times {10}^{-11}\text{ N}\cdot {\text{m}}^2{\text{kg}}^{-2})(333000\times 5.98\times {10}^{24}\text{ kg})}{(23500\times 6.38\times {10}^6\text{ m})}}=42.2\text{ km/s}$

Conclusion:

Thus, speed needed by an ejected particle at a distance from the Sun equal to Earth’s distance from the sun to escape the solar system is $42.2\text{ km/s}$ .