Problem 49 - Physics for Scientists and Engineers with Modern Physics

Question

(II) Determine the escape velocity from the Sun for an object \((a)\) at the Sun's surface \(\left(r=7.0 \times 10^{5} \mathrm{~km}\right.\), \(\left.M=2.0 \times 10^{30} \mathrm{~kg}\right),\) and \((b)\) at the average distance of the Earth \(\left(1.50 \times 10^{8} \mathrm{~km}\right)\). Compare to the speed of the Earth in its orbit.

Step-by-Step Solution

Answer

The escape velocity at the Sun's surface is approximately 618,000 m/s, and at Earth's distance is approximately 42,000 m/s. Earth's orbital speed is 29,800 m/s.

Step 1: Determine Escape Velocity Formula

The formula for escape velocity (v) from a celestial body is given by \( v = \sqrt{\frac{2GM}{r}} \), where \( G \) is the gravitational constant \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \), \( M \) is the mass of the celestial body, and \( r \) is the distance from the center of the celestial body.

Step 2: Calculate Escape Velocity at Sun's Surface

Substitute \( M = 2.0 \times 10^{30} \, \text{kg} \) and \( r = 7.0 \times 10^5 \, \text{km} = 7.0 \times 10^8 \, \text{m} \) into the escape velocity formula. Thus, \( v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 2.0 \times 10^{30}}{7.0 \times 10^8}} \). Calculate \( v \) to find the escape velocity at the Sun's surface.

Step 3: Escape Velocity at Sun's Surface Calculation

Carrying out the calculations, \( v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 2.0 \times 10^{30}}{7.0 \times 10^8}} = \sqrt{3.81 \times 10^{11}} \approx 6.18 \times 10^5 \, \text{m/s} \). The escape velocity from the Sun's surface is approximately \( 618,000 \, \text{m/s} \).

Step 4: Calculate Escape Velocity at Earth's Average Distance

Use the same formula with \( r = 1.50 \times 10^8 \, \text{km} = 1.50 \times 10^{11} \, \text{m} \). Substitute \( M = 2.0 \times 10^{30} \, \text{kg} \) and calculate \( v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 2.0 \times 10^{30}}{1.50 \times 10^{11}}} \).

Step 5: Escape Velocity at Earth's Average Distance Calculation

Computing the above expression, \( v = \sqrt{\frac{2 \times 6.674 \times 10^{-11} \times 2.0 \times 10^{30}}{1.50 \times 10^{11}}} = \sqrt{1.89 \times 10^{4}} \approx 4.2 \times 10^4 \, \text{m/s} \). The escape velocity at Earth's average orbital distance from the Sun is approximately \( 42,000 \, \text{m/s} \).

Step 6: Compare to Earth's Orbital Speed

The average orbital speed of Earth is approximately \( 29,800 \, \text{m/s} \). Comparing this to the escape velocities calculated, Earth's speed is less than the escape velocity at its average distance from the Sun (\( 42,000 \, \text{m/s} \)), meaning it cannot spontaneously escape the Sun's gravitational field.

Key Concepts

Gravitational Constant

The gravitational constant, often denoted as \( G \), is a key value in the universe that describes the strength of the gravitational force. It's a fundamental component of the law of universal gravitation, which relates the force of attraction between two masses to their separation distance. The gravitational constant \( G \) has a value of approximately \( 6.674 \times 10^{-11} \, \text{m}^3 \, \text{kg}^{-1} \, \text{s}^{-2} \). This value might seem small, but it plays a crucial role in calculations involving celestial bodies like planets and stars.

• \( G \) is used to calculate gravitational forces exerted by celestial bodies.
• It defines how much gravitational "pull" a massive object, like the Sun, has on other objects.
• Its small magnitude reflects the relative weakness of the gravitational force compared to other fundamental forces in nature.

Understanding \( G \) is essential for exploring how objects interact on an astronomical scale, influencing orbits and escape velocities.

Celestial Mechanics

Celestial mechanics is the branch of astronomy that deals with the motions of celestial objects. It involves the application of physics, particularly Newton's laws, to explain the movements within our solar system and beyond.

• It explains the orbits of planets around the Sun, moons around planets, and even spacecraft trajectories.
• By understanding these mechanics, scientists can predict the positions and paths of celestial bodies.
• The study helps in the calculation of escape velocity, which is the speed needed for an object to break free from a celestial body's gravitational pull.

The calculations of escape velocity in the exercise rely on principles derived from celestial mechanics. These principles help us understand complex phenomena such as why planets remain in orbit, the dynamics of comet paths, and even the feasibility of sending spacecraft to other planets. Using celestial mechanics, we can determine various attributes, like orbital speed and period, which are vital for spacecraft navigation and astronomical observation.

Orbital Speed

Orbital speed is the velocity a celestial object requires to maintain a stable orbit around another object. This speed depends on the gravitational pull of the larger mass and the distance from it. For Earth, this is about \( 29,800 \, \text{m/s} \), as given in the exercise.

• Orbital speed ensures that an object remains in a consistent path around a celestial body.
• It must be carefully calculated for satellites and space missions to ensure they stay in orbit over the desired path.
• Depending on the distance from the celestial body, the speed will change; closer orbits require faster speeds.

In the exercise, Earth's orbital speed is compared to the escape velocity required to leave the Sun's gravitational influence. Despite seeming high, Earth's speed is not enough for it to drift away into space. Understanding orbital speed is crucial in celestial mechanics, as it helps astronomers and engineers design efficient paths for space-bound vehicles and maintain satellite operations at optimal positions.