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Ch. 08 - Conservation of Energy
Giancoli Douglas - Physics for Scientists and Engineers 5th edition
Giancoli Douglas5th editionPhysics for Scientists and EngineersISBN: 9780137488179Non è quello che usi tu?Cambia libro di testo
Tutti i libri di testoGiancoli Douglas 5th editionCh. 08 - Conservation of EnergyProblema 44b
Capitolo 8, Problema 44b

Determine the escape velocity from the Sun for an object at the average distance of the Earth (1.50 x 10⁸ km). Compare (give factor for each) to the speed of the Earth in its orbit.

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Step 1: Recall the formula for escape velocity, which is derived from the conservation of energy. The escape velocity \( v_{e} \) is given by \( v_{e} = \sqrt{\frac{2GM}{r}} \), 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 Sun (\( 1.989 \times 10^{30} \ \text{kg} \)), and \( r \) is the distance from the Sun (\( 1.50 \times 10^{11} \ \text{m} \), converted from kilometers).
Step 2: Substitute the known values into the escape velocity formula. This involves plugging in \( G \), \( M \), and \( r \) into \( v_{e} = \sqrt{\frac{2GM}{r}} \). Ensure all units are consistent (e.g., meters, kilograms, seconds).
Step 3: To compare the escape velocity to the Earth's orbital speed, recall the formula for orbital speed \( v_{o} \), which is \( v_{o} = \sqrt{\frac{GM}{r}} \). Notice that the escape velocity is related to the orbital speed by \( v_{e} = \sqrt{2} \cdot v_{o} \).
Step 4: Calculate the factor by which the escape velocity exceeds the Earth's orbital speed. Using the relationship \( v_{e} = \sqrt{2} \cdot v_{o} \), the factor is simply \( \sqrt{2} \), which is approximately 1.414.
Step 5: Conclude that the escape velocity from the Sun at Earth's average distance is approximately 1.414 times the Earth's orbital speed. This relationship is a direct consequence of the mathematical connection between escape velocity and orbital speed.

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Escape Velocity

Escape velocity is the minimum speed an object must reach to break free from the gravitational pull of a celestial body without any additional propulsion. It depends on the mass of the body and the distance from its center. For the Sun, this velocity can be calculated using the formula v = √(2GM/r), where G is the gravitational constant, M is the mass of the Sun, and r is the distance from the Sun.
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Escape Velocity

Gravitational Force

The gravitational force is the attractive force between two masses, described by Newton's law of universal gravitation. It states that the force is directly proportional to the product of the two masses and inversely proportional to the square of the distance between their centers. This force is crucial in determining the escape velocity, as it dictates how strongly the Sun pulls on objects at a given distance.
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Orbital Speed

Orbital speed is the velocity required for an object to maintain a stable orbit around a celestial body. For Earth, this speed is approximately 29.78 km/s at an average distance from the Sun. Comparing escape velocity to orbital speed helps understand the dynamics of celestial motion, as an object must exceed escape velocity to leave the Sun's gravitational influence.
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