College Physics II – Mechanics, Sound, Oscillations, and Waves

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$v_e$

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College Physics II – Mechanics, Sound, Oscillations, and Waves

Definition

$v_e$ represents the escape velocity, which is the minimum velocity an object must have to break free from the gravitational pull of a celestial body and escape its gravitational field. This term is crucial in understanding the concepts of gravitational potential energy and total energy.

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5 Must Know Facts For Your Next Test

  1. The escape velocity is independent of the object's mass, and only depends on the mass of the celestial body and the distance from its center.
  2. The formula for escape velocity is $v_e = \sqrt{\frac{2GM}{r}}$, where $G$ is the gravitational constant, $M$ is the mass of the celestial body, and $r$ is the distance from the center of the body.
  3. If an object's velocity is less than the escape velocity, it will remain bound to the gravitational field of the celestial body and may enter a stable orbit.
  4. The escape velocity of the Earth is approximately 11.2 km/s, which means an object must reach this speed or greater to break free from the Earth's gravitational pull.
  5. The concept of escape velocity is crucial in understanding the launch of spacecraft, as well as the behavior of celestial bodies and their interactions within a gravitational system.

Review Questions

  • Explain how the escape velocity, $v_e$, is related to the concept of gravitational potential energy.
    • The escape velocity, $v_e$, is the minimum velocity an object must have to overcome the gravitational potential energy of a celestial body and escape its gravitational field. If an object's velocity is less than $v_e$, it will remain bound to the gravitational potential well of the body and may enter a stable orbit. The escape velocity represents the point at which the object's kinetic energy is equal to the negative of its gravitational potential energy, allowing it to break free and leave the gravitational influence of the body.
  • Describe how the concept of total energy is related to the escape velocity, $v_e$.
    • The total energy of an object in a gravitational field is the sum of its kinetic energy and gravitational potential energy. At the escape velocity, $v_e$, the total energy of the object is zero, meaning that its kinetic energy is exactly equal to the negative of its gravitational potential energy. This balance of energies allows the object to escape the gravitational field of the celestial body. If the object's velocity is less than $v_e$, its total energy will be negative, and it will remain bound to the gravitational field, potentially entering a stable orbit.
  • Analyze the factors that influence the escape velocity, $v_e$, and explain how changes in these factors would affect the value of $v_e$.
    • The escape velocity, $v_e$, is determined by the mass of the celestial body, $M$, and the distance from the center of the body, $r$, as described by the formula $v_e = \sqrt{\frac{2GM}{r}}$. If the mass of the celestial body increases, the escape velocity will also increase, as the gravitational pull of the body becomes stronger. Conversely, if the distance from the center of the body increases, the escape velocity will decrease, as the object is farther away from the source of the gravitational field. These relationships demonstrate how changes in the mass and radius of a celestial body can significantly impact the value of the escape velocity, which is a critical parameter in understanding the dynamics of gravitational systems.

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