The escape velocity of a body depends upon mass as

General
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the escape velocity of a body depends upon mass as

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LectureNotes said the escape velocity of a body depends upon mass as

Answer:

To accurately explain how the escape velocity of a body depends on mass, let’s delve into the physics behind escape velocity. Escape velocity is the minimum speed needed for an object to break free from the gravitational attraction of a massive body without further propulsion. It is crucial to note that the escape velocity itself does not directly depend on the mass of the object trying to escape but rather on the mass of the celestial body from which it is escaping.

The formula for escape velocity (v_e) is derived from equating the kinetic energy of the object to the gravitational potential energy, and it is given by:

v_e = \sqrt{\frac{2GM}{R}}

where:

  • v_e is the escape velocity,
  • G is the universal gravitational constant (6.674 \times 10^{-11} \, \text{m}^3 \text{kg}^{-1} \text{s}^{-2}),
  • M is the mass of the celestial body,
  • R is the radius of the celestial body from its center to the point of escape.

Key Points

  1. Dependence on Mass of Celestial Body (M):

    • The escape velocity increases with the mass of the celestial body. Hence, larger and more massive planets or stars exert a stronger gravitational pull, requiring a higher speed to escape their gravity.

  2. Independence from Mass of the Escaping Object:

    • The formula shows no dependence on the mass of the object attempting to escape. This implies that whether the object is a small satellite or a large rocket, the required escape velocity remains the same for a particular celestial body.

Detailed Explanation

  1. Gravitational Force and Potential Energy:

    • The gravitational force acting on an object of mass m near a celestial body of mass M is given by Newton’s law of gravitation:

      F = \frac{GMm}{R^2}

    • The gravitational potential energy (U) at a distance R from the center of the celestial body is:

      U = -\frac{GMm}{R}

  2. Kinetic Energy and Energy Conservation:

    • The kinetic energy (K) of an object of mass m moving with velocity v is:

      K = \frac{1}{2}mv^2

    • At escape velocity, the kinetic energy must be equal to the magnitude of the gravitational potential energy to “escape” the gravitational field:

      \frac{1}{2}mv_e^2 = \frac{GMm}{R}

  3. Solving for Escape Velocity:

    • Rearranging the equation to solve for v_e while cancelling out the mass m of the object:

      v_e^2 = \frac{2GM}{R}
      v_e = \sqrt{\frac{2GM}{R}}

    • From this, it is evident that v_e is dependent on the mass M and the radius R of the celestial body, but not on the mass m of the escaping object.

Final Answer:
The escape velocity of a body depends on the mass M of the celestial body from which it is escaping but is independent of the mass of the escaping object. The exact relationship is given by the formula v_e = \sqrt{\frac{2GM}{R}}.