derive the escape speed
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Answer:
Derivation of Escape Speed
The derivation starts with the initial gravitational potential energy at the given altitude and the initial kinetic energy of the object. This total initial energy is then compared with the sum of the potential and kinetic energies at an infinite separation, in order to determine the escape velocity equation
The Law of Conservation of Energy states that the total energy of a closed system remains constant. In this case, the closed system consists of the two objects with the gravitational force between them and no outside energy or force affecting either object.
Thus the total final energy—potential energy plus kinetic energy—must equal the total initial energy:
TEi = TE∞
KEi + PEi = 0
Substitute values:
mve2/2 − GMm/Ri = 0
Add GMm/Ri to both sides of equation:
mve2/2 = GMm/Ri
Solve for ve2:
ve2 = 2GM/Ri
Take the square root of each expression to get:
ve = ± √(2GM/Ri)
Considering our gravitational convention for direction, ve is upward or away from the other object and is thus negative:
ve = − √(2GM/Ri)
The derivation of the gravitational escape velocity of an object from a much larger mass is achieved by comparing the potential and kinetic energy values at some given point with the values at infinity, applying the Law of Conservation of Energy.
The equation for the gravitational escape velocity is:
object from a much larger mass is achieved by comparing the potential and kinetic energy values at some given point with the values at infinity, applying the Law of Conservation of Energy.
The equation for the gravitational escape velocity is:
ve = − √(2GM/Ri)
Taking altitude into account, the equation can be written as:
ve = − √[2GM/(r + h)]
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ve = − √[2GM/(r + h)]
Explanation: