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What is escape velocity and how do we get the formula vesc = root of 2GM/R in Class 10 Gravitation?

GradeHunt_TeamGradeHunt Team · · Maharashtra Board 10th

I'm in SSC (Maharashtra Board), Science Part 1, chapter Gravitation. I understood g and free fall, but escape velocity is confusing me. Why do we take total energy as zero at infinity, and does escape velocity depend on the mass of the rocket? Our teacher said 11.2 km/s for earth but I can't get the derivation in my head.
escape velocitygravitationssc science part 1maharashtra board

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Escape velocity is the minimum velocity with which an object must be thrown up from the surface of a planet so that it never falls back. The Balbharati Std X textbook (Science and Technology Part 1, chapter Gravitation) derives it with the law of conservation of energy. Why infinity? Gravitational force becomes zero only at an infinite distance. So to escape, the object must just reach infinity and stop there. At infinity its kinetic energy is 0 and its potential energy (-GMm/r with r infinite) is also 0. Total energy E2 = 0. On the surface of the earth (mass M, radius R): Kinetic energy = 1/2 m vesc^2 Potential energy = -GMm/R Total energy E1 = 1/2 m vesc^2 - GMm/R Set E1 = E2: 1/2 m vesc^2 - GMm/R = 0 vesc^2 = 2GM/R vesc = root of (2GM/R) Since g = GM/R^2, this can also be written as vesc = root of (2gR). For the earth: root of (2 x 9.8 x 6.4 x 10^6) = root of (1.2544 x 10^8) = 1.12 x 10^4 m/s = 11.2 km/s. Does it depend on the rocket's mass? No. The mass m cancels out in the step above. A 1 kg ball and a 500 tonne rocket need the same 11.2 km/s. Worked example: a planet has 9 times the mass of the earth and 4 times its radius. Find its escape velocity. vesc(planet) / vesc(earth) = root of (Mp/Me x Re/Rp) = root of (9 x 1/4) = 3/2 vesc(planet) = 11.2 x 1.5 = 16.8 km/s A bigger mass raises escape velocity and a bigger radius lowers it. The moon's radius is smaller than the earth's, which alone would raise it, but its mass is only about 1/81 of the earth's, so its escape velocity is just about 2.37 km/s (the textbook's own solved example). Tip for ratio sums: write the ratio formula first and cancel G before putting in numbers. It saves a lot of calculation.
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