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A particle of mass m is thrown upwards from the surface of the earth, with a velocity u. The mass and the radius of the earth are, respectively, M and R. G is gravitational constant and g is acceleration due to gravity on the surface of the earth. The minimum value of u so that the particle does not return back to earth, is

1. (2GM / R)^1/2
2. (GM / R)^1/2
3. (gR)^1/2
4. (2gR)^1/2

Correct answer: (2GM / R)^1/2

Solution

The minimum velocity required for a particle to escape Earth's gravitational field is the escape velocity, given by \( v_{escape} = \sqrt{\frac{2GM}{R}} \). This is derived from equating the kinetic energy to the gravitational potential energy at the surface of the Earth. Hence, the correct answer is \( \sqrt{\frac{2GM}{R}} \).

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