A particle is projected radially outwards from the surface of the Earth. Its initial speed is exactly equal to the orbital speed of a satellite revolving very close to the Earth's surface. If air resistance is neglected, the maximum height reached by the particle, measured from the surface of the Earth, will be ( R is the radius of the Earth):

Step-by-step solution

The orbital speed of a satellite very close to the Earth's surface is given by v₀ = GM R . Applying conservation of mechanical energy between the surface of the Earth and the maximum height h (where the kinetic energy becomes zero): K_i + U_i = K_f + U_f 1 2 mv₀^2 - GMm R = 0 - GMm R+h Substituting v₀^2 = GM R : 1 2 m ( GM R ) - GMm R = - GMm R+h GMm 2R - GMm R = - GMm R+h - GMm 2R = - GMm R+h Equating the denominators, we get: R+h = 2R h = R Thus, the maximum height reached from the surface of the Earth is R . Ans