A particle is released from rest at a height R above the surface of the Earth, where R is the radius of the Earth. It falls along a straight line towards the Earth and enters a narrow tunnel passing through the center of the Earth. Let g be the acceleration due to gravity at the surface of the Earth. The speed of the particle when it reaches the center of the Earth is

Step-by-step solution

Let M and R be the mass and radius of the Earth, respectively. The particle is released from rest at a height h = R , so its initial distance from the center of the Earth is r = 2R . The initial total mechanical energy is E_ i = - GMm 2R The gravitational potential at the center of a solid sphere of mass M and radius R is V_ c = - 3GM 2R . Thus, the final total mechanical energy of the particle at the center of the Earth, where its speed is v , is E_ f = - 3GMm 2R + 1 2 mv^2 By conservation of mechanical energy, E_