Final answer:
When a particle initially bound to a planet's surface with binding energy equivalent to half its kinetic energy is dropped through a tunnel to the center of the planet, its speed at the center is the same as its initial speed at the surface, assuming no energy losses.
Step-by-step explanation:
The question deals with the binding energy of a particle on the surface of a planet and what happens to its speed when it is dropped through a tunnel that passes through the center of the planet. Using the conservation of energy, we can deduce that when the particle reaches the planet's center, its kinetic energy will be at its maximum since the gravitational potential energy is zero at this point. Given that the binding energy on the surface was 1/2 mv² (where v is the initial speed) and assuming the conservation of total mechanical energy, at the center of the planet, the observation can be made that all initial potential energy has been converted to kinetic energy.
Therefore, when the particle reaches the planet's center, its speed will be v (option a), assuming no energy losses along the way. The reasoning behind this is that the binding energy at the surface already accounts for half of the kinetic energy, which would equate to the total energy in the system, meaning when the particle falls to the center, the available potential energy has been completely transformed into kinetic energy, thus conserving the total energy as 1/2 mv².