Question

The binding energy of a particle of mass m with a planet, when it is on the planet's surface, is 201mV². A tunnel is dug along a diameter of the planet, and the particle is dropped into it from the surface. When the body reaches the center of the planet, what is its speed?

Answer 1

The particle's speed at the center of the planet is obtained by equating the initial binding energy to the kinetic energy at the center, resulting in the particle's speed being the square root of 402 times the square of the initial speed on the surface.

Step-by-step explanation:

The question deals with the concept of binding energy and the change of potential energy when a particle is moved from the surface of a planet to its center. According to the law of conservation of energy, the initial binding energy on the surface will be converted into kinetic energy when the particle reaches the center of the planet. Given that the binding energy is 201mV², we can equate this to the kinetic energy at the center of the planet (½mv²).

Thus, we have:

1. Initial binding energy: 201mV²
2. Kinetic energy at the center: ½mv²

We can set these two equal to each other to find the final speed v:

201mV² = ½mv²

Cancelling the mass m from each side, we get:

402V² = v²

Taking the square root of both sides, we obtain:

v = √(402V²)

This provides the speed of the particle when it reaches the center of the planet.