Potential energy of particle in gravitational field of disk

[Oomph!]

Homework Statement

I have a particle of mass m. The particle is moving in direction of axis z because of the gravitational force of a homogeneous circular disk of mass M and radius a. There is a formula for gravitational force of the disk on the picture.

Task:
1) Find the formula of potentional energy for z>0 (formula is on the picture).
2) We release the particle in z=4a/3. Find the velocity of impact on disk (formula is on the picture).

Homework Equations

Relevant equation is the formula for force of the disk. I just have to work with that.

The Attempt at a Solution

So, the first task is simple. I just have to integrate the force and i will get the formula of potentional energy. However, I don't understand why is important that z>0. And what about the constant? If I integrate, there is a constant C. I can just say, that C=0?

The second task is more difficult for me. I think that I have to compare kinetic energy (E=0.5mv²) and the potencional energy from first task. I have to substitute the z=4a/3 . I thought that the potentional energy at the beginning and the kinetic energy of impact on disk is same. However, probably not, because my formula was little bit different. Please, could you tell me what I have to do better?

Thank you.

 

[mfb]

The formula for the force is valid for z>=0 only, you would need a different formula to generalize it.

The integration constant should be chosen to get U->0 for large z.

The second part works via conservation of energy. What is the total initial energy? What is the total final energy?

 

[Oomph!]

Now I know what was wrong. The particle has also potentional energy when it impact. So the energy at the beginning was just potentional for z=4a/3. But when the particle impacted on the disk, the total energy was kinetic and also potentional for z=0. Then I have correct result. However, how could I say that z=0? The particle falls to the center of the disk?

 

[mfb]

The whole disk is at z=0, not just the center. The given formulas assume the particle moves along the symmetry axis, however, so it will hit the center of the disk.