Real gravitational potential energy to kinetic energy

1. Jun 15, 2013

omiros

This isn't a homework problem, but something that has being buzzing in my head.

I've been thinking about it cause at huge distances from earth g changes and also in what happens for really dense objected and small things could change too.

The real equation for gravity no matter which body we are talking about with a spherical symmetry, is Fg=GmM1M2/(r+s)2,
with r = the radius of the main body and s = the distance of the body from the surface of the other one.

Let's consider M1the main one and M2the one that can be viewed as a particle. We know that ΔK = -∫s0Fds , however the force is chaning all through the 'free fall' and makes things more complicated.

The solution that I find is Fg=GmM1M2s/(r2+rs)

Where do I need the help?
I don't know if this is the right solution and I also don't know how to interprete the limiting cases, like limr→0 and lims→0 at the same time(in the center).
What are all the 3 different equations that describe the variance of the force?
Also could someone transform the energy equations if both bodies are 'big' but with different radius(not same r or mass)?
What energy do they have before they collide?
What would happen if they were like earth? (not spherical, with equations)

2. Jun 15, 2013

Khashishi

Your integral is incorrect. The integral of -1/(r+s)^2 is 1/(r+s)

3. Jun 15, 2013

omiros

I am sorry. That is my mistake. I meant ΔK not Fgwith limits s and 0

Last edited: Jun 15, 2013
4. Jun 15, 2013

Staff: Mentor

This does not change the fact that your integral is wrong.

That does not work if both objects have a finite size. The denominator has to be the distance between the central points of the objects. It is convenient to use r for this distance, if you use other definitions you have to adjust the denominator to get the same result.
As long as the spherical objects do not overlap, this formula is true independent of the sizes and distances of the objects.

You need a multipole expansion or numerical simulations.

5. Jun 16, 2013

omiros

Did you try to solve it before you say it was wrong? If you simplify the two fractions that you get, this is the result.

6. Jun 16, 2013

Staff: Mentor

Oh sorry, I was confused as you used s for integration limits and as variable at the same time.
Okay, the result looks good.
Where is the point in taking the limit s->0 (i.e. not lifting the body at all) then? The limit for r,s->0 will depend on the way they go to zero.