>Suppose that we are getting our force from gravity. Like any other force, we need double to accelerate the 10kg object.

OK.

>Luckily, the force of gravity increases whenever the mass goes up. So 10kg is being pulled on more than 5kg.

Yes...

>But we need more force to pull the 10kg object.

We need more force to push a 10kg object, not pull it...

>So, the conclusion is that the 5kg and 10kg object will be moved at the same rate.

Now you're talking about velocity.. speed. v = d/t. So you're saying that the masses will pull towards each other at the same speed regardless of the mass, which is completely different than "as the mass increases, the rate of acceleration DECREASES"...

Your assertion is: the rate at which one object approaches the other object due to gravity is independent of the first object's mass.

v = d/t, a = v/t

F1 = m1*a1

F = G*m1*m2/d^2

Now suppose that F = F1, then you would be correct, I suppose.

F = F1 = m1*a1 = G*m1*m2/d^2 , so: v1 = t * G*m2/d^2

Can you say that F1 = F? Obviously if you do this you will be confusing the frame of reference. If F1 = F, then F2 = F, and F1=F2:

F1 = F2

m1*a1 = m2*a2

But we know that a1 = a2 (and v1=v2), and to say that m1=m2 is ridiculous. Ifthe point of reference was simply the object in question (e.g. F1=F2=F), NOT the other object:

1) The first object's velocity towards its own position is not dependant on the mass of the first object, but rather the second object's mass.

2) The second object's velocity towards its own position is not dependant on the second object's mass, but rather the first object's mass.