Time to collide of hammer vs Moon, feather vs Moon, considering that the Moon is attracted by them (barycenter) I am an extreme novice in Physics, I am also a beginner in Physics Stack Exchange, and I'm not fluent in English, so please bear with me, consider my question with indulgence.
I request indulgence especially because $99\text{%}$ of my question is a well-known question with an evident and well-known answer, but the important thing of the question is the remaining $1\text{%}$ which I will try to explain the best I can.
This well-known question is:

"On the Moon, drop a feather and a hammer, will they touch the ground at the same time?"

(and the question is the same on Earth if we exclude the air density/shapes of bodies that fall).
This question is always answered by the strict affirmation that "yes they do", and confirmed by astronaut David Scott experiment during Apollo $15$ mission.
I agree with that.
I agree that it's a strictly equal time, if we consider that the Moon is not attracted by the feather or the hammer.
It may sound obvious to many people that the Moon is not attracted by so much light objects, but theoretically it is because these objects have a mass, even if the feather/hammer mass is around 1E-23 times of the Moon mass. (It's like the Earth which is 100 times more massive than the Moon, so the Earth is also attracted and the effect is its "oscillation" around the barycenter of Earth/Moon which is at approximately $4,670$ to $6,380\ km$ away from the geographical center of the Earth.)
And here comes my question, which may look identical, but there is a change in the conditions and in the enunciation, especially I say "body" instead of "Moon", and I say "mutually attracted" rather than "one object falls on the body" so that you better understand how my question is 1% different from the previous question:

"In space vacuum, a feather and a body are both initially positioned at the same distance and with no relative movement, and are mutually attracted because of their masses and will collide in a given amount of time. Let's repeat the same experiment with a hammer instead of the feather, will they collide in exactly the same amount of time?"

(even if there's a tiny difference of 1E-20 second, I consider that it's not the same time)
My assumption is that the hammer and the feather will be both attracted identically by the body, BUT at the same time the body will be more attracted by the hammer than by the feather (because the hammer is heavier than the feather). So, the body and the hammer will collide quicker than the body and the feather.
Am I right or not? Where am I wrong?
Of course, my question is related to the first well-known question, and my assumption is that there is a difference, although extremely small, maybe 1E-20 seconds, but theoretically it's different because the Moon is attracted. What do you think? Do you know any document which talks about that? (I searched a lot but I couldn't find anything)
Thank you very much for your interest in this question.
 A: Say we drop an object of mass $m$ and it falls owing to its attraction to an object of mass $M$. The force on either object is
$$
f = G M m / r^2
$$
where $r$ is the distance between the centre of one object and the centre of the other. Therefore the object whose mass is $m$ gets an acceleration
$$
a_m = G M / r^2 .
$$
Meanwhile the object of mass $M$ is experiencing a force of the same size, so it gets an acceleration
$$
a_M = G m / r^2 .
$$
So the two objects are accelerating towards one another. At any given time the distance between them is $r$, and this distance is changing (getting smaller). Its second rate of change is
$$
\frac{d^2 r}{d t^2} = a_m + a_M = \frac{G(M+m)}{r^2}
$$
Now your question concerns one moon of mass $M$ and two different objects (hammer and feather). Let's call their masses $m_h$ and $m_f$. We will do three experiments.
Experiment 1. Drop a hammer on the moon, without any feather.
The relative acceleration of moon and hammer is
$$
a_1 = G(M + m_h)/r^2 .
$$
Experiment 2. Drop a feather on the moon, without any hammer.
The relative acceleration of moon and feather is
$$
a_2 = G(M + m_f)/r^2 .
$$
We find $a_1$ a tiny bit larger than $a_2$, so indeed the hammer hits the moon in less time than the feather if the two experiments are done separately.
But usually the two experiments are done at the same time.
Experiment 3. Drop hammer and feather together, on the moon.
Now the total thing pulling on the moon is both hammer and feather together. In this case the acceleration of the moon is
$$
a_{\rm moon} = G(m_h + m_f)/r^2
$$
and the accelerations of the other two objects are
$$
a_{\rm hammer} = GM/r^2, \;\;\;\;\; a_{\rm feather} = GM/r^2 
$$
(and they also have a slight acceleration towards one another). The main point now is that theses two accelerations, $a_{\rm hammer}$ and $a_{\rm feather}$, are the same. The moon is meanwhile also accelerating a tiny bit. But hammer and feather have the same motion as each other. Therefore when dropped side by side the hammer and the feather hit the moon together at exactly the same time.
Your intuition was valid when applied to experiments 1 and 2, but not for experiment 3. But now I will adjust the experiment a little. Suppose the hammer and feather are dropped at the same time, but not very close to one another. For example, they might be a few kilometres apart, or even on opposite sides of the moon. In this case the moon will accelerate somewhat more towards the hammer than towards the feather, so in this case your intuition is right and the hammer hits the moon very slightly before the feather.
