In "absolute" time, all of my questions must be answered.
Only an approximate answer can be obtained without correct answers to my questions.
Based on a two body equation the above issues would exceed or be significant relative to the movement the moons CG relative to either object. (we are talking about trillionths of a meter at a few meters of a second) A mountain would cause deflection from an perfect line between the two CG's greater then the moons CG would move toward the object.
All this ass-ide...
The approximate answer is "equal time".
MarkCB750, you are overcomplicating the problem.
Objects move toward their common center of mass (assuming no angular motion). The more the combined masses are (with all else equal), the shorter the collision time. This is intuitive. Imagine the same object, if dropped on earth, takes a shorter amount of time than if dropped on the moon (if done in a vacuum, i.e., no air resistance). Remember seeing the astronauts bouncing around in 'slow motion' on the moon? This is because the moon is less massive. And hence, 'g' is smaller.
G, the gravitational constant, is constant. However, 'g', the gravitational
acceleration differs depending on the masses involved (and the distance between them). On earth, 'g' is about 9.8 m/s
2. On the moon, it is about 1.62 m/s
2. However, both of these values assume a very small
test mass relative to the planet's mass. If 'm', the mass of the object being dropped, was quite large, then we can no longer assume m << M (i.e., that 'm' is negligible), and we must adjust the value of 'g' accordingly. Of course, the 'g' we use is only applicable to near-surface calculations, and 'g' changes as we move farther away from the earth. But this is neither here nor there. You should be able to see my point regarding how the masses involved can change 'g'.
Now, as long as 'm' is not zero, it has an effect,
no matter how small that effect is. The question is not what is within the limits of measurability. The question was an ideal thought experiment meant to illustrate that objects move toward their common center of mass (assuming no angular motion). The more the combined masses are (with all else equal), the shorter the collision time.
The "engineer" answer is therefore, "approximately equal," because he only works in the regime of what can be measured.
The "mathematician" answer is therefore, "more massive object (with all else being equal) has the shorter collision time."
The "physicist," well... he, like the engineer, prefers to work with things that are measurable, but he would concede that, theoretically, the more massive hammer (with all else being equal) would hit first even though it might be beyond our capability to measure it.
Now we only need the priest answer and the rabbi answer to round it out.
And again, this only applies when they are released separately, not together. Together, as Galileo showed over 300 years ago, they take the same amount of time, because it is like dropping
one object with a mass equal to the combined masses of the hammer and feather.
Perhaps, when I have more time, I'll post the mathematical solution for the collision time between two massive objects, if released from a set distance with no angular momentum, and effected only by their mutual gravitational force. You'll see that both masses end up in the solution, meaning that the collision time depends on the masses involved. We normally neglect the smaller mass because it is soooo much smaller than the larger one (e.g., the mass of the hammer is much
much smaller than the mass of the moon), but neglecting the smaller mass is fudging, and the more accurate answer is the one where we don't throw out the smaller mass.
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