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It is well known that the rate of fall in the vacuum on the Earth is equal for all bodies, e.g. a feather and a gold bar, so on the moon as well. However, the moon has according to less mass also less gravity. As far as I know the duration of fall from equal height is about 6 times longer on the moon (please correct me).

I am wondering how long the fall time would be, when the moon would fall from 100 meters onto the earth? Or from the other point of view: How long would the earth need to fall onto the moon? How long would a feather or a hammer need on the earth and on the moon? (without any air friction of course)

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    $\begingroup$ Possible duplicates here and here. $\endgroup$
    – rob
    Dec 27, 2018 at 0:41
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    $\begingroup$ A couple things, (1) yes there is a tide force, or force gradient that will stretch and rip the body, (2) the CoM will be quite far from the surface. The CoM of the moon cannot be 100m from the surface of the earth. If you are simply curious about the uniformity of gravity that is connected to the hypothesis (fact?) that inertial and gravitational mass are equivalent. Hence acceleration due to gravity is independent of the mass of the body being acted upon. $\endgroup$
    – user196418
    Dec 27, 2018 at 0:42
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    $\begingroup$ Ok, one of the linked questions (physics.stackexchange.com/questions/3534/…) seems to answer my questions. The fall should be in fact "shorter" since both object fall onto each other, thus the way of the fall gets shorter during the fall time. The way of the moon to the original point of the earth would last as long as a feather would need. But the collision is earlier since the earth has moved to the moon too. $\endgroup$ Dec 27, 2018 at 0:55
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    $\begingroup$ Don't heavier objects actually fall faster because they exert their own gravity? is one of my favourite questions on this site. BTW, stuff doesn't fall 6 times slower on the Moon compared to in a vacuum on Earth. The formula is $s=\frac 12at^2$, where $a$ is the acceration & $s$ is the distance. The gravitational acceleration at the Moon's surface is about $\frac 16$ that of Earth, so for a fall of the same distance, the time is about $\sqrt 6$ times longer on the Moon, a factor of around 2.46. $\endgroup$
    – PM 2Ring
    Dec 27, 2018 at 1:12
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    $\begingroup$ Conclusion: So actually in the proper meaning of the word it is wrong to say: 'Every object regardless of the mass falls in the same time to the ground' since also the earth moves against the object depending of the object's mass. This is just negligible when little masses are involved. $\endgroup$ Dec 27, 2018 at 1:38

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The statement that all objects fall at the same rate on the earth (more accurately, with the same acceleration) is an idealization that relies on two approximations. Both approximations are excellent for objects of humanly manageable size and mass near the surface of the earth, but both approximations fail for something with the size and mass of the moon.

The first approximation is regarding the size of the falling object. This approximation assumes that the objects have negligible size compared to the scale over which the gravitational field varies. Near the surface of the earth, the "acceleration of gravity" on the far side of a 1-meter diameter ball is almost identical to the "acceleration of gravity" on the near side, so the statement applies quite accurately for objects of humanly manageable size.

However, this approximation is not true at all for something the size of the moon that is nearly touching the earth (say, with only 100 meters between their surfaces), because then the force on the moon due to the earth's gravity is much less on the far side of the moon than it is on the near side of the moon.

The second approximation is regarding the mass of the falling object. The statement that all objects fall at the same acceleration on the earth also assumes that the acceleration of the earth due to the gravitational field of the falling object is negligible. This approximation isn't very good for the earth-moon system even when they're far away from each other. If the earth and moon started at rest at some large distance away from each other, they would both accelerate toward each other (because the moon pulls on the earth, too), thus decreasing the distance between them more quickly than if only one of them were accelerating. This more rapid decrease in distance causes an more rapid increase in their mutual attraction, which would not occur if either body were replaced by a feather.

A mathematical analysis of this second point is given in another post:

Don't heavier objects actually fall faster because they exert their own gravity?

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  • $\begingroup$ You are pointing out many more aspects than I had seen without much physical knowledge. My intuition and rational thoughts told me, that can not be generalized as usually done. Many thanks for this pleadings! $\endgroup$ Dec 27, 2018 at 1:03
  • $\begingroup$ While your link gives physical formulars, your pleadings make it better understandable in a popular scientific way - for people like me. :) ["that I had not seen" I wanted to say in my previous comment.] $\endgroup$ Dec 27, 2018 at 1:06
  • $\begingroup$ "...also assumes that the gravitational force that the falling object exerts on the earth is negligible." This could never be an assumption as it violates Newton's 3rd law. This is not what's happening. The disparity in masses causes a disparity in acceleration not force. Thus one can assume the larger object does not move. This also does not contradict the fact that the instantaneous acceleration of a feather at some distance from the Earth's center is the same a a point as massive as the moon at the same distance. I'd recommend editing your answer to address the first point. $\endgroup$
    – user196418
    Dec 27, 2018 at 2:08
  • $\begingroup$ @ggcg Yikes, you're right! Don't know what I was thinking. I edited the answer. Thank you very much for noticing that! $\endgroup$ Dec 27, 2018 at 3:44
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    $\begingroup$ @Quasimodo'sclone Another user (ggcg) noticed that one of the sentences in my answer was badly worded, specifically in the paragraph about the mass of the falling object. I edited it to fix the mistake. Just wanted to let you know so I don't cause any misconceptions. $\endgroup$ Dec 27, 2018 at 3:47

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