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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)

Chiral Anomaly
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Pinke Helga
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  • Strange question. When you ask " how long the fall time would be, when the moon would fall from 100 meters onto the earth?" are you asking about the true moon as an extended object or a point mass = the moon's mass? –  Dec 27 '18 at 00:30
  • That's a good point. Sure it will make a difference since some of the mass is in much further distance. I did not take that into account. So what distance should I choose to make this negligible? The point I want to understand is if there is really no difference how much is the mass falling onto earth or if this is valid only for relatively small masses compared to the earth - and first of all: How can it be explained. So I guess we could assume a point of mass. – Pinke Helga Dec 27 '18 at 00:38
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    Possible duplicates here and here. – rob Dec 27 '18 at 00:41
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    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. –  Dec 27 '18 at 00:42
  • @ggcg Yes I am confused since a mass of the earth should fall as fast as a feather onto the moon as well as the mass of the moon should fall as fast as a feather onto the earth. Obviously the one excludes the other (or my abstract thinking is too bad). So I cannot solve this conflict. – Pinke Helga Dec 27 '18 at 00:46
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    Ok, one of the linked questions (https://physics.stackexchange.com/questions/3534/dont-heavier-objects-actually-fall-faster-because-they-exert-their-own-gravity) 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. – Pinke Helga Dec 27 '18 at 00:55
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    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. – PM 2Ring Dec 27 '18 at 01:12
  • @PM2Ring Yes, something with 1/6 was in my memory - in fact it's the acceleration. I have suspected that I was wrong. My assumption that the relative fall speed of two objects (relative to each other) cannot be the same if regarded only one of the masses has been confirmed. I understand the the fall speed of different point masses relative to a fixed point in the room is equal. This little excursion helped me to extend my mind. – Pinke Helga Dec 27 '18 at 01:22
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    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. – Pinke Helga Dec 27 '18 at 01:38
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    Indeed! On a related note, when computing the orbits of moons, we can usually pretend that the planet is fixed, and the moon revolves around the planet's center. But when the moon has significant mass relative to the planet, as is the case with our Moon, that approximation doesn't work so well, and we need to compute the orbits of the planet & moon about their barycenter. – PM 2Ring Dec 27 '18 at 01:50

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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?

Chiral Anomaly
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  • 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! – Pinke Helga Dec 27 '18 at 01:03
  • 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.] – Pinke Helga Dec 27 '18 at 01:06
  • "...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. –  Dec 27 '18 at 02:08
  • @ggcg Yikes, you're right! Don't know what I was thinking. I edited the answer. Thank you very much for noticing that! – Chiral Anomaly Dec 27 '18 at 03:44
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    @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. – Chiral Anomaly Dec 27 '18 at 03:47