I Dropped objects hitting the ground at the same time?

[Ty Wendland]

okay, so I’ve had this random thought. We have all been told that objects fall to the ground at the same speed, even if they have different masses. While it’s true that any two objects, regardless of mass, will accelerate towards Earth at the same speed, that doesn’t mean the Earth is accelerated towards the object at the same speed. Heavier objects have a higher gravitational force, and because of Newton’s Third Law, the Earth expirences the same force towards the object. Because Earth will accelerate towards heavier objects faster, heavier objects dropped in a perfect vacuum will hit the ground before a lighter one.

 

[jedishrfu]

It’s more than we have been told, it’s been measured to great accuracy. It’s known as the Galileo experiment. What’s not known if Galileo actually performed it but others have and things fall at the same rate unless air resistance becomes a factor as when feathers are used.

https://en.m.wikipedia.org/wiki/Galileo's_Leaning_Tower_of_Pisa_experiment

Often when we try to think our way through these things without fully understanding the math, we get it wrong and that’s what has happened here.

Here’s a simple discussion on Newton’s law

 

[Ty Wendland]

It’s more than we have been told, it’s been measured to great accuracy. It’s known as the Galileo experiment. What’s not known if Galileo actually performed it but others have and things fall at the same rate unless air resistance becomes a factor as when feathers are used.

https://en.m.wikipedia.org/wiki/Galileo's_Leaning_Tower_of_Pisa_experiment

Often when we try to think our way through these things without fully understanding the math, we get it wrong and that’s what has happened here.

Here’s a simple discussion on Newton’s law

Okay, this is all true, but consider the following scenario.

You have 2 planets will identical mass and no other forces are acting on the objects is the gravitational force between them. We will name these planets P1 and P2. Because the objects have the same mass as Earth, the objects accelerate to then will be 9.81 m/s^2. So P2 will be falling towards P1 9.81m/s^2. But P1 will be falling towards P2 at the same rate as well. The net rate that the the two planets accelerate towards each other will be 19.62m/s^2 because each object has the same mass and each object is expirencing the same force. Because the objects we experiment with are so small relative to the mass of Earth, the net acceleration towards one another is extremely close to 9.81m/s^2, but the large the mass gets relative to the mass of earth, the higher the net acceleration will be.

 

[Ty Wendland]

 

[Bandersnatch]

You're right, @Ty Wendland . The adage 'all objects fall at the same rate' is an approximation, valid only when one of the two masses is much larger than the other.

 

[Ty Wendland]

You're right, @Ty Wendland . The adage 'all objects fall at the same rate' is an approximation, valid only when one of the two masses is much larger than the other.

Just a random thought I had and wanted feedback on! Thank you!

 

[davenn]

Just a random thought I had and wanted feedback on! Thank you!

I'm still concerned that you may think that 2 objects that are both much smaller mass that say the Earth, don't hit the ground at the same time ?
They do, the experiment has been done and measured countless times
Astronauts on one of the moon landings even did it on the moon
The classic hammer and feather experiment

On Earth in an atmosphere, the hammer WILL hit first because of the air resistance the feather encounters, mainly because of its shape

The experiment was done in a vacuum chamber to negate air resistance and both the hammer and feather hit the ground together

Prof. Brian Cox did an experiment at a NASA vacuum chamber with a bunch of feathers and a bowling ball

https://www.bing.com/videos/search?...7E3D58E6A738E5D18D5B7E3D58E6A738E5D&FORM=VIREDave

 

[sophiecentaur]

I'm still concerned that you may think that 2 objects that are both much smaller mass that say the Earth, don't hit the ground at the same time ?

Me too.
There is a great temptation to look at these things from the point of view that says "Science got it wrong". I feel that is unhealthy because it neglects the basic authority of what is, in modern times, a very well justified body of work. In all matters Scientific, the actual numbers count and you have to take account of that before thinking that someone has found a loophole. Galilieo's (thought?) experiment shows that 'things fall down with the same acceleration'. That has been demonstrated to be true to greater and greater accuracy over the years.
What happens between bodies of near equal masses (planets etc.), follows the exact same rules and the same basic equations until General Relativity comes into it. You just have to look at the real (quantitative) consequences of that.

The attractive force between two masses m and M is always F = mMG/d². (G is the universal gravitational constant and not g, d is the distance between the centres of mass of the two objects and both objects are actually heading for their mutual centre of mass )The resulting acceleration of the two bodies will be F/m and F/M. Newton tells us that. Clearly the larger mass will accelerate less than the smaller mass. F/m is the same for any value of m and it only depends on the 'other' mass M.

So why is it that that Earth seems to be different from the falling ball? Simply because the two accelerations are so different and the Earth's surface accelerates so little, relative to the centre of mass of the two. One easily measurable effect of what happens between two no-dissimilar masses is the wobble of the Earth as the Moon orbits it. The Centre of mass of the two (the Barycentre) is about 4500km from the Earth's centre, which shows that the Earth is pulled into 'an orbit' by the gravity of the Moon. There are no convenient stationary large masses up there to observe how they fall together so we have to imply things from orbiting objects but the centripetal acceleration of the Earth as it orbits round the barycentre is (2Πf)² r which comes to about 30μm/s². How much less is the acceleration of the Earth when you drop a 1kg mass - or a 10kg - or a 1000kg? That's why we can say the Earth stays still for the Galileo experiments and why we can use g all over the place.

 

[Nugatory]

Because Earth will accelerate towards heavier objects faster, heavier objects dropped in a perfect vacuum will hit the ground before a lighter one.

Physics being a quantitative science, it's worth calculating the magnitude of this effect and comparing it with all the other experimental uncertainties in any measurement of the time it takes a dropped object to hit the ground. (For example... suppose someone on the other side of the Earth happens to fire off an antiaircraft gun while you're running your experiment? And just where is Jupiter right now?)

As for whether all of this means that it is wrong to say that things "fall at the same speed regardless of their weight"? It's hard to improve on Isaac Asimov's take on this sort of question: https://chem.tufts.edu/answersinscience/relativityofwrong.htm

 

[A.T.]

The net rate that the the two planets accelerate towards each other will be 19.62m/s^2.

The same is true for two small object falling on exactly opposite sides of the Earth.

 

[Lord Jestocost]

Because Earth will accelerate towards heavier objects faster, heavier objects dropped in a perfect vacuum will hit the ground before a lighter one.

If I remember correctly: Relative to earth, a heavier object would in principle accelerate faster and thus hit the ground after a shorter time compared to a lighter one. This doesn't hold, however, if the the body originates from the Earth itself. All terrestrial bodies falling individually fall at exactly the same rate in an ideal scenario.

 

[russ_watters]

I'm still concerned that you may think that 2 objects that are both much smaller mass that say the Earth, don't hit the ground at the same time ?
They do, the experiment has been done and measured countless times

As worded the logic of the OP looks fine (with the possible exception of the last sentence), it just doesn't include numbers. I would hope and assume the OP recognizes that a hammer isn't going to accelerate the Earth very much.

Also, there is a trick to this question depending on the wording in that two objects simultaneously dropped to Earth will impact at the same time, but if dropped separately the fall times will be (very, very slightly) different.

 

[Ty Wendland]

Me too.
There is a great temptation to look at these things from the point of view that says "Science got it wrong". I feel that is unhealthy because it neglects the basic authority of what is, in modern times, a very well justified body of work. In all matters Scientific, the actual numbers count and you have to take account of that before thinking that someone has found a loophole. Galilieo's (thought?) experiment shows that 'things fall down with the same acceleration'. That has been demonstrated to be true to greater and greater accuracy over the years.
What happens between bodies of near equal masses (planets etc.), follows the exact same rules and the same basic equations until General Relativity comes into it. You just have to look at the real (quantitative) consequences of that.

The attractive force between two masses m and M is always F = mMG/d². (G is the universal gravitational constant and not g, d is the distance between the centres of mass of the two objects and both objects are actually heading for their mutual centre of mass )The resulting acceleration of the two bodies will be F/m and F/M. Newton tells us that. Clearly the larger mass will accelerate less than the smaller mass. F/m is the same for any value of m and it only depends on the 'other' mass M.

So why is it that that Earth seems to be different from the falling ball? Simply because the two accelerations are so different and the Earth's surface accelerates so little, relative to the centre of mass of the two. One easily measurable effect of what happens between two no-dissimilar masses is the wobble of the Earth as the Moon orbits it. The Centre of mass of the two (the Barycentre) is about 4500km from the Earth's centre, which shows that the Earth is pulled into 'an orbit' by the gravity of the Moon. There are no convenient stationary large masses up there to observe how they fall together so we have to imply things from orbiting objects but the centripetal acceleration of the Earth as it orbits round the barycentre is (2Πf)² r which comes to about 30μm/s². How much less is the acceleration of the Earth when you drop a 1kg mass - or a 10kg - or a 1000kg? That's why we can say the Earth stays still for the Galileo experiments and why we can use g all over the place.

I don’t think that objects accelerate to Earth at different speeds, it’s a proven fact that everything accelerated to Earth at 9.81m/s^2. What I’m focused on is the time it two objects, the Earth and some dropped object, to come into contact. Obviously the mass difference between a 1kg object and a 10kg object compared to the Earth is nearly identical, but there is still a difference. In my scenario I’m saying there are no other external forces acting on the two objects besides gravity( which I know is extremely unrealistic). It’s
more of a hypothetical thought experiment than anything I would apply to the real world.

 

[PeroK]

I don’t think that objects accelerate to Earth at different speeds, it’s a proven fact that everything accelerated to Earth at 9.81m/s^2. What I’m focused on is the time it two objects, the Earth and some dropped object, to come into contact. Obviously the mass difference between a 1kg object and a 10kg object compared to the Earth is nearly identical, but there is still a difference. In my scenario I’m saying there are no other external forces acting on the two objects besides gravity( which I know is extremely unrealistic). It’s
more of a hypothetical thought experiment than anything I would apply to the real world.

If you do the maths then two objects of non-negligible masses $m_1$ and $m_2$ will collide in the same time that a small object of negligible mass takes to fall to a large object of mass $m_1 + m_2$.

More generally you could search online for the two body gravitational problem and the concept of "reduced mass".

 

[OmCheeto]

And just where is Jupiter right now?

Jupiter? From my calculations, even Pluto has a greater effect on me than I have on the Earth.

Earth's acceleration towards me: 1.23E-22 m/s²
My acceleration towards Pluto: 2.59E-14 m/s² (rough minimum distance)

Factor: ≈200 million​

Fun thought experiment.

It’s more of a hypothetical thought experiment than anything I would apply to the real world.

Thank you! I've always wondered about this. Never bothered to do the maths till now.

 

[Ty Wendland]

Jupiter? From my calculations, even Pluto has a greater effect on me than I have on the Earth.

Earth's acceleration towards me: 1.23E-22 m/s²
My acceleration towards Pluto: 2.59E-14 m/s² (rough minimum distance)

Factor: ≈200 million​

I guess that really shows how silly it is to calculate the amount I cause the Earth to accelerate towards me when an object billions of kilometres away has a much greater affect on me than I do the Earth!

 

[jkxzgyk]

Given any difference is due to the Earth accelerating towards the object, wouldn't both objects still hit the ground at exactly the same time (if dropped side by side) because the Earth's acceleration would be towards both objects, not just the heavier one?

 

[Ty Wendland]

Given any difference is due to the Earth accelerating towards the object, wouldn't both objects still hit the ground at exactly the same time (if dropped side by side) because the Earth's acceleration would be towards both objects, not just the heavier one?

I guess I worded what I was trying to say incorrectly. You are absolutely right. I was thinking more that these “experiments” were done separately, and the times for the objects of different mass were measured separately without influencing one another. I guess this is why I posted this, is so others could find the flaws in my logic!

 

[Bandersnatch]

Given any difference is due to the Earth accelerating towards the object, wouldn't both objects still hit the ground at exactly the same time (if dropped side by side) because the Earth's acceleration would be towards both objects, not just the heavier one?

If we're talking about completely negligible, but in principle present effects, then we also have the direction of the force from each of the objects being different. This means that the Earth will fall a bit sideways, away from the line equidistant from the two objects, and towards the heavier object.
This in turn - continuing with the unmeasurable physics theme - will mean that due to the surface of the Earth curving away more for the lighter object, it will have a larger distance to cover, and will fall after the heavier one.

Just to hammer in the obvious - for the m1, m2 << M case, any such difference will be swamped by orders of magnitude-larger errors in measuring instruments, timing of the release, the fact that no floor follows exactly the equipotential surface of the planet. And pretty much anything else one can think of.

But if one imagines the masses involved as being on a sliding scale, going all the way from m1, m2 << M to m1, m2 >> M, at some point the differences in the rate of falling become important and easy to spot. After all, if we 'drop' the Mars and the Sun on the Earth from 1 AU, they won't hit the ground at the same time.

 

[ZapperZ]

Okay, this is all true, but consider the following scenario.

You have 2 planets will identical mass and no other forces are acting on the objects is the gravitational force between them. We will name these planets P1 and P2. Because the objects have the same mass as Earth, the objects accelerate to then will be 9.81 m/s^2. So P2 will be falling towards P1 9.81m/s^2. But P1 will be falling towards P2 at the same rate as well. The net rate that the the two planets accelerate towards each other will be 19.62m/s^2 because each object has the same mass and each object is expirencing the same force. Because the objects we experiment with are so small relative to the mass of Earth, the net acceleration towards one another is extremely close to 9.81m/s^2, but the large the mass gets relative to the mass of earth, the higher the net acceleration will be.

You may want to start by reading this:

https://www.physicsforums.com/insights/why-is-acceleration-due-to-gravity-a-constant/

Zz.

 

[russ_watters]

Given any difference is due to the Earth accelerating towards the object, wouldn't both objects still hit the ground at exactly the same time (if dropped side by side) because the Earth's acceleration would be towards both objects, not just the heavier one?

I pointed that out in Post #12...

 

[FallenApple]

Einstein's Equivalence Principle: Imagine two different valued masses stationary in an elevator and the ground accelerating up towards it.

 

[256bits]

I guess I worded what I was trying to say incorrectly. You are absolutely right. I was thinking more that these “experiments” were done separately, and the times for the objects of different mass were measured separately without influencing one another. I guess this is why I posted this, is so others could find the flaws in my logic!

Works for any object or assemblage of terrestrial origin whether dropped separately or together, whether you want to measure to the nano second, pico second or whatever. One will find the value g.
Adding in an object from extra terrestrial origin changes the scenario, and the drop time will decrease..
Drop time is the time for the objects to reach the centre of mass.
I think you have noted that correctly in you post #13. about the two planets.

 

[OldYat47]

Re: Comment #11. At a human scale two objects of different masses will initially accelerate at exactly the same rate. Suppose A is twice as massive as B. Acceleration = Force / Mass. The force on A is (mA X g). The acceleration of A = [(mA X g) / mA]. mA cancels out and leaves you with g. Same as object B. Suppose A and B are spheres of identical size. Air resistance increases as the square of the speed. As air resistance increases the acceleration of the object slows and eventually reaches zero at terminal velocity. The force of drag would be the same for both objects at the same velocities. The accelerating force would be (mass X g) - (drag). So a less massive object's acceleration will decrease more quickly and its terminal velocity will be lower than a similar but more massive object.

 

[Lord Jestocost]

At a human scale two objects of different masses will initially accelerate at exactly the same rate.

From a fundamental perspective regarding the effective - let's say - "closing acceleration", this would only be the case if the two objects are of terrestrial origin.

 

[OldYat47]

From a fundamental perspective regarding the effective - let's say - "closing acceleration", this would only be the case if the two objects are of terrestrial origin.

That's what was implied by "at a human scale" although the analysis works on any planet with an atmosphere. More to come on a "non-human" scale as soon as I edit it more. Too lengthy now.

 

[OldYat47]

We have two spheres of similar size and identical mass in otherwise empty space. The gravitational force attracting them to each other is:
F = G*(m1+m2)/d. Since the masses are identical we could write that as:
F = (G*2*m)/d

Acceleration of each sphere:
Am1 = [(G*2*m)/d]/m1

Am2 = [(G*2*m)/d]/m2

Closing acceleration = Am1+Am2.
Aclosing = 2*(G*2m/d)/m, or A closing = (4*G/d)

Now suppose the mass of m2 is twice the mass of m1. We can substitute (2*m1) for m2.
F = G*(3m)/d

Acceleration of each sphere:
Am1 = [G*(3m)/d]/m

Am2 = [G*(3m/d]/(2*m)

Simplifying and adding those two together to get the closing acceleration:
A closing = (3*G/d) + (3/2*G/d) = 4.5*(G/d).

So for these conditions the lighter object will collide before the heavier object. Hope that's right, it's early.

 

[jbriggs444]

F = G*(m1+m2)/d.

Try $F=\frac{Gm_1m_2}{d^2}$

 

[OldYat47]

Too early to think, apparently, but the conclusion is the same.
We have two spheres of similar size and identical mass in otherwise empty space. The gravitational force attracting them to each other is:
F = G*(m1*m2)/d^2. (Using ^ as indication of exponent). Since the masses are identical we could write that as:
F = (G*m^2)/d^2

Acceleration of each sphere:
Am1 = [(G*m^2)/d^2]/m1 or Am1 = (G*m)/d^2, and since the masses are identical,

Am2 = (G*m)/d^2

Closing acceleration = Am1+Am2.
Aclosing = 2*(G*m)/d^2

Now suppose the mass of m2 is twice the mass of m1. We can substitute (2*m1) for m2.
F = G*(2m^2)/d^2

Acceleration of each sphere:
Am1 = [G*(2m^2)/d^2]/m or Am1 = 2*(G*m)/d^2

Am2 = [G*(2m^2/d^2]/(2*m) or Am2 = (G*m)/d^2

Aclosing = Am1 + Am2 = 3*(G*m)/d^2

So for these conditions the lighter object will collide before the heavier object.

 

[jbriggs444]

So for these conditions the lighter object will collide before the heavier object.

You are trying to claim that mass 1 collides with mass 2 before mass 2 collides with mass 1? That does not seem very reasonable.

 

[OldYat47]

Of course not. Collisions are simultaneous events. And there is a basic fault in my original logic. I'm pretty rusty, it seems. I guess I should be consoled that I caught it first.

A true comparison is when the force and distance are the same but the masses of the objects change. That is, the product of mass 1 and mass 2 is the same in both cases.
So let m1 = 2 and m2 =2.
F = (G*4)/d^2, the acceleration of each mass is (G*2)/d^2, and
A closing = (G*4)/d^2

Now let m1 = 1 and m2 = 4
F doesn't change, = (G*4)/d^2
But the accelerations do change:
Am1 = (G*4)/d^2
Am2 = (G)/d^2, and
A closing = (G*5)/d^2
So in cases where the gravitational force is equal, where masses are equal closing acceleration is lower than where masses are unequal. The greater the difference in masses, the greater the closing acceleration.

 

[Tom Kunich]

okay, so I’ve had this random thought. We have all been told that objects fall to the ground at the same speed, even if they have different masses. While it’s true that any two objects, regardless of mass, will accelerate towards Earth at the same speed, that doesn’t mean the Earth is accelerated towards the object at the same speed. Heavier objects have a higher gravitational force, and because of Newton’s Third Law, the Earth expirences the same force towards the object. Because Earth will accelerate towards heavier objects faster, heavier objects dropped in a perfect vacuum will hit the ground before a lighter one.

Gravity is different and can be calculated as GMm/r^2 where G is the gravitational constant, M is one mass, m the other and r the distance between them. Because in the real world the ratio of M to m is so huge you cannot measure differences of acceleration in a gravitational field with the skin friction of an atmosphere.

 

[OldYat47]

Here's some numbers expanding on what Mr. Kunich said above. What happens if you have one really big mass and two much smaller unequal masses at the same height?
Suppose we have two object with masses m1 and m2 = 4*m1. We have a third object with mass m3 = 10^24*m1. The objects are spherical, m1 and m2 are some distance d above the surface of m3, and there is no atmosphere.
We drop m1 onto m3.
Attractive force = [(G * (10^24 * m1) * m1)] / (d^2) or
Attractive force = (G * 10^24 * m1^2) / (d^2)

Acceleration of m1 = (G * 10^24 * m1) / (d^2)
Acceleration of m3 = (G * m1) / (d^2)
Closing acceleration (m1 to m3) = [G * m1 * (10^24 + 1)] / (d^2)

We drop m2 onto m3.
Attractive force = [(G * (10^24 * m1) * (4 * m1)] / (d^2) or
Attractive force = (G * (40^24 * m1^2) / (d^2)

Acceleration of m2 = (G * 10^24 * m1) / (d^2)
Acceleration of m3 = (G * 4 * m1) / (d^2)
Closing acceleration (m2 to m3) = [G * m1 * (10^24 + 4)] / (d^2)

So m2 and m3 would collide in less time than m1 and m3, but the difference in time is tiny (certainly beyond our capacity for measurement) until d gets quite large, and m3 moves a relatively tiny amount. This is different from my previous example because the attractive forces are different in this case.

 

[Sudha Sarita]

Dropped objects hit the ground same time. Objects which are of point size with respect to Earth are attracted towards ground due to force of gravitation. Force of gravitation is given by formula F = G\frac{m_1m_E}{R^2}. When an object of mass m_1 is dropped means it has zero initial velocity. It experiences an acceleration towards Earth which is given by a = \frac{F}{m_1}. Depending on the mass of every object there can be variation in acceleration towards Earth over a wide range. Assume that all objects are dropped from the same height then on the basis of equation h = ut + 1/2 (at^2), time can vary over a range. I have not considered force due to air resistance.

 

[OldYat47]

Sudha Sarita, no, they don't hit the ground at the same time, the heavier object hits first. The time difference is immeasurably small, but it's there.
The algebra is not complex, but perhaps I could have included the non-simplified factors.
We have two objects, m1 and m3. m3's mass is 10^24 times m1's mass or (10^24 * m1). The attractive force is [G * m1 * m3 / d^2].

So the attractive force = G * [(10^24 * m1) * m1] / (d^2) = G * (10^24 * m1^2) / (d^2)

The acceleration of m1 is the attractive force divided by m1 = [G * (10^24 * m1^2) / (d^2)] / m1 = G * (10^24 * m1) / (d^2) = (G * m1 / d^2) * (10^24)

The acceleration of m3 is the attractive force divided by m3 = [G * (10^24 * m1^2) / (d^2)] / (10^24 * m1) =
(G * m1 / d^2) * (1)

The closing acceleration is the sum of those two accelerations, (G * m1 /d^2) * (10^24 + 1)]

Similar for m2 and m3.

 

[ZapperZ]

Dropped objects hit the ground same time. Objects which are of point size with respect to Earth are attracted towards ground due to force of gravitation. Force of gravitation is given by formula F = G\frac{m_1m_E}{R^2}. When an object of mass m_1 is dropped means it has zero initial velocity. It experiences an acceleration towards Earth which is given by a = \frac{F}{m_1}. Depending on the mass of every object there can be variation in acceleration towards Earth over a wide range. Assume that all objects are dropped from the same height then on the basis of equation h = ut + 1/2 (at^2), time can vary over a range. I have not considered force due to air resistance.

This is very confusing and contradictory.

First of all, for m<<M, where M is mass of the Earth, then "g" is a constant. That is what we use in intro General Physics.

However, if you want to deal with this in general by stating that "... there can be variation in acceleration towards Earth over a wide range ...", then you can no longer use "h = ut + 1/2 (at^2)", because that kinematical equation was derived under the condition that acceleration is a constant!

Zz.

 

[OldYat47]

There can be variations in acceleration "towards Earth". Suppose we have two Earths. The mutual attraction is greater than g.
G * (Earth's mass * Earth's mass) / d^2 = g * (Earth's mass^2) / d^2.
The acceleration of each Earth = G * (Earth's mass) / d^2
The closing acceleration is the sum of both accelerations, in this case the closing acceleration = 2 * G * (Earth's mass) / d^2
If one of the objects is (Earth's mass / 2) the closing acceleration is greater. If you are sitting on "our" Earth the acceleration you observe will be different in the two cases.

 

[Ty Wendland]

There can be variations in acceleration "towards Earth". Suppose we have two Earths. The mutual attraction is greater than g.
G * (Earth's mass * Earth's mass) / d^2 = g * (Earth's mass^2) / d^2.
The acceleration of each Earth = G * (Earth's mass) / d^2
The closing acceleration is the sum of both accelerations, in this case the closing acceleration = 2 * G * (Earth's mass) / d^2
If one of the objects is (Earth's mass / 2) the closing acceleration is greater. If you are sitting on "our" Earth the acceleration you observe will be different in the two cases.

So I guess you agree with my initial hypothesis?

 

[RandyD123]

Not sure if this applies but I saw this on YouTube:

 

[OldYat47]

Yes and no. The heavier object will hit first, but Earth's movement will "contribute" only proportionally to the relative magnitudes of the masses. Imagine a planet identical to Earth, but with zero atmosphere. Take two objects identical except for mass. Take them to the same height and drop them. The more massive object will hit the "Earth2" first, but the difference in time between the two impacts will be in proportion to the mass of the objects relative to the mass of Earth2 [(In the case of my last example, The difference between (10^24 + 1) and (10^24 + 4), a really small difference]. The total travel distance of Earth2 before impact will be similarly proportional by comparative masses to the movement of the other mass.

 

[Janus]

Yes and no. The heavier object will hit first, but Earth's movement will "contribute" only proportionally to the relative magnitudes of the masses. Imagine a planet identical to Earth, but with zero atmosphere. Take two objects identical except for mass. Take them to the same height and drop them. The more massive object will hit the "Earth2" first, but the difference in time between the two impacts will be in proportion to the mass of the objects relative to the mass of Earth2 [(In the case of my last example, The difference between (10^24 + 1) and (10^24 + 4), a really small difference]. The total travel distance of Earth2 before impact will be similarly proportional by comparative masses to the movement of the other mass.

If you drop the objects individually and at different times, then the more massive object will take slightly less time to impact. However, if you drop them simultaneously, and side by side, then they will impact at the same time. The Earth cannot have two different values of acceleration over the same interval. Instead, it will accelerate towards the falling objects at a rate that is due to the sum the masses of the objects.

 

[berkeman]

If you drop the objects individually and at different times, then the more massive object will take slightly less time to impact. However, if you drop them simultaneously, and side by side, then they will impact at the same time. The Earth cannot have two different values of acceleration over the same interval. Instead, it will accelerate towards the falling objects at a rate that is due to the sum the masses of the objects.

And with this clarification restated, I think the question has been answered well. Thanks everybody for a surprisingly interesting discussion of a topic that seems obvious on the surface of it, but requires a careful problem definition to give a correct answer.