Devil's Advocate - Myth 1? Gravity and 2 objects falling.

[DevilsAdvocate]

Hello and welcome to my first post.

 

I am the Devils Advocate. I do thought experiments that are just plain evil. I start with something wrong and twisted and then try and persuade you everything you know is wrong and twisted.But with no further delay... my first topic:

Myth 1? Two objects fall identically due to gravity regardless of mass.

 

Really? Interesting.You can do this one at home.

 

1. Take an object, an egg will do, and drop it from your outstretched hand. Observe its fall. Now clear it up.

 

2. Now take an imaginary egg up to the moon on a quick field trip. Do the same experiment. Drop it from the same height. Let's call it a metre in both cases. It falls to the ground, a little slower, but in a similar manner. Lets say it takes x seconds to fall on earth and 1.5x seconds on the moon. The figures don't matter, but it takes longer on the moon.

 

Do you agree so far? YES?

 

3. Tell me why it takes longer for the egg to fall to the floor on the moon?-Because the earth has more mass than the moon- Therefore the moon's gravity is weaker than the earth's- Therefore the force making the egg fall on the moon is weaker, it goes slower and it takes longer.

 

Correct? YES?

 

4. But if Uncle Albert taught us anything, it's to look at the bigger picture! Although within our normal frame of reference, gravity means attraction of things downwards towards earth, it doesn't actually mean that.

 

Gravity is the force of attraction, (if that's what it is - see a later discussion, but here we're just debunking this Newtonian myth for the moment), between 2 masses. Between them.

 

What does that mean? It means that each object attracts the other. So as well as the earth attracting the apple, the apple (to a miniscule extent) attracts the earth. Not very much obviously. But then Eve only had a little nibble and see where that got us. Apples are potent. Try one?

 

5. So you've dropped an egg on your kitchen floor and you've dropped one up on the moon. The one on the moon took longer to fall (and you didn't have to clean it up). And we agreed that the reason for this slower fall on the moon was the larger mass of the earth.

 

Now I want to see what happens if we get rid of the egg and now do a control experiment with just the moon and the earth.

 

First I'd like you to drop the moon onto the earth, from exactly a metre high. Thanks. I never liked Colorado anyway.

 

Second, I'd like you to drop the earth on the moon from a height of exactly one metre.

 

Thirdly I'd like you to (you may have just done this twice already) suspend the two lumps, earth and moon, a metre away from each other in space and 'drop' them towards each other.

 

Watch what happens in each (should be pretty similar).My simple question is: do the earth and the moon fall together within the normal time for the earth (x seconds), for the moon (1.5x seconds) or longer, or shorter?

 

6. Hopefully you went for shorter. That is the correct answer.The earth and the moon fall together quicker than the earth and a set of car keys.Therefore if you drop the moon and a set of keys off the leaning tower of Pisa at the same time, the earth and the moon will get together before the keys get to first base.

 

7. Newton's gravity equation:

 

Shows the force of attraction between two masses, m1 and m2. If you increase the mass of one, you increase the force. So why is there this myth that objects fall at the same rate regardless of mass? The difference is tiny tiny tiny obviously, because the earth is the major influence on the force. But there is a difference. You drop a heavier object and the masses involved are larger, which means the forces are larger, which means the fall is faster.

 

8. There is one further point to add. To do with our frame of reference. Whenever something is taken from the mass of the earth and then dropped back onto it, the total masses involved in working out the gravity are the same as if any object on earth was taken from earth and dropped on the earth. Because it is a closed system.

 

You drop a tank on the earth, the total masses involved are that of the earth and that of the tank, which was just removed from the earth. So if you do that and then drop a mug on the earth, the total masses involved will be the same in each case, because the earth in the first case has the mass of the earth (including the mug, but minus the tank) and that is being combined with the tank. In the second, the earth has the mass of the tank included, but is missing the mug, but the earth and the mug's masses are combined to work out the fore of attraction anyway.

 

So if you have two identical objects and you drop the first on the earth, then you get the second one from the moon where it was being stored, and you drop the second one.... it will fall ever so slightly faster than the first, even though the objects were identical. Because the total masses involved are slightly higher and therefore the gravity of earth would have increased slightly.

AM I RIGHT? OR ARE THEY WRONG?

Edited November 30, 2010 by DevilsAdvocate

[John Cuthber]

I tried this sequence of experiments, but I couldn't get them to work.

I dropped an egg from one metre but I noticed that once I was no longer holding the egg and the earth apart, they both moved.

The egg went down, but the earth moved up. So, the egg never travelled the whole metre.

I tried it on the moon too. The moon, being lighter, moved up even more than the earth did.

Finally, when I dropped the moon onto the earth, neither the moon nor the earth actually travelled the whole metre.

At this point, I gave up because I didn't feel like doing the maths

 

Tell me, when you did this, did the earth move for you?

[swansont]

If you are using the distance from the earth to the moon, then you are not using an inertial reference frame for your measurement. You have to use a meter stick fixed in some frame; the CoM frame, where things start at rest, is probably most convenient.

[D H]

AM I RIGHT? OR ARE THEY WRONG?

You are long winded is what you are. I can't tell, given that long-wided diatribe, whether you are tilting at windmills or arguing something that has been known since Newton's time.

 

One consequence of the equivalence principle is that from the perspective of an inertial observer, different test objects placed at the same location will fall with the same acceleration independent of the test object's masses, internal structure, or composition. If you are arguing with this you are tilting at windmills. The equivalence principle stands as one of the most accurately tested hypotheses of all of physics. See http://einstein.stan...Cent-PW2005.pdf, for example.

 

If you are arguing that, ignoring air resistance, the amount of time it takes for a 1 milligram spec of dust to fall to the Earth is not the same as the amount of time it takes for a 10 million gram boulder to fall the same distance, that is a "well, duh". Newton's third law, and Newton's law of gravitation, says that the Earth also accelerates toward the spec of dust or boulder. (I'll ignore that Newton's laws are not quite the correct description of nature.) When you drop a spec of dust or a boulder toward the Earth and observe the acceleration from the surface of the Earth you are not an inertial observer. The Earth, and hence you, are accelerating toward the spec of dust / boulder.

 

Taking this into account, the acceleration of one body of mass m toward another body of mass M is G(M+m)/r². In the case of the spec of dust versus the boulder, this refinement doesn't amount to much at all. The acceleration with respect to the surface of the Earth experienced by the dust particle at the same distance from the center of the Earth as the boulder are the same to 18 decimal places. This is a very tiny effect. Here's a thought experiment that illustrates how tiny this effect is. Assume the ability to time the fall of some object with extreme precision. First we'll place a 1 milligram spec of dust at the bottom of an evacuated chamber, lift it 4.9 meters high, and time its fall. Now do the same with a 24 meter diameter spherical boulder with a mass of 10 million kilograms. Which will take the shorter amount of time to fall? The answer: The spec of dust. (Hint: the center of mass of the boulder is 12 meters from the surface of the boulder.)

 

That the gravitational acceleration of one body toward another is G(M+m)/r² can become significant in describing the motions of planets, binary stars, etc. Kepler's laws do not account for the mass of the planets. Newton's laws do. Kepler's third law describes the motion of Jupiter to only about three decimal places because Jupiter's mass is about 1/1000 that of the Sun. This effect is also critical in explaining the birth of planets. One a planetesimal reaches a significant mass it will orbit its star slightly faster than will the neighboring dust particles. The planetesimal will sweep a path through the dust, growing in size as it does so.

[Mr Skeptic]

This is why we scientists so love the equations, they tell you what is really going on. The things you have said in words can all be deduced from the equations, including your finding an unobservably slight exception to things falling the same speed on earth. Note that Earth's mass is constantly increasing due to falling space dust and micrometeorites, but earth is very heavy and these slight increases won't add up to a noticeable effect, at least not when compared to air resistance or turbulence in the air.

[ydoaPs]

[math]F=ma[/math]

 

[math]F_g=G\frac{Mm}{r^2}[/math]

 

[math]ma=G\frac{Mm}{r^2}[/math]

 

[math]a=G\frac{M}{r^2}

[/math]

It looks like the mass whose acceleration is being examined dropped out.

Edited November 30, 2010 by ydoaPs

[Janus]

 

 

6. Hopefully you went for shorter. That is the correct answer.The earth and the moon fall together quicker than the earth and a set of car keys.Therefore if you drop the moon and a set of keys off the leaning tower of Pisa at the same time, the earth and the moon will get together before the keys get to first base.

 

 

Nope, and here's why. First off, we'll replace the keys and moon with balls of equal size, but of the same respective masses of each. This makes it a fair race, as both the centers of the objects and their surfaces start at an equal distance from the surface of the Earth. It is true that if you drop the moon-massed sphere from the tower of Pisa, it will hit the ground sooner than if you separately drop the key-massed sphere from the same height. However, this is because the closing acceleration is equal to the sum of the falling acceleration of the sphere towards the Earth and the falling acceleration of the Earth towards the sphere. Since the Earth will fall towards the moon-massed sphere faster, they will hit sooner, even though both spheres fall towards the Earth at the same speed. However, when you drop them together, side by side, the Earth can't fall towards the moon-massed sphere without also falling towards the keys-massed sphere at the same rate. In fact, what happens is that the Earth falls toward the pair in response to the sum of their masses and the pair will hit the ground together just a hair sooner than the moon-massed sphere did by itself.

[D H]

I like my answer better: The keys.

[DevilsAdvocate]

Thank you for your braintime.

 

It's such a nice thing about the internet as opposed to the classroom that the ones who haven't got it are usually the ones to speak up. Much more effective teaching wise.

 

A literalist may assume from my vivid descriptions that I'm talking about dropping things on earth. I'm not really. That's just fun. I'm talking about dropping things in a perfect spherical world of theory.

 

And in this world (very much like in our world)...

 

Unfortunately you cannot drop two objects from exactly the same place via exactly the same space and therefore the earth will approach the heavier faster, admittedly imperceptibly to all but the most gigantically miniscule pedant, such as [drum roll...] the Devil's Advocate

 

The serious point is that what is often inferred from the lovely idea that a feather and a ton weight drop at the same rate without air resistance is that mass is not an important factor in gravity.

 

It is.

 

It's just that the total mass of the two objects involved (the faller and the fallen towards in most cases) is what counts. And here on earth the fallen towards always outmasses the faller by a gigantic ratio.

 

You who have commented know this already and may see it as too obvious to bother pointing out. Which is exactly why I've pointed it out, because you're obviously the ones who explain things around here.

 

Ta-dah!

 

Thankyou.

 

Devil's Advocate

[D H]

How about if you apply some of your braintime? Stop with the BS, do the math.

[Cap'n Refsmmat]

It's just that the total mass of the two objects involved (the faller and the fallen towards in most cases) is what counts.

No, not in a Newtonian world.

 

Consider two masses, m₁ and m₂, a distance r apart. They are gravitationally attracted to each other with corresponding forces:

 

[math]F_1 = G \frac{m_1 m_2}{r^2}[/math]

 

[math]F_2 = G \frac{m_2 m_1}{r^2}[/math]

 

Now, we know that [imath]F=ma[/imath]:

 

[math]F_1 = m_1 a_1 = G \frac{m_1 m_2}{r^2}[/math]

 

[math]F_2 = m_2 a_2 = G \frac{m_2 m_1}{r^2}[/math]

 

Cancelling...

 

[math]a_1 = G \frac{m_2}{r^2}[/math]

 

[math]a_2 = G \frac{m_1}{r^2}[/math]

 

Their accelerations are independent of their own masses, depending only on the mass of the other object. So if I drop two objects of different masses, their acceleration in Earth's gravity will be identical.

 

Now, it's true that the Earth moves as well as the dropped object, but it is not true that the objects fall any faster or slower.

[D H]

So if I drop two objects of different masses, their acceleration in Earth's gravity will be identical.

In Devil's Advocates favor, Cap'n, you are assuming an inertial observer. To an observer attached to either of the two masses, the observed acceleration of the relative acceleration between the two masses is G(m₁+m₂)/r².

 

DA appears to be thinking that he (I am assuming DA is a male; most crackpots are) has made a great discovery. Not at all. In his disfavor, he has not done the math. So, DA, some challenges. Both thought experiments involve spherically symmetric, non-spinning objects with no atmosphere. Assume Newton's law of gravitation.

 

Case 1: Two planets of equal density but different radii are falling straight toward one another. A tiny space bug of inconsequential mass sits on the surface of the smaller planet at the point directly between the two centers of mass. Will the bug be smashed by the collision or will it at some point fall free of the smaller planet, land on the larger planet prior to the collision, and be able to scurry off to safety?

 

Case 2 (my previous example): A 10 million kilogram, 24 meter diameter boulder and a 1 milligram spec of dust are resting on the surface of a planet with radius R=6378 km and mass M such that GM/R² is 10 m/s². The boulder and dust spec are lifted from the surface such that the distance between the bottom of the boulder and the surface of the Earth and the distance between the spec of dust and the surface of the Earth are each 5 meters. The two objects are released simultaneously. Which hits first?

 

 

Don't just use words. Do the math.

Edited December 1, 2010 by D H

[John Cuthber]

In the spirit of this thread (i.e. pointless nonsense) I will point out that a milligram is actually rather heavy for a speck of dust.

[swansont]

In the spirit of this thread (i.e. pointless nonsense) I will point out that a milligram is actually rather heavy for a speck of dust.

 

And one might think it rude to point that out. All of the other specks of dust already make fun of him. It's a glandular thing.