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(io9)   Remember Galileo dropping two different-sized cannonballs off the leaning tower of Pisa to prove objects of different mass fall at the same rate? Yeah, about that   (io9.com) divider line 98
    More: Interesting, Pisa, mass fall, interest rates  
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10713 clicks; posted to Geek » on 03 Feb 2013 at 6:25 PM (1 year ago)   |  Favorite    |   share:  Share on Twitter share via Email Share on Facebook   more»



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2013-02-04 01:05:43 AM

DON.MAC: Smeggy Smurf: I always figured the object with the greater mass fell faster because the more massive object's additional gravity helped things along

If you have two items of the same size and weight but one has a denser core, it will fall faster because pull of he side of the higher density areas is closer to the center line in theory.  It would take some impressive equipment to do that experiment and get any useful data since a fall of meters might have fall times that differ far less than a trillionth of a second.


I'd guess the difference would be something on the order of octillionths of a second.  And would probably be dwarfed by the effects of sporadic thermal vibrations.
 
2013-02-04 01:06:41 AM

doglover: Actually science journalism is almost dead in America.


Fortunately, we have Science News, which I discovered thirty years ago and have subscribed to ever since. It's damn fine science journalism, and it has something for everyone. Not only are the stories in-depth and well-written, they still seem to employ copy-editors, so I don't feel the urge to strangle someone every few sentences.
 
2013-02-04 01:40:16 AM
His first experiment failed badly.
img832.imageshack.us
 
2013-02-04 01:42:22 AM
So if

upload.wikimedia.org

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.
 
2013-02-04 01:43:18 AM
I am old enough to remember watching the Apollo demonstration live in realtime: a heavy hand tool versus an eagle feather, dropped in lunar gravity. They hit at the same time. I always wondered if the ramps used by Gallileo with his balls had a difference in friction effects while rolling down the incline. Notice how I didn't snicker once while saying that.
 
2013-02-04 01:46:51 AM

jigger: So if



you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.


So much this. Congratulations on being one of the few non-crackpots in this thread.
 
2013-02-04 02:02:45 AM

aerojockey: DON.MAC: Smeggy Smurf: I always figured the object with the greater mass fell faster because the more massive object's additional gravity helped things along

If you have two items of the same size and weight but one has a denser core, it will fall faster because pull of he side of the higher density areas is closer to the center line in theory.  It would take some impressive equipment to do that experiment and get any useful data since a fall of meters might have fall times that differ far less than a trillionth of a second.

I'd guess the difference would be something on the order of octillionths of a second.  And would probably be dwarfed by the effects of sporadic thermal vibrations.


And half a billion other things that would effect it more....  like thermal noise leaving the falling spheres.

I think the last great experiment that showed that gravity may not be as well understood as we thought was the Pioneer Anomaly but like all good science, it lead to hard questions and when those questions were answered, the answer was "yep what Galileo,Newton and Einstein said"
 
2013-02-04 02:03:04 AM

Zavulon: jigger: So if

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.

So much this. Congratulations on being one of the few non-crackpots in this thread.


Nope.  The two objects would accelerate at exactly the same rate (at first).

For object 1, the accerlation is given by:

a1 = Fg / m1 = (G * m1 * mE / r^2) / m1 = G * mE / r^2

For object 2, the acceleration is given by:

a2 = Fg / m2 = (G * m2 * mE / r^2) / m2 = G * mE / r^2

That is, the acceleration is independent of the object's mass.

The thing that will make the heavier object fall (very so slightly) faster is that the Earth is also drawn to the objects, and the Earth will move slightly faster toward the heavier object than the lighter one, thus decreasing the radius faster on the heavier object than the lighter object, meaning the heavier object starts to accelerate faster.
 
2013-02-04 02:07:31 AM

RexTalionis: Without reading article, no Galileo never did that. He slowed down the fall by rolling both cannonballs off an incline so he can actually observe it.


He did neither, it was a thought experiment based on known principles of seigecraft, which the Italians were some of the best in the world at using.

Basically "I drop two equally-sized cannonballs next to each other, they fall in a certain time.  Now I tie the two balls together with some rope.  Do the balls now fall faster?  No, that's not how it works.  Ergo, increasing mass does not increase falling speed.
 
2013-02-04 03:27:38 AM
If you were to perform this experiment in real life from a decent height with two perfectly smooth balls with equal sizes and different masses, the heavier one will in fact fall faster. As previously stated, the two objects will feel different forces of attraction that result in equal accelerations due to gravity. However, air resistance has a noticeable effect over a few stories. You see, the more massive sphere has greater inertia to overcome the damping effect of air resistance. The result is that the more massive sphere actually accelerates a bit faster and reaches a higher terminal velocity.


/Of course, we could also account for things like the transversal distance between the two spheres due to their mutual attraction
//But let's avoid that rabbit hole
 
2013-02-04 04:19:35 AM

jigger: So if

[upload.wikimedia.org image 200x140]

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.


came to post this. the balls pull on the earth at the same time the earth pulls on the balls (ow my balls) so the heavier ball would pull the earth towards itself slightly more than the smaller ball. dropped at the same time the balls would hit at the same time as they have a combined pull on the earth but dropped independently the heavier ball would hit in a faster time. Of course this ignores air resistance.
 
2013-02-04 04:42:28 AM

DON.MAC: jack21221: DON.MAC: If you have two items of the same size and weight but one has a denser core, it will fall faster because pull of he side of the higher density areas is closer to the center line in theory.  It would take some impressive equipment to do that experiment and get any useful data since a fall of meters might have fall times that differ far less than a trillionth of a second.

Gauss's law disagrees with you. Unless you can break out some math, I do believe that the theoretical difference in the rate of falling is 0.

Hmmm... let me grab a copy of Dr Parkinson's Gravity Probe B paper and copy and paste the math.... oh... I don't seem to have enough room.

I'm guessing that the zero you talk about is a good round number.  I'm guessing the real difference would be something on the order of one test sphere having one more atom than the other but my second guess would be the ratio of the difference would be something in the order of 1 to 10-75.


Uh. Gauss's law is Gauss's law. Unless you can explain a way "the field lines exiting a surface don't add up to the charge inside", you can't break it. And that law is backed by some significant math, not just physical observations
 
2013-02-04 06:11:06 AM
The Earth is a closed system, so no net external effect - see  MythBusters Episode 77: Birds in a Truck

/You're welcome Professor
 
2013-02-04 07:04:27 AM
There's a really interesting article to be written about the amazing fact that gravitational mass appears to be identical to inertial mass (Einstein's equivalence principle), although nobody has a compelling explanation of why this should be -- which is one reason physicists like to speculate about whether there are corner cases where they aren't -- but this isn't it.

For instance, I read an interesting paper about the Unruh effect recently that proposes an actually physically plausible mechanism for violation of the Equivalence Principle at very small accelerations. The math is way beyond me, but the authors speculate that the effect might be large enough to account for at least some of the anomalous acceleration of gravity conventionally attributed to Dark Matter. Good stuff.
 
2013-02-04 07:26:30 AM
Everybody knows that Custer died at the battle of Little Big Horn. What this book presupposes is....maybe he didn't?
 
2013-02-04 07:34:30 AM
Ugh, take an orbital dynamics course already.  This isn't even close to a new idea.  Yes, gravity works between all bodies and not just the orbital order we super-impose on it.  Hint:  watch the CoG of the earth and moon combined (somewhere in the crust) relative to the distance from the sun.  Yes, this means we're slightly closer to the sun during a full moon.
 
2013-02-04 08:08:21 AM

MayoSlather: In later experiments I conclusively proved Galileo wrong by dropping a paper airplane and a dumbbell off of a local Walmart.


How did you get the manager on the roof?
 
2013-02-04 08:13:55 AM

DON.MAC: Hmmm... let me grab a copy of Dr Parkinson's Gravity Probe B paper and copy and paste the math.... oh... I don't seem to have enough room.

I'm guessing that the zero you talk about is a good round number.  I'm guessing the real difference would be something on the order of one test sphere having one more atom than the other but my second guess would be the ratio of the difference would be something in the order of 1 to 10-75.


You didn't have the room to link to the paper with a page reference? You're doing an awful lot of guessing. I'm telling you that outside of the surface of an object, the gravity acts exactly as if all of the mass were concentrated at the center of mass, regardless of internal composition. This is true both in Newtonian gravity and in GR. That's how I learned it. If you have something difference, you need to post a reference and stop "guessing."
 
2013-02-04 08:15:10 AM
Well I was going to come in and dazzle you all with science and physics while showing that mass does affect acceleration, just not to a noticable amount in reference to objects on the earth, but it looks like its already been covered.

So... uh.. hi.
 
2013-02-04 08:23:52 AM

lordargent: The Nordtvedt effect has been tested, and so far no evidence has been found that the more massive Earth is falling towards the sun faster than the moon. If there is an effect, it's very slight.

So an unproved theory, wake me up when you have actual data that supports the theory.


Actually, it's not even a theory, in the scientific sense.
 
2013-02-04 08:26:11 AM

MontanaDave: lordargent: The Nordtvedt effect has been tested, and so far no evidence has been found that the more massive Earth is falling towards the sun faster than the moon. If there is an effect, it's very slight.

So an unproved theory, wake me up when you have actual data that supports the theory.

What's more, he's from Montana State. Bobcats are inherently untrustworthy.

/Maroon and Silver blood.
//GO GRIZ!


Hey, how did the Griz do in the I-AA playoffs this year?  Oh yeah...

/NDSU grad, go Bison!!
//I know a player from the Griz who is now in the NFL...
 
2013-02-04 08:35:20 AM

LrdPhoenix: Well, of course it does.  This is kinda like how Newton was mostly right unless you're traveling at relativistic speeds and what not.  For objects of extremely low mass, like everything we deal with every day, the pull exerted on the Earth is so negligible that it really isn't even necessary to consider it, thus everything accelerates at about the same rate, +- 10-30 m/s or whatever the actual error would be.  It's really the mass difference between the two objects that matter when disregarding the smaller mass, so the Earth or Moon compared to the Sun is like a 20 pound cannonball or a 10 pound cannonball compared to Earth.

Otherwise, the actual equation for acceleration due to gravity is F= G * ((m1 * m2) / r2), which makes it quite obvious that both masses effect acceleration.  If a second Earth mass planet magically popped up nearby, the rate at which it would accelerate towards the Earth would be at least twice that of a much less massive object, like the cannonball.  Well, really, this second planet would accelerate towards the earth at the usual rate we're used to, but the Earth would also accelerate towards this second planet by the same amount, thus from the reference point of the Earth (and all reference points must be valid), the second planet would be accelerating twice as fast as a much less massive object.


Except, F=ma (Newton's second law)
So, m1a = G*((m1 *m2)/r2), or a = G*m2/r2
 
2013-02-04 08:38:00 AM

jigger: So if

[upload.wikimedia.org image 200x140]

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.


Acceleration is proportional to mass too, though, so as the force increases the force required to provide the exact same acceleration increases at the exact same rate.
 
2013-02-04 08:42:31 AM

dready zim: jigger: So if

[upload.wikimedia.org image 200x140]

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.

came to post this. the balls pull on the earth at the same time the earth pulls on the balls (ow my balls) so the heavier ball would pull the earth towards itself slightly more than the smaller ball. dropped at the same time the balls would hit at the same time as they have a combined pull on the earth but dropped independently the heavier ball would hit in a faster time. Of course this ignores air resistance.


This would not happen in any scale we could measure accurately.  You'll notice that you're ignoring the effects of the moon in this.  Why?  Because it's on a scale so immeasurably small that it's irrelevant.
 
2013-02-04 09:32:35 AM
That wasn't me, Mr. Woodman, that was Crazy Delaney.

No, no, Kotter.No, no. No, Delaney was the maniac who dropped two teachers of different weights from the gymnasium window to see if they hit the ground at the same time. He said it was a physics experiment.

www.dvdtalk.com
 
2013-02-04 09:35:11 AM
LrdPhoenix:Otherwise, the actual equation for acceleration due to gravity is F= G * ((m1 * m2) / r2), which makes it quite obvious that both masses effect acceleration.

You're trolling, right?

F = ma = G * ((m1 * m2) / r2)

Divide by the mass you're finding the acceleration for and you end up with the same term for "a" no matter the mass of the object you're dropping.
 
2013-02-04 09:35:29 AM

meanmutton: dready zim: jigger: So if

[upload.wikimedia.org image 200x140]

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.

came to post this. the balls pull on the earth at the same time the earth pulls on the balls (ow my balls) so the heavier ball would pull the earth towards itself slightly more than the smaller ball. dropped at the same time the balls would hit at the same time as they have a combined pull on the earth but dropped independently the heavier ball would hit in a faster time. Of course this ignores air resistance.

This would not happen in any scale we could measure accurately.  You'll notice that you're ignoring the effects of the moon in this.  Why?  Because it's on a scale so immeasurably small that it's irrelevant.


What if a gnat flies over one of the objects. Wouldn't that slow it down?
 
2013-02-04 09:39:45 AM

meanmutton: LrdPhoenix: Well, of course it does.  This is kinda like how Newton was mostly right unless you're traveling at relativistic speeds and what not.  For objects of extremely low mass, like everything we deal with every day, the pull exerted on the Earth is so negligible that it really isn't even necessary to consider it, thus everything accelerates at about the same rate, +- 10-30 m/s or whatever the actual error would be.  It's really the mass difference between the two objects that matter when disregarding the smaller mass, so the Earth or Moon compared to the Sun is like a 20 pound cannonball or a 10 pound cannonball compared to Earth.

Otherwise, the actual equation for acceleration due to gravity is F= G * ((m1 * m2) / r2), which makes it quite obvious that both masses effect acceleration.  If a second Earth mass planet magically popped up nearby, the rate at which it would accelerate towards the Earth would be at least twice that of a much less massive object, like the cannonball.  Well, really, this second planet would accelerate towards the earth at the usual rate we're used to, but the Earth would also accelerate towards this second planet by the same amount, thus from the reference point of the Earth (and all reference points must be valid), the second planet would be accelerating twice as fast as a much less massive object.

Except, F=ma (Newton's second law)
So, m1a = G*((m1 *m2)/r2), or a = G*m2/r2


You failed to read my last sentence.
 
2013-02-04 09:42:09 AM

StrangeQ: LrdPhoenix:Otherwise, the actual equation for acceleration due to gravity is F= G * ((m1 * m2) / r2), which makes it quite obvious that both masses effect acceleration.

You're trolling, right?

F = ma = G * ((m1 * m2) / r2)

Divide by the mass you're finding the acceleration for and you end up with the same term for "a" no matter the mass of the object you're dropping.


You also failed to read my last sentence.
 
2013-02-04 09:49:43 AM

LrdPhoenix: You also failed to read my last sentence.


As long as you're willing to state that the Sun is accelerating away from the stationary Earth, you're correct. In fact, the math works out the same if you assume the Earth is completely stationary and the entire universe is rotating around it. It's just that most people aren't happy doing that.

In any case, that's not what this article is about.
 
2013-02-04 10:11:17 AM

LrdPhoenix: StrangeQ: LrdPhoenix:Otherwise, the actual equation for acceleration due to gravity is F= G * ((m1 * m2) / r2), which makes it quite obvious that both masses effect acceleration.

You're trolling, right?

F = ma = G * ((m1 * m2) / r2)

Divide by the mass you're finding the acceleration for and you end up with the same term for "a" no matter the mass of the object you're dropping.

You also failed to read my last sentence.


Okay, then, here you go.  How much, exactly, is the Earth moving by you dropping something?  Let's go ahead and assume that you're going to drop something that is 1 million kg from a distance of one meter.  How much is the Earth going to move?

1 million kg has a weight of ~10 million N (assuming 10 m/s/s for gravitational acceleration) and thus that's the force that the item would exert on the Earth (equal and opposite and all that).

Using 10 m/s/s for gravitational force at the surface of the Earth (rounding makes math easier), you end up with about 10 million N of force applied by the object against the Earth.  10 MN sounds like a lot, right?  Well, the Earth is about 6x1024kg, so we go with 10x107N / 6x1024kg = acceleration of 1.7x10-17m/s/s.

Dropping an object one meter takes about .45 seconds (sqrt(2*1m/10m/s/s)) = sqrt (.2) = .45s.

d = 1/2 * a * t2 = .5 * 1.7x10-17 * .2 = 1.7x10-18m.

Please check my math because doing math in a window like this is a pain.

That said, 1.7x10-18m is tiny.  For comparison, a proton has a diameter of about 10-15m.  So, you are looking at a difference in position of something 1/1,000th the size of a proton.
 
2013-02-04 10:12:48 AM
This sounds like something out of the Annals of Improbable Research.  Guy has an idea, guy tests idea, guy finds out idea is worthless.

/Wouldn't the "gravitational self-energy" cancel each other out? Ultimately the only thing that matters is the gravitational force between the two objects.  Force (or acceleration) is a vector, after all.
 
2013-02-04 10:32:13 AM

meanmutton: LrdPhoenix: StrangeQ: LrdPhoenix:Otherwise, the actual equation for acceleration due to gravity is F= G * ((m1 * m2) / r2), which makes it quite obvious that both masses effect acceleration.

You're trolling, right?

F = ma = G * ((m1 * m2) / r2)

Divide by the mass you're finding the acceleration for and you end up with the same term for "a" no matter the mass of the object you're dropping.

You also failed to read my last sentence.

Okay, then, here you go.  How much, exactly, is the Earth moving by you dropping something?  Let's go ahead and assume that you're going to drop something that is 1 million kg from a distance of one meter.  How much is the Earth going to move?

1 million kg has a weight of ~10 million N (assuming 10 m/s/s for gravitational acceleration) and thus that's the force that the item would exert on the Earth (equal and opposite and all that).

Using 10 m/s/s for gravitational force at the surface of the Earth (rounding makes math easier), you end up with about 10 million N of force applied by the object against the Earth.  10 MN sounds like a lot, right?  Well, the Earth is about 6x1024kg, so we go with 10x107N / 6x1024kg = acceleration of 1.7x10-17m/s/s.

Dropping an object one meter takes about .45 seconds (sqrt(2*1m/10m/s/s)) = sqrt (.2) = .45s.

d = 1/2 * a * t2 = .5 * 1.7x10-17 * .2 = 1.7x10-18m.

Please check my math because doing math in a window like this is a pain.

That said, 1.7x10-18m is tiny.  For comparison, a proton has a diameter of about 10-15m.  So, you are looking at a difference in position of something 1/1,000th the size of a proton.


And you failed to read the rest of my paragraph.
 
2013-02-04 10:35:27 AM

LrdPhoenix: meanmutton: LrdPhoenix: StrangeQ: LrdPhoenix:Otherwise, the actual equation for acceleration due to gravity is F= G * ((m1 * m2) / r2), which makes it quite obvious that both masses effect acceleration.

You're trolling, right?

F = ma = G * ((m1 * m2) / r2)

Divide by the mass you're finding the acceleration for and you end up with the same term for "a" no matter the mass of the object you're dropping.

You also failed to read my last sentence.

Okay, then, here you go.  How much, exactly, is the Earth moving by you dropping something?  Let's go ahead and assume that you're going to drop something that is 1 million kg from a distance of one meter.  How much is the Earth going to move?

1 million kg has a weight of ~10 million N (assuming 10 m/s/s for gravitational acceleration) and thus that's the force that the item would exert on the Earth (equal and opposite and all that).

Using 10 m/s/s for gravitational force at the surface of the Earth (rounding makes math easier), you end up with about 10 million N of force applied by the object against the Earth.  10 MN sounds like a lot, right?  Well, the Earth is about 6x1024kg, so we go with 10x107N / 6x1024kg = acceleration of 1.7x10-17m/s/s.

Dropping an object one meter takes about .45 seconds (sqrt(2*1m/10m/s/s)) = sqrt (.2) = .45s.

d = 1/2 * a * t2 = .5 * 1.7x10-17 * .2 = 1.7x10-18m.

Please check my math because doing math in a window like this is a pain.

That said, 1.7x10-18m is tiny.  For comparison, a proton has a diameter of about 10-15m.  So, you are looking at a difference in position of something 1/1,000th the size of a proton.

And you failed to read the rest of my paragraph.


By the way, the reason I know that is that you just restated my first paragraph but added in actual math.
 
2013-02-04 10:53:27 AM

jack21221: You didn't have the room to link to the paper with a page reference? You're doing an awful lot of guessing. I'm telling you that outside of the surface of an object, the gravity acts exactly as if all of the mass were concentrated at the center of mass, regardless of internal composition. This is true both in Newtonian gravity and in GR. That's how I learned it. If you have something difference, you need to post a reference and stop "guessing."


You can find examples that show it isn't that simple in Newtonian gravity with tidal forces and non of the gravity measuring satellites would work.  It is true for the average of the mass, but the forces have a different effect as the density varies.

The copy of the paper I have is more than a decade old and half way around the world (and is on real paper so it isn't easy to copy and pasteable).
The gravity detector in those satellites works by having slight differences forces of the leading spin side of the gyroscope vs the back side which will slow it down or speed it up based on the relativistic changes in gravity.  If you care to integrate the over all the mass in the earth and the two falling sphere of different density but the same mass, you will find a non-zero number that is useless as I expect to to be something on the order of 50 orders of magnitude smaller than any other number commonly used in physics.

You could also do the 10^75 ^ (10^75) summations of the standard two body gravity equation for all the atoms in both spheres of two different densities. The result is will also be so close to zero that it is immaterial but it will not be exactly zero.  But that would assume you have a computer that can cope with numbers of that precision that are that small without massive rounding problems.

That sat was designed by the guy who worked out the relativistic issues of clocks for the GPS system and came up with a the 12 degree polar vector polynomial that every GPS system uses to figure out how to direct drivers to drive into rivers.
 
2013-02-04 11:34:15 AM

ecmoRandomNumbers: My whole science education has be a LIE! a LIE!


Apparently so has your English education.
 
2013-02-04 11:52:35 AM

vinniethepoo: MayoSlather: In later experiments I conclusively proved Galileo wrong by dropping a paper airplane and a dumbbell off of a local Walmart.

How did you get the manager on the roof?


He used a barometer?
 
2013-02-04 12:35:17 PM

Neondistraction: King Something: Galileo Figaro!

Magnifico!


I hate you both.


Beazbulb blah blah blah
 
2013-02-04 01:16:46 PM

AbiNormal: Next you're gonna tell me an apple didn't fall on Newton's head. BLASPHEMER!!


No, silly it was an apple shot off his head with an arrow.
 
2013-02-04 03:03:16 PM

jack21221: LrdPhoenix: You also failed to read my last sentence.

As long as you're willing to state that the Sun is accelerating away from the stationary Earth, you're correct. In fact, the math works out the same if you assume the Earth is completely stationary and the entire universe is rotating around it. It's just that most people aren't happy doing that.


I am quite happy to say that, because as Einstein pointed out, you can never tell what is stationary and what is moving, as it's all relative to the point of reference, and all points of reference are equally valid.  Person A says Person B is traveling at him at 100 km/h, Person B says Person A is traveling towards him at 100 km/h, Person C says that Person A and B are traveling towards each other at 50 km/h each, and they all must be correct.  What's more,  A would say B's total mass and energy has been increased by X due to his velocity, B would say that A's total mass and energy has been increased by X due to his velocity, and C would say that both A and B's total mass/energy has been increased by X/2 each, and they also must all be correct, at least inside their own reference frames.
 
2013-02-04 03:24:01 PM

LrdPhoenix: I am quite happy to say that, because as Einstein pointed out, you can never tell what is stationary and what is moving, as it's all relative to the point of reference, and all points of reference are equally valid.  Person A says Person B is traveling at him at 100 km/h, Person B says Person A is traveling towards him at 100 km/h, Person C says that Person A and B are traveling towards each other at 50 km/h each, and they all must be correct.  What's more,  A would say B's total mass and energy has been increased by X due to his velocity, B would say that A's total mass and energy has been increased by X due to his velocity, and C would say that both A and B's total mass/energy has been increased by X/2 each, and they also must all be correct, at least inside their own reference frames.


Choosing a reference frame appropriately makes the math a hell of a lot simpler, though.

In any case, that's not what this article is about.
 
2013-02-04 03:55:36 PM

DON.MAC: You can find examples that show it isn't that simple in Newtonian gravity with tidal forces and non of the gravity measuring satellites would work. It is true for the average of the mass, but the forces have a different effect as the density varies.


So... would it be true that Gauss's law would hold for the idealized situation of a completely rigid body in a static configuration? But once you have a real body with finite size, finite rigidity, and finite speed of propagation of the gravitational force, you get these tiny corrections?
 
2013-02-04 06:02:07 PM
"Hey, Galileo...'member that time you had those cannonballs and did stuff?  That was cool."

chrisfarley.jpg
 
2013-02-04 06:05:32 PM

czetie: So... would it be true that Gauss's law would hold for the idealized situation of a completely rigid body in a static configuration? But once you have a real body with finite size, finite rigidity, and finite speed of propagation of the gravitational force, you get these tiny corrections?


I would say so.  But these tiny corrections are so small as to not be measurable at all.  They are also so damn small that if you found the ratio fit in to a constant in a formula, it might be the smallest constant in modern physics.  I expect something (not this) also is why nearly everything in the observable universe is spinning in the same direction as everything local to it where local can be a huge number.
 
2013-02-04 06:56:52 PM

aerojockey: Zavulon: jigger: So if

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.

So much this. Congratulations on being one of the few non-crackpots in this thread.

Nope.  The two objects would accelerate at exactly the same rate (at first).

For object 1, the accerlation is given by:

a1 = Fg / m1 = (G * m1 * mE / r^2) / m1 = G * mE / r^2

For object 2, the acceleration is given by:

a2 = Fg / m2 = (G * m2 * mE / r^2) / m2 = G * mE / r^2

That is, the acceleration is independent of the object's mass.

The thing that will make the heavier object fall (very so slightly) faster is that the Earth is also drawn to the objects, and the Earth will move slightly faster toward the heavier object than the lighter one, thus decreasing the radius faster on the heavier object than the lighter object, meaning the heavier object starts to accelerate faster.



The a1 and a2 above are only for one direction. If you do the same thing with the mass of the Earth you'll get two different accelerations of the Earth toward the smaller objects. Well, ok. I guess I was thinking of the "acceleration" as the change in the rate at which the two objects approach one another, but I guess that's not the way to think about it.

But you know, I didn't even think about the effect of the two small objects being right next to each other and falling toward the Earth at the same time. Since the Earth is so much larger and the forces holding it together are so much stronger than gravity, especially the gravity of two small balls, that there really only would be one acceleration for both balls, and there wouldn't even be an imperceptible difference in the accelerations. Unless the Earth immeasurably deformed differently under where each ball was falling. No, I don't see it happening, so I guess the acceleration of the two balls, if dropped simultaneously side-by-side, would be identical (and not just identical as we can measure).
 
2013-02-04 07:03:17 PM

LrdPhoenix: I am quite happy to say that, because as Einstein

Lorentz pointed out, you can never tell what is stationary and what is moving, as it's all relative to the point of reference, and all points of reference are equally valid.
 
2013-02-05 04:17:04 PM

GameSprocket: meanmutton: dready zim: jigger: So if

[upload.wikimedia.org image 200x140]

you have different forces for different masses. The acceleration depends on the masses of both objects, therefore the acceleration is different for a heavier ball and a lighter ball. Of course, if we're talking about two object easily held in the hand versus the Earth, then the difference would be negligible to the point of almost being nonexistent. But, technically, they would fall at "different" rates.

came to post this. the balls pull on the earth at the same time the earth pulls on the balls (ow my balls) so the heavier ball would pull the earth towards itself slightly more than the smaller ball. dropped at the same time the balls would hit at the same time as they have a combined pull on the earth but dropped independently the heavier ball would hit in a faster time. Of course this ignores air resistance.

This would not happen in any scale we could measure accurately.  You'll notice that you're ignoring the effects of the moon in this.  Why?  Because it's on a scale so immeasurably small that it's irrelevant.

What if a gnat flies over one of the objects. Wouldn't that slow it down?


yes it would, as was said, not really measurable but still by a small amount. My example deliberately missed out all the bodies in the universe except earth and two balls to simplify and show a concept. Sometimes it is really important to notice that there is a difference between zero and really really small.
 
2013-02-05 06:53:29 PM

jack21221: I'm telling you that outside of the surface of an object, the gravity acts exactly as if all of the mass were concentrated at the center of mass, regardless of internal composition.

This is true both in Newtonian gravity and in GR.

Charge too (and magnetism too, until a monopole is observed IIRC).  Works basically for any inverse-square law...

brachiopod: He used a barometer?


heh - but you offer the barometer in exchange for the manager telling you the height of the bldg - he won't go up there. Unless he takes you up on the offer, and you then proceed to tell him that you think he's wrong and will prove it by precisely timing how long it takes it to hit the ground when you drop it.
 
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