Could Galileo's experiment have been wrong?

[RESmonkey]

**NOTE: This is theory. Not proven. Galileo is still 'correct' right now, haha. Title is more of a pun. What the article is about: there may be some tiny tiny tiny thing to gravity that would not be equal on two different mass objects...one *could* hit the ground earlier than the other. Some weak weak weak force is the cause of it. Gotta find it; it will, in some way I don't know of, help string theory.

http://science.nasa.gov/headlines/y2007...ay_equivalenceprinciple.htm?list844236

Enjoy.

 

[lyssword]

he was close enough

 

[eLiu]

Within the accuracy of the simultaneous release, the objects were observed to undergo the same acceleration and strike the lunar surface simultaneously, which was a result predicted by well-established theory, but a result nonetheless reassuring considering both the number of viewers that witnessed the experiment and the fact that the homeward journey was based critically on the validity of the particular theory being tested.

Hahahahaha

 

[skreet]

Sigh, can't get to the link

 

[JonB]

Here's a similar (and working) link:

http://www.redorbit.com/news/space/9402...inciple_was_galileo_correct/index.html

 

[Karot]

Well the only other forces that we know to exist besides gravity are the strong force, weak force, and EM.

 

[tank171]

It is to my understanding that all mass has its own gravity. The formula is F= (Gravitational constant) times (mass 1) times (mass 2) / (distance)^2. So a heavier object would have more pull than a lighter one. This amount of gravity is miniscule, so it would not be easy to spot by the naked eye, and it would be hard to get two objects to drop at the same exact time.

Just my 2 cents.

 

[f95toli]

Yes, but F=ma which means that the mass of the object that is falling cancels out from the equation. The rest of the equation (the mass of the earth and the distance to the centre of mass) can be combined into a constant g=9.8 m/s^2, which is why ALL objects regardless of mass accelerate at that rate when they fall to the ground on earth.

 

[DrPizza]

I know I'm overthinking this somehow, and overlooking something...

And, I agree that all objects near the earth's surface would have the same acceleration. But (and maybe I'm overlooking something because I'm tired; I never had this particular thought before) because forces are equal and opposite, that is, the force of gravity exerted on the object by the earth is equal and opposite in direction to the force on the earth exerted by the object, wouldn't the earth's acceleration upward toward the falling object be just slightly greater in the case of a more massive object - actually decreasing the amount of time to impact by a miniscule amount?

 

[tank171]

This is not quite about the main topic, but, DrPizza, .999999... doesnt equal 1. It equals .999999.... But, .999999..., when converted to a fraction is 1. There is a difference between the decimal and fraction form.

 

[CycloWizard]

Originally posted by: tank171
This is not quite about the main topic, but, DrPizza, .999999... doesnt equal 1. It equals .999999.... But, .999999..., when converted to a fraction is 1. There is a difference between the decimal and fraction form.

*PUNT*

 

[Biftheunderstudy]

Although in theory the factor could be tiny, the current reigning champ says that they are equal. Not just close but identical. As DrPizza said, the little m's cancel and the acceleration only depends on the gravitating object.

One reason I can think of to ensure this is that even if this tiny force were one part in say 10^20, things like stars which have 10^30+ kilograms and galaxies would have vastly different orbital properties--I think this is something we would have noticed.

That is, of course, assuming that this force acts at large distances and on large objects.

 

[grant2]

Originally posted by: Biftheunderstudy
One reason I can think of to ensure this is that even if this tiny force were one part in say 10^20, things like stars which have 10^30+ kilograms and galaxies would have vastly different orbital properties--I think this is something we would have noticed.

1) Theorists believe the new force would be at least ten million million (10^13) times weaker than gravity. ... i.e., any "vastly orbital property" would still be miniscule in cosmic terms.

If a galaxy's orbit were off by a few meters, how could we possibly determine that?

2) This theory cannot be tested without comparing 2 objects with (nearly) identical gravity affecting them. Since by definition large objects are very far apart, any differences in their orbit cannot be conclusively attributed to this theorized force, vs. regular gravity from nearby bodies.

 

[Looney]

Originally posted by: tank171
This is not quite about the main topic, but, DrPizza, .999999... doesnt equal 1. It equals .999999.... But, .999999..., when converted to a fraction is 1. There is a difference between the decimal and fraction form.

OMG not this again!

 

[MrDudeMan]

Originally posted by: tank171
This is not quite about the main topic, but, DrPizza, .999999... doesnt equal 1. It equals .999999.... But, .999999..., when converted to a fraction is 1. There is a difference between the decimal and fraction form.

go back to off topic please. in this forum it is expected to have a firm grasp on math which you clearly do not.

 

[Howard]

Originally posted by: tank171
This is not quite about the main topic, but, DrPizza, .999999... doesnt equal 1. It equals .999999.... But, .999999..., when converted to a fraction is 1. There is a difference between the decimal and fraction form.

I vote we put you on the treadmill and see how far we can get you.

 

[Howard]

Originally posted by: DrPizza
I know I'm overthinking this somehow, and overlooking something...

And, I agree that all objects near the earth's surface would have the same acceleration. But (and maybe I'm overlooking something because I'm tired; I never had this particular thought before) because forces are equal and opposite, that is, the force of gravity exerted on the object by the earth is equal and opposite in direction to the force on the earth exerted by the object, wouldn't the earth's acceleration upward toward the falling object be just slightly greater in the case of a more massive object - actually decreasing the amount of time to impact by a miniscule amount?

Yes, but the way I see it, we don't measure the time it takes for the object to hit the ground, but rather the time it takes for the object to travel the distance between itself and the ground when the motion begins.

 

[Shoal07]

Is it possible we?re forgetting the relativistic constant? Physics theory and formula (Galileo, Newton, even Einstein) only work at low speeds. As an object accelerates then we have to add in the relativistic constant in order to stay accurate. It is omitted in low level physics courses because it is unnecessary. It went unnoticed by Newton and others because they had no concepts of the speeds at which it applies. However, even at low speeds you?ll still be off by the tiny amount of the relativistic constant. I?d post it here, but I have my physics book at home. It?s something like take pretty much any basic equation, and to make it true at high speeds, add ?1 / sqrt(1 - V^2 / C^2)?... or something.

 

[Sc4freak]

Originally posted by: f95toli
Yes, but F=ma which means that the mass of the object that is falling cancels out from the equation. The rest of the equation (the mass of the earth and the distance to the centre of mass) can be combined into a constant g=9.8 m/s^2, which is why ALL objects regardless of mass accelerate at that rate when they fall to the ground on earth.

That's not true.

Theoretically, if you drop a feather and a hammer on the moon, the hammer will hit the ground first. While the moon is exerting the same force on the feather and the hammer, the hammer has a slightly higher gravitational pull than the feather - and will pull the moon up towards it.

The effect is miniscule (probably too miniscule to measure) but theoretically it should be there.

 

[CycloWizard]

Originally posted by: Sc4freak
That's not true.

Theoretically, if you drop a feather and a hammer on the moon, the hammer will hit the ground first. While the moon is exerting the same force on the feather and the hammer, the hammer has a slightly higher gravitational pull than the feather - and will pull the moon up towards it.

The effect is miniscule (probably too miniscule to measure) but theoretically it should be there.

The upward force applied by the object on the moon is absolutely negligible simply because this correction is on the order of the ratio of the mass of the brick compared to the mass of the moon.