February 5th, 2007, 05:28 AM
If the Earth pulls a big mass with a greater force than so how when dropping two objects with different masses from the same height, they reach the ground at the same instance????????
MOD EDIT: This is the second oversized post I have seen of yours today, if I come across any more I will trash them - irrespective of their value - you may then re-submit in a more 'polite' form.
February 5th, 2007, 05:40 AM
Earth does not pull greater masses with a larger force. Earth's gravitational pull is the same regardless of object size, shape or weight. So both masses will fall towards the earth at the same rate and at the same time.
Oh dear...
1) All objects (independant of mass) accelerate towards the ground at the same rate -0 that does NOT mean they hit the ground at the same time.
IF you could hold a mass the same as that of the earth and drop it, it would accelerate at 9.8m/s/s BUT the earth would move towards the object at the same rate - thus the closing (combined) acceleration would be around 19.6m/s/s - so acceleration is constant for all objects BUT the closing speed is different.
2) when you drop something the earth also experiences a 'pull' toward the object in free fall. This however is so slight that it can be calculated but not measured - have a look at 'The hammer and the Feather' article in physics - it seems one of our members proved this and had the work published - though sadly he no longer seems to partake of this forum...
And to the original poster you too are wrong - "at the same instant" is an immeasurable quantity - it implies infinite accuracy. One should say 'appear to strike the ground at the same moment'
So they hit the ground at the same time do they? drop a hammer and a feather and see..... 8) - if you mean 'without an atmosphere' or 'in ideal conditions' then please state!
cause of newtons old formulas
F=m<sub>1</sub>a
F=Gm<sub>1</sub>m<sub>2</sub>/r²
m<sub>2</sub> is earth
m<sub>1</sub> is the object putting the formulas you get:
m<sub>1</sub>a=Gm<sub>1</sub>m<sub>2</sub>/r²
a=Gm<sub>2</sub>/r²
doesnt anyone read newtonian physics anymore?
Try it with two large bodies like earth and venus and you see the problem.
Zelos's maths is corrct BUT it only calculates the acceleration of the smaller mass, - you have to remember that ALL objects [masses] however small also have gravity - they pull the larger mass towards them as well.
Zelos, the newtonian mathematics you refer to assume the larger mass is fixed and immovable.
Zelos try your calculation on two earth sized masses as Farsight suggests and use the WHOLE formula next time.
@Zelos:
This formula of Newton is only accurate if mass1 is relatively small.
And even then it is NOT exactly right.
It is like almost all his other laws which are good in small instances, but not in the big ones.
Maybe THAT's why nobody reads it anymore.
not really it only assumes the the change of earth position is so tiny it can be ignoredOriginally Posted by Megabrain
if i have 2 giant masses i get both of them accelerating equally fast against each other
wow really? the formula often works for great masses aswell since it took alot of time before they noticed the error in merceries orbit.It is like almost all his other laws which are good in small instances, but not in the big ones.
Maybe THAT's why nobody reads it anymore.
the newtonian fysisk is accapteble in none extreme cases. newtonian bases gravity simulator is capable to simulate the entire solersystem perfect except a few differenses in merceries orbit cause of relativity. quite fun watching them
Hi
First, sorry Mr moderator I did not konw that writing in big size is something not good!!!!
It seems that there is a good debate here .
What I meant is this :
When we drop 2 objects of different masses from the same height above the Earth and since the gravitional field strength which the force exerted by the Earth on 1 kg is a constant value the gravitional force which is the force exerted by the Earth on the bigger mass object will be greater since it it has a bigger mass.
Does not that mean the bigger mass will reach the ground before the smaller one?????
If yes, this is not coinsiding with what we studied in schools that all masses will fall towards the Earth with same acceleration because same acceleration means that the chang in speeds of both objects in unit time is the same which means they should reach the Earth at the same time since we are dropping them with 0 m/s???!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!
I think it is obvious now
waiting for replies.
February 6th, 2007, 07:44 AM
Oh and Almirza,
What was with that big font ? I almost got sucked into my monitor due to its graviatational pull !!
Anyway, okay put in simple terms then. All objects of mass have gravity. When you drop 2 objects towards the earth with different masses. The one with the most mass will hit the earth before the one with the lighter mass. but only fractionally so. whilst the earths graviatational pull on both objects is a constant and is the same for both objects, the gravitational pull of the heavier object on the EARTH is larger therefore both the heavier object and the earth will meet first before the earth and the lighter object does. If this were not true then everything on the earth, regardless of mass would weigh the same in newtons.
So I am right.
What cause me to be confused that I studied as I remember in school that when Galilio dropped 2 different masses towards the Earth, they reached the Earth at the same time.
February 8th, 2007, 04:10 AM
Yes that is correct. Although if Galileo did not do this experiment in a vacuum then im suprised it worked. Anyhow....
In the time of Galileo we did not have the technology to record such a small difference. (And probably still dont now) Although we do have the maths.
February 8th, 2007, 04:54 AM
Ah Leo!Oh and Almirza,
What was with that big font ? I almost got sucked into my monitor due to its graviatational pull !!
Anyway, okay put in simple terms then. All objects of mass have gravity. When you drop 2 objects towards the earth with different masses. The one with the most mass will hit the earth before the one with the lighter mass. but only fractionally so. whilst the earths graviatational pull on both objects is a constant and is the same for both objects, the gravitational pull of the heavier object on the EARTH is larger therefore both the heavier object and the earth will meet first before the earth and the lighter object does. If this were not true then everything on the earth, regardless of mass would weigh the same in newtons.
You have obviously read the same paper, I saw it published too and was fascinated - I noted that the Feather is travelling faster than the hammer when it impacts the surface, yet why does it , if travelling faster, strike the ground after the hammer? :wink:
Well as i think the reason for the falling of hammer first and feather second is the air resistance. It depends upon the density and space occupied.
If this were to be done in vaccum the hammer and feather would fall at the same time because gravitational force of earth is constant irrespective of the mass of the object.
Whoa, no, guys.
If you drop a large mass and a small mass together they hit the ground at the same time. That's because together they are effectively a single even larger mass.
You have to drop them separately to notice any difference, and if you're on Earth doing this, you won't notice any difference because it's infinitesimal.
However if you used an asteroid instead of the earth, another asteroid as your large mass, and a pebble as your small mass, you would notice the difference.
Then you can't ignore it, and you realise why a=Gm2/r² is an approximation that warrants a little caution.
Oh, to answer the original question: the larger mass takes more pulling.
hey how can a feather and a hammer reach the ground at the same time when dropped from a same height together.
Farsight,
I don't know quite how to put this and I don't really want to defend it but here it is anyway - and you can take it that there is a mathematical proof -Which has been published, which I have read, and with which I agree:
In an ideal experiment where there are three and only three objects to consider, where each is uniformly dense and of identical matter, where the ratios of the mass's are equivalent to the Moon, A hammer and a feather. Then in releasing the hammer and feather each from an identical height, at a finite instant in time, will produce the following effects,
1) Each mass (inlucing the moon) will accelerate towards the combined centre of gravity of the other two.
2) The terminal velocity of the feather will be greater than that of the hammer.
3) The hammer will contact the moon before the feather.
Can you post me a link megabrain?
February 8th, 2007, 04:41 PM
Actually, if they are dropped at the SAME time as suggested then the moon will be pulled towards both of the objects combined gravitational pull therefore both objects WILL hit the moon as the SAME time.
Could youjust jot down the maths for that - I'd love to compare it with the descriptions I have here -
February 8th, 2007, 04:58 PM
No Maths.
Think.
Right.....The moons graviatational pull on ANY falling body will be the same. Okay. Thats the moons gravity out of the way....
....now drop two different objects of differing masses AT THE SAME TIME towards the moon, and both objects WILL hit the moon at the SAME TIME.
However, if you drop them one after the other then yes, then you will see that the object of heavier mass takes less time to fall because it has pulled the moon towards the object more than the object of less mass.
Nah, load of crap imagine dropping them on diametrically opposite sides of the moon the moon will move towards the hammer - the feather will be 'chasing' the moon - now imagine it at right angles, similar effect - there's a sin function in there I'll wager, so as you drop from closer to the hammer the effect is less but never zero. - ergo the hammer will always strike first - but the difference will be down in the 'ten to the minus twenties' - it's a maths problem - you cannot do it without maths. and it's a '3-body problem' - so there's gonna be lot's of maths! - it's 'rate of change' so it's gonna be loaded with calculus.
February 8th, 2007, 05:22 PM
Hmmm You have a point. Unless of course you drop a feather and at the same time drop a heavier yet hollow ring AROUND the feather. Only then will they both strike at EXACTLY the same time if they are both dropped at the same time.
Now what of this problem: You drop two objects of 100% IDENTICAL mass at 100% EXACT OPPOSITE sides of the moon. What happens then ? to the moon that is ?
Suppose you drop two hammers 1 either side of the feather - what then smart ass :wink:
But if you insist - The moon, upon realising the gravity of the situation, jumps out of the way?
Sorry mate dropping your ring around the feather in a vacuum is not a good idea!
Well I'll tell you - the moon accelerates towards the combined centre of gravity of the two hammers, which exactly corresponds to the direction of the feather - it falls in a straight line (as the force from the two hammers) is cancelled - result - feather hits first!
Yes.
I want a link for that too if u can.
OK I get it. But I'm not convinced the hammer hits first if both it and the feather start on the same side of the moon. Let's say they're metres apart, a thousand metres above the moon. Both accelerate towards the moon at the same rate, and the moon accelerates a little towards their combined centre of mass. Then in addition, on top of that, the feather accelerates towards the hammer and vice versa. This gives them an extra velocity vector on top of the moonwards acceleration, so it sounds to me like they hit the moon at the same time. I think. Duh. Hmmn.
Careful,
The acceleration towards the moon is constant [ignoring the moons change of position] the attraction to each other is acceleration in a different plane - the vector sum may mean they are travelling at differing terminal velocities, but their paths are curved...
It has to increase, really. That's how gravity works. I know tidal forces usually get chucked out as insignificant, but you just can't have a uniform gravitational field. If you don't have some gradient, and some tidal force, things won't fall down.
I do wish you had added the [ignoring the moon's change of position] even though I should have said 'relative' position.It has to increase, really. That's how gravity works. I know tidal forces usually get chucked out as insignificant, but you just can't have a uniform gravitational field. If you don't have some gradient, and some tidal force, things won't fall down.
It has to increase, regardless of the moon's changing position. That's how gravity works. I know tidal forces usually get chucked out as insignificant, but you just can't have a uniform gravitational field. If you don't have some gradient, and some tidal force, things won't fall down.
Aw, it's maybe not relevant though Megabrain.
Yes I know, I just failed to recognise the actual point - the effect is there, has to be considered yet since identical to both objects can be cancelled.
February 10th, 2007, 05:28 PM
I have just thought of something. A large body such as Jupiter would warp space-time greater than the earth, so it gravity would be stronger, yes ?
But what of a neutron star, whoose size is less than that of the moon, surely it should warp space-time less and thus have weaker gravity. But we know that this is not the case. Does anyone have an explanation for this ?
I think you will find if it is any larger than a football (or three abouts) it will have a mass greater than the moon!
The more massive a nuetron star is, the smaller it is!
Yeah like all the atoms collapse, all the electrons are 'expelled' in a flash of light that outshines the galaxy it is in!
Really ?
I thought the electrons, due to gravity, were forced into the nucleus of the atoms, each electron combining with each proton to form a neutron. Thus, a neutron star is ONLY made of neutrons, hence its name.
I think you are referring to a supernova ?
You need to go back a page. A nuetron star may be left after a supernova. Thats where the electrons, go out in a flash.
February 11th, 2007, 12:41 PM
I don't think of it as warping space-time, Leo. I think of it as putting space under tension. Imagine the surface of a rubber sheet. You grap hold of a square foot of it and tie it into a knot. The surrounding rubber sheet is under tension. If you tie the knot tighter and smaller there's more tension in the surrounding rubber. Now move up a dimension and imagine you're inside a whole block of rubber tying your knots. That's my loose mental model for how it works, a bit of an analogy I know, but it kinda gets the picture across.
March 1st, 2007, 01:41 AM
JESUS, the kid asked a simple question with an obvious answer and you guys have to go and complicate things bringing acceleration of the earth and general relativity into this. (zelos gave the answer, but not very helpfully)
if you want to raise other issues fine, but atleast answer his question first.
In addition obviously he is talking about conditions of no atmosphere. His question is doesnt even point to that and specifically states what his problem is: why since force on the objects are proportional to their mass, do they reach the earth at the same time.
Answer: because the acceleration is inversly proportional to the objects masses. so as force increases due to mass, more force would be required to accelerate the mass to the same amount. so acceleration remains constant.
mathematically:
F=mpg where m is the mass of the object and g=Gxmass of the earth
a=F/m
therefore a=pg which is independant of the objects mass.
you ignored that force and acceleration are not the same thing
Sohy,
He got the correct answer in the second reply.
yes, your post is very correct, but I do not agree that it answers his question. He has a very specific issue, and your post while giving a very "correct" answer, does not help him understand where his mistake lies. It doesnt point it out to him in a clear way.
maybe i am wrong and your answer did make him understand his mistake, but i guess only he can clear that up.
Sometimes they never return to the topic, to acknowledge whether their question has been answered to their satisfaction. Some sort of debate might produce the answer in the form he can understand. By the way, welcome to the forum, have a 'nice stay'.
yeah thanks, keeps me from being bored at work, law wasnt for me i dont think. is there any professional physicists on this forum?
Not as such. I am a retired Design development Engineer with a long history in the physics of electronics, and an interest in all things scientific, I can usually come up with the answer (just sometimes wish I could explain it in lay terms a bit better). Most oif the other forums seem to be the same - oops I think 'William' is very well placed to correctly answer most things in physics, (have a look at some of his replies), though where the ol bugger is these days....
That is interesting you should say that. My wife is a trainee solicitor, she too regrets "getting into" law. What sort of law do you specialise in ?
Anyway...as for farsight's rubber sheet analogy. Yes I see what you are saying but space by its very nature is made from mostly nothing. So how can you put a vacuum under tension. It sounds to me as if you are doing your best to apply the laws of matter to the laws of space. Unless of course the vacuum of space is filled with a kind of "potential" energy.
Ive just had a thought.........What if the "space" that we occupied was the only space that we ever COULD occupy. What I mean is, when we travel "through"space; we are NOT actually travelling through space, it is OUR particular space that we are "locked" into, or that our matter particles are locked into that is moving and that the rest of the space is moving to fill our void slightly. Does anyone see what im getting at ?
What do you think matter is made out of, leo? Check out pair production, where you can make matter out of a high-energy photon. Which is basically just a wibble in space. And you can put matter under tension, even though it's mostly made out of space.
Nope!Ive just had a thought.........What if the "space" that we occupied was the only space that we ever COULD occupy. What I mean is, when we travel "through"space; we are NOT actually travelling through space, it is OUR particular space that we are "locked" into, or that our matter particles are locked into that is moving and that the rest of the space is moving to fill our void slightly. Does anyone see what im getting at ?
Ouch! this must be the most disillusioning thread ever. I can only explain it by assuming most of you are too young to have been through high school. Or maybe someone in high school has terribly failed you: your teacher!
Let's see if we can get consensus on this:
Two objects of very different arbitrary material, shape and mass are dropped at the same time from the same altitude above Earth inside a tower with perfect vaccum. They fall by the exact same distance before hitting the ground. Assume that interaction between the objects and the influence of the tower walls are negligible compared to the effect of the Earth's gravity. The initial altitude and fall distance are very small compared to the radius of the Earth, but large compared to the object size.
Do these two objects, within measurable accuracy
-- experience the same force (yes/no)
-- experience the same acceleration (yes/no)
-- hit the ground at the same time (yes/no)
Each object will accelerate by the same rate towards the earth.
Each object will experience forces from both the other object and the earth.
The object will each hit the ground at a different times, the difference is in all sense, impractical to measure.
Do these two objects, within measurable accuracy:
Experience the same force?
No.
Experience the same acceleration?
Yes.
Hit the ground at the same time?
Yes.
This doesn't quite hold if you repeat the experiment dropping the two objects at different times. But you won't notice any measurable difference conducting this on earth in a tower.
However you would notice a difference if you had masses of 1m, 10m, and 100m in free space. If you "drop" the 1m and 10m masses the 100m "earth" mass will also be attracted towards a net 11m mass. These collide with the 100m mass and "hit the ground" at the same time. If however you repeat the experiment with the 1m mass, then the 10m mass, they will collide with the 100m mass after different elapsed times.
Now, what's all this too-young high-school stuff?
And when you drop a hammer and feather the earth bgins to accelerate towards the cog of both of them. If you care to all the maths (three body problem) you will discover that the hammer and fether land at different times even though they both accelerate towards the earth at the same rate. It's purely maths though and immeasurable.
but you are not taking into effect the aero dynamic properties of the objects (although it probably wouldn't have any effect upon the outcome)
Neither did I take into account the passing butterfly and the effects of the downdraught of it's tiny wings. - We are talking Vacuum, Ideal conditions, that is ignoring all other factors not listed in the equations for three body motion.
I would actually really like to see a feather falling in a vacuum !!
I bet its awe-inspiring.
I have seen a hammer and feather dropped on stage, the feather definitely hit first, the trick was that a very fine thread was attached which ran down through the floor to an apparatus which reeled in the feather a precise distance, the hammer similarly was suspended, it was so unexpected a trick that when it happened it was really hilarious, it was at a company 'do'.
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