# Do all objects really fall at the same velocity to the ground in vacuum?

Discussion in 'Physics & Math' started by pluto2, Sep 23, 2012.

1. ### pluto2Registered Senior Member

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For example a gas atom and a bowling ball will both fall at the same velocity to the ground in vacuum?

Because if not, then something is seriously wrong with our understanding of the gravitational force.

3. ### Robittybob1BannedBanned

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Well it does and we know it does or else we would not have a thing called air pressure. How do you think air pressure comes about?

5. ### OnlyMeValued Senior Member

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Stanford News Service

The force of gravity is the same for atoms and baseballs

7. ### pluto2Registered Senior Member

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I have no idea but it is probably due to gravity.

Anyway, sorry for my pseudointellectual hunger for metaphysical discussion.

8. ### Motor Daddy☼☼☼☼☼☼☼☼☼☼☼Valued Senior Member

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Let's just ASSUME the Earth has the mass of 5,974,200,000,000,000,000,000,000 kg.

We are going to compare impact times of two different objects when dropped from an exact height of 16.087 feet.

Object A has a mass of 1 kg
Object B has a mass of 10 kg

Using the formula A=(L-S)/R2

Object A has a "A value" of 371,368,185,491,390,563,809,286.93976503

Object B has a "A value" of 371,368,185,491,390,563,809,286.38030708

A previous test was done with object A. It was determined that object A took exactly 1 second to impact the ground when dropped from a height of 16.087 feet, which is an acceleration of 32.174 ft/sec^2.

That means an “A value” of 1, has an acceleration of
32.174/371,368,185,491,390,563,809,286.93976503=

.000000000000000000000086636392822469954136118658002141 ft/sec^2

If you multiply that by Object B’s “A value”, you find an acceleration of 32.173999999999999999999951530582 ft/sec^2 for object B.

Now, let’s look at the time it takes for each object to hit the ground, when dropped from the 16.087 feet.

Object A- 1.0000000000000000000000000000000 seconds
Object B- 1.0000000000000000000000007532389 seconds

9. ### OnlyMeValued Senior Member

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No need to appologise for anything. It sounded like an honest question to me.

As long as they are honest, there are no stupid questions, just stupid answers.

Had I not remembered an article associated with the link I gave earlier, you might had gotten a long winded one from me... It is not unheard of.

Check out this link. It is a straight forward answer, Standford News Release The force of gravity is the same for atoms and baseballs

10. ### Aqueous Idflat Earth skepticValued Senior Member

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I think any difficulty you may have with the concept merely arises from common experience, in air, in which light objects tend to float, or experience air resistance, whereas an object as heavy as a bowling ball would not be noticeably slowed by air.

In a vacuum, the force = mass x acceleration. The bowling ball and the atom have substantially different mass, so the gravitational force is substantially different, but the acceleration is exactly the same: 9.8 m/s[sup]2[/sup]. Since they experience the same acceleration, and since acceleration = rate of change of velocity, then at each instant their velocities will be exactly the same.

11. ### wlminexBannedBanned

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The difference in 'rate' is due to air friction on the objects relative to their surface area. Bigger objects have larger surface area . . . ergo, more air is displaced (> friction) as it falls, relative to a smaller surface area object. Also, less dense material (e.g., a feather vs a more dense bowling ball) is more subject to air effects. In a vacuum (i.e., no air resistance of other air effects), the rate is the same for both.

12. ### AlphaNumericFully ionizedModerator

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It's a sign that you don't do any actual physics or engineering that you're giving ridiculously long decimal expressions. Any competent physicist or engineer knows that you cannot have predictions with more significant figures than the worst input. For example, if you only measure the value of g to 3 significant figures, $g = 9.81 m/s^{2}$ and you only measure height to 2 significant figures, $h = 1.0 m$ then you cannot give the time it takes for such a dropped object to hit the ground to more than 2 signficant figures, ie 0.45 seconds. If you used a calculator to solve $s = \frac{1}{2}gt^{2}$ then you'll get 0.451524.... You cannot trust your answer to such a degree because you didn't measure your inputs accurately enough to have such a level of precision. In your example you give the mass of the Earth to less than 10 significant figures yet your outputs differ much much further down than 10 significant figures. If you wanted to compute such things to that many places you'd need to consider a great many additional contributions to the results, down to the size and shape of the objects due to the Earth's gravitational field not being constant with height and even local gravitational anomalies.

I've explained this to you before but obviously you didn't listen or didn't understand. You might think giving more decimal places makes it look like you're being more competent but in fact it's showing the opposite.

13. ### RealityCheckBannedBanned

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Hi AN.

About that physicist/engineering dictum (bolded) you just pointed out to Motor Daddy.

In cases like PI etc, the 'output values' can be computed to a never-ending number of decimal places from 'input' numbers having 'exact values' not requiring any decimal places for their representation.

Should such cases justify a 'caveat' for that dictum you pointed out, or is it somehow a different context altogether?

14. ### Robittybob1BannedBanned

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Why be sorry, I was only asking a question like you were!

15. ### SyneSine qua nonValued Senior Member

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You must have missed where he said "predictions". Pi is a calculation, not a prediction.

16. ### AlphaNumericFully ionizedModerator

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Pi is a mathematically defined construct, we do not measure it via experiment but determine it perfectly via geometry and calculus. Extracting an arbitrarily accurate decimal representation is then straightforward. Predictions/results depending on quantities we measure, not mathematically define, have the problem I just explained. In fact your bringing up of pi leads into another example. Suppose you have something you believe to be circular shaped. You can easily measure its diameter using a straight ruler. Suppose you measure it to be 1 metre, accurate to 1cm, ie you declare it to be 1.00 metres. What is the circumference? Well we all know $C = 2\pi r = \pi D$ so we can determine the circumference by multiplying by pi our measurement. How many significant figures do we go to in our value of pi? Well what I explained in my last post applies. There's no point going to 20 significant figures in your value for pi when you're only going to 3 significant figures in the number you're multiplying with it. Therefore saying pi ~ 3.142 is going to result in just as accurate a prediction of the circumference as using 3.14159265358979323846264338327950288419716939937510... . Everything past the 3rd significant figure is pointless to include because it is 'swamped' by the noise/error introduced by only measuring your 1 metre to an accuracy of 1 centimetre. Obviously for us today with calculators there's no extra work using a highly accurate value of pi but in the days of slide rules and doing it by hand it would be important to note the fact you don't need to go past 3.142 in such a model.

In the case of MD's post he goes to more than 24 significant figures. That's a trilliionth of a trillionth. If you're going to such levels of accuracy, making comments about how values change in their 26th significant figure, then you need to make sure you've included ALL contributions for ALL effects which are going to influence any of the preceding digits. If you're measuring to 24 significant figures then you need to include how the gravitational force on the ground is different from an object dropped from 1 metre. It's even different across the length of the object. The latitude of the experiment will have an effect due to angular momentum contributions, much like going around a bend makes you experience a g-force. Even ignoring how the Earth isn't a perfect sphere or how mountains and oceans perturb the gravitational field, doing an experiment at a pole compared to on the equator will involve variations in something like the 5th significant figure. Relativistic corrections need to be applied, as they are trillions of times larger than the 26th significant figure. The radius of the visible universe is approximately $1.2 \times 10^{26}$ metres so MD's 'precision' is like knowing the radius of the visible universe to an error of less than 1 metre.

Hell, you cannot even use the calculator on your computer to compute variations in the 24th significant figure! IEEE standards for number encoding in 'double precision' only goes to 15 significant figures! For people doing extremely precise modelling, such as fractal patterns, you have to use custom or non-standard number handling, such as double double precision or quad precision! This is something computer scientists have to consider. For example, mathematically $x^{2}-y^{2} = (x+y)(x-y)$ for all x,y. However, if x and y are very close but very large it is preferential to use the right hand version, as you get a small number times a large number, while the left hand side has a HUGE number minus a very close HUGE number, the difference might be in the 10th significant figure, beyond the ability of double precision to measure, so you'll at best get back 0 or at worst a random answer. As such it's not even clear how much of MD's 'discrepancy' is real and how much is just noise from computer roundings.

The mindless application of calculators, copying down precisely, to the last digit, what they give you is never a good sign. It's highly correlated with not understanding the algebra which lead to such expressions. Funny how hacks often complain the mainstream mindlessly follow the mathematics, ignoring the physics, when people like MD wield a calculator in such a way....

17. ### RealityCheckBannedBanned

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Actually, I didn't miss it. That was one of the reasons I wanted clarification as to 'context' of what AN posted. Thanks anyway for your I trust well meant response. Cheers.

18. ### RealityCheckBannedBanned

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Hi AN.

Thanks for your comprehensive clarification/info. Much appreciated.

Hmmm. I do understand about desired accuracy etc. However, in particle physics and other spheres where we deal with just 'the numbers' (probability etc etc), the 'calculation' IS the 'prediction', isn't it? And they go to great decimal accuracy in some cases, don't they, simply because small initial 'values' may have large eventual consequences (butterfly effect and all that).

That is how I saw MD's 'calculation', as a purely numerical 'prediction/manipulation to the necessary accuracy/decimal places which showed that their may be an 'inconsistency' or 'difference' which shows up only when the exercise is taken to great enough decimal places (irrespective of the validity of the measurements involved....like just using the numbers AS pure numbers for illustrative purposes and not actually involving any 'measurements' of input values per se....just as you say, purely mathematical/geometrical exercise, not actually involving a real measured 'circle' for Pi calculation/prediction to be output).

Anyhow, I read MD's exercise as not actually involving real measured quantities, just hypothetical pure numbers for illustrative purposes about something that shows up only at the most extreme accuracy of 'observation', else it is not 'apparent' in the usual convenient treatments/accuracies used generally.

That's my reason for asking. From an entirely different angle, I see now. Thanks for that, much appreciated!

19. ### ElectricFetusSanity going, going, goneValued Senior Member

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I think the ridiculous decimal points was the demonstrate that yes indeed objects of different mass do fall at different speeds: the more massive the object the more its own gravity enhances it acceleration towards earth, or two objects are pulling on each other the earth and the object of interest, its just the object of interest needs to be the size of a small asteroid to be able to reasonably measure its gravitational contribution to that falling.

20. ### SyneSine qua nonValued Senior Member

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Prediction without observation does not address physical phenomena, and is not physics. In particle physics, even though you use probability, you still must address that to some physical system where the accuracy of the initial measurement limits that of any subsequent prediction.

21. ### wlminexBannedBanned

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Good point EF! . . . . .hadn't really thought about the 'mutual gravitational attraction' by two different masses . . . . probably is negligible for small masses (e.g., feathers and bowling balls).

22. ### prometheusviva voce!Moderator

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It's straight forward to work out what the gravitational force between two objects is, especially if they are low in mass and density like feathers and bowling balls, as you can use Newton's gravitational law: $F = \frac{G m_1 m_2}{r^2}$. For example, set the masses and the distance between them (r) to one say to work out the gravitational force between 2 bags of sugar one meter apart, and it immediately follows that F is numerically equal to G. G is Newton's gravitational constant and so the force is $6.67 \times 10^{-11} N$ -in other words, a tiny force.

23. ### Motor Daddy☼☼☼☼☼☼☼☼☼☼☼Valued Senior Member

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Precisely. The numbers were shown to be what they are as to show that there IS a difference in the time it takes for two different mass objects to fall the same distance of 16.087 feet to the earth. I am in no way saying that is what was actually measured. I'm actually saying the opposite, that in order to measure the difference in time of impact, the mass difference would have to be phenomenal! That is my point! When science can measure to that great of an accuracy then let me know. Til' then I'll keep believing that the LESS MASSIVE object takes less time to fall 16.087 feet to the earth than the more massive object.