# Object falls

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1. May 22, 2015

### Stephanus

Dear PF Forum,
I am wondering about object fall speed.
Aristotle had formulated that object drops relative to their mass. But Galileo(?) suggested the other way when he dropped object with different mass (supposedly from Pisa).
And astronaut had demonstrated on the moon that object regardless of their mass drop at the same speed. Okay, this is Newton, right?
Considering this.
Two objects.
A. A pebble
B. 1000 tonnes sphere rock.
If they are dropped at, say, 49 KM above ground, assuming earth has no atmosphere, they will reach ground at 100 seconds, is that true?
What about A (a pebble) and J (a very massive sphere rock, say, at Jupiter mass) are dropped at 49 KM, will A and J fall at the same SPEED?
Or, the earth will actually FALL to J at 24.79M2?
Was Aristotle right?
So, back to the original question.
(A) A pebble and (B) 1000 tonnes rock drop at 49 KM, will B reach ground in 99.99999.... seconds?

2. May 22, 2015

### davenn

OK show us your working so we know you are on the right track
what formula did you use ?
how did you get 100 sec ?

do you understand what causes them to drop .... actually more specifically, mutually attract ?

cheers
Dave

3. May 22, 2015

### Stephanus

An object that is accelerated by earth gravity 9.8 m/sec2 for 100 sec will move ½at2
So,
½ x 9.8 x 1002 =
(½ x 9.8) x 10,000 =
4.9 x 10,0000 = 49 000 M = 4.9 KM

Jupiter surface gravity according to Wikipedia is 24.79

"Mutually atract" is the best explanation after all.Binary stars orbiting each other, and Earth/Moon orbiting each other and eventualy their centre mass orbits the sun. And for that matter, the sun also orbits it's centre mass against Sun - Earth - Moon.

1. So, was Aristotle actually correct, although intuitively?
2. Do we use Newton, because Aristotle opinion/philosophy will be very difficult to calculate for small object againts massive objects, such as a pebble againts earth?

4. May 22, 2015

### davenn

not quite right for a start you were assuming the time where time is what you wanted to find out

also look at your original post, you were saying 49 km NOT 4.9 km

you know g and you know the distance so now work out the time

Time taken for an object to fall distance [PLAIN]http://upload.wikimedia.org/math/d/d/b/ddbe49d9a0e9607bddcdbc9c3f84aff1.png: [Broken]

try that and see what you get

now after all that .... notice the mass of the object isn't involved
So what does that tell you about falling objects of 2 different masses ?

Dave

Last edited by a moderator: May 7, 2017
5. May 23, 2015

### Stephanus

Oh, no. 49,000M is FORTY NINE THOUSANDS KM. Typo.

Okayy, no mass there. Just time and distance. t is time, d distance and g is distance/time2
Thanks Dave, I got it.

6. May 23, 2015

### davenn

no, 49,000 metres = 49 km

7. May 23, 2015

### Stephanus

Arghhhhh. My English! and my Math
yes, FORTY NINE KILO METRES!
Kilo, Hecto, Deca, Metre, Deci, Centi, Milli
Micro for a millionth.
Nano for a billionth and there's angstrom, too. Forgot how much an angstrom is.
I'm no physicist, I know the calculations, it's just that intuitively I think Aristotle was right after all. Altough the difference is very negligible.
Okayy, there's no mass in

But what about "g", how can we derive G? G is derived from Newton constant, right? Depending of the earth Mass, G is 9.8 m/sec2

Steven

8. May 23, 2015

### davenn

no not quite right

g and G are 2 different beasts and used for different calculations
small g isn't used in Big G calculations, nor visa versa

the value of g does vary around the earth due to the fact that the earth's interior and surface vary in density from location to location

but using the mean value for g of 9.8 m/sec2 suffices for most calculations

G is a constant ..... approximately 6.674×10−11 N⋅m2/kg2 and denoted by upper case letter G

note in G we are not talking about an acceleration as we are with lower case g

G is a universal gravitational constant, also known as Newtons Constant, and used in the calculations of gravitational force between objects

Dave

Last edited: May 23, 2015
9. May 23, 2015

### Stephanus

Hi Dave,
I'm quoting my question. Again this iritating typo!
What I mean is.
But what about "g", how can we derive g? g (first letter capital) is derived from Newton constant, right? Depending of the earth Mass, g is 9.8 m/sec2

Steven

10. May 23, 2015

### theodoros.mihos

Take a "no mass" string of lengh l, hung of something about 50g (grammars of mass), move your mobile phone to time counter and take the time that this pendulum take for 30-40 complete vibrations. The g computed by:
$$g = \frac{4\pi^2}{T^2}l$$
where T is the time for one vibration.

From Newton's constant:
$$F=-G\frac{Mm}{R^2} \Rightarrow g=\frac{F}{m} = -G\frac{M}{R^2}$$

11. May 23, 2015

### xAxis

g depends on mass, not on Newton's constant. It also depends on distance. Aristotel wasn't right because he thought speed of falling object is proportional to mass, which is not true. All bodies fall with exactly equal speeds in inertial reference frame. Even with your Jupiter-weight stone this will be true :)

12. May 23, 2015

### Staff: Mentor

Note, that if you really did mean 49,000 km, g won't be constant throughout the entire trip because the distance is too large. 9.81 m/s2 is only the average on the surface of the earth. It makes the quantatative question harder to answer, but doesn't change the qualatative one.

13. May 23, 2015

### Stephanus

Thanks every body, thanks xAxis for the answer,
I'm not saying that Aristotle is wrong, it's not proportional of course, but would it be just tiny different speed, not exactly, with different mass?
Newton would have said that the the stone dropped on earth at 9.8m/sec2. But Einstein would have said that the earth actually dropped to the Jupiter mass stone?

14. May 23, 2015

### Stephanus

Argghh, again my typo causes wrong answer. It's fory nine KM Russ_watters

15. May 23, 2015

### theodoros.mihos

There is a difference about -350m, but you say if this is problem.

16. May 23, 2015

### Stephanus

Wow, thanks Theodoros Mihos.
But, if (a big if). If we can find massless string, if there's no air friction and there's no friction in the edge of the string. It will be very accurate. But I wonder if the pendulum can vibrate 30 times, even if you swing in on the moon. Otherwise, it's Carnot machine all over again.

17. May 23, 2015

### mrspeedybob

The same force that accelerated the pebble towards the earth also accelerates the earth towards the pebble. Since the earth is much larger the acceleration is very small and the earth does not move much during the time that the pebble is falling. The huge boulder will require a much larger force to accelerate, thus, the earth also experiences a larger force and accelerates more. Therefore, the earth moves a greater distance towards the boulder while the boulder is falling then it moves towards the pebble when it is falling, therefore, the boulder makes contact with the earth in a slightly shorter time then the pebble, as long as you conduct these 2 experiments 1 at a time.

If you drop both the pebble and the boulder at the same time then they will contact the earth at the same time since the earth would be moving under the combined gravitational influence of the boulder and the pebble.

As you noted in your original post, if you scale the boulder up to the size of Jupiter then the boulders gravity becomes the dominant factor.

18. May 23, 2015

### Stephanus

Thanks MrSpeedyBob.
That's the answer that I'd like to know. I mean the verification of my idea. I suspected that the boulder will be faster, but if I drop them AT THE SAME TIME, they both will reach the earth right at the same time right. But conducting the 2 experiments one at a time would be impossible to find the right answer on earth. Atmosphere and there are many other movement on earth, ocean waves, cars, cloud, wind, all of them can't give the correct answer. But for all practical purpose, we'll consider they fall at the same speed, is that right?
Of course I'm not trying only to find answer that just fit my thought (if not theory). It's just that I need verification of my idea.
And thanks to all of you that taking effort to answer me.

Last edited: May 23, 2015
19. May 23, 2015

### theodoros.mihos

Try it

20. May 24, 2015

### davenn

NO it doesn't as I have stated several times the masses of the falling objects is irrelevant hence why it isn't included in the formula !!

Dave