Object falls

  • #1
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Main Question or Discussion Point

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?
Thanks for any answer.
 

Answers and Replies

  • #2
davenn
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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?
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
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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
Thanks Dave for the answer.
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

do you understand what causes them to drop .... actually more specifically, mutually attract ?
"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?
Thanks Dave for the answer.
 
  • #4
davenn
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Thanks Dave for the answer.
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

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

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

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

Time
edb9059105668247cbf033dc6815b858.png
taken for an object to fall distance [PLAIN]http://upload.wikimedia.org/math/d/d/b/ddbe49d9a0e9607bddcdbc9c3f84aff1.png: [Broken]

3832419694a712f0b3521feebb1a0eff.png


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
 
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  • #5
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[FONT=Courier New said:
4.9 x 10,0000 = 49 000 M = 4.9 KM[/FONT]
Oh, no. 49,000M is FORTY NINE THOUSANDS KM. Typo.

3832419694a712f0b3521feebb1a0eff.png
Okayy, no mass there. Just time and distance. t is time, d distance and g is distance/time2
Thanks Dave, I got it.
 
  • #6
davenn
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Oh, no. 49,000M is FORTY NINE THOUSANDS KM. Typo.

no, 49,000 metres = 49 km
 
  • #7
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no, 49,000 metres = 49 km
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

3832419694a712f0b3521feebb1a0eff.png

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
Thanks for your correction, again.

Steven
 
  • #8
davenn
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Stephanus said:
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
no not quite right :smile:

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
 
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  • #9
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Hi Dave,
I'm quoting my question. Again this iritating typo!
3832419694a712f0b3521feebb1a0eff.png

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
Thanks for your correction, again.
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

Thanks for your correction and additional information, Dave.

Steven
 
  • #10
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:
$$ F=-G\frac{Mm}{R^2} \Rightarrow g=\frac{F}{m} = -G\frac{M}{R^2} $$
R = earth radius
 
  • #11
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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

Thanks for your correction and additional information, Dave.

Steven
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
russ_watters
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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
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Aristotle 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 :)
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?
About Jupiter - weight stone.
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? :smile:
 
  • #14
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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.
Argghh, again my typo causes wrong answer. It's fory nine KM Russ_watters :smile:
 
  • #15
There is a difference about -350m, but you say if this is problem.
 
  • #16
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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.
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.:smile:
 
  • #17
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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
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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
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.
 
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  • #19
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.
Try it :wink:
 
  • #20
davenn
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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 :)
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 !!
you even contradicted yourself throughout your statement

Dave
 
  • #21
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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 !!
you even contradicted yourself throughout your statement

Dave
What masses? I sad mass. I was talking about g as acceleration. Basically I just restated your:
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
I guess I should have sad which mass I refered to
 
  • #22
Drakkith
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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
No, it is not. Let's work through this problem from the top.

The general form of the equation for the force of gravitation is: F = GM1M2/r2
This means that the force, which is the same magnitude for both objects, is proportional to the mass of the objects, and inversely proportional to the distance between them.
The acceleration of each object depends upon the force applied divided by the mass, or: A = F/M
Rearranging the equation gives us : F = MA

Since gravity is a force, we can substitute MA into the original equation for gravitation: MA = GM1M2/r2
Now, notice that we have 2 different masses on the right side. That means we have to choose which object we are talking about. Let's find the acceleration for M2 and say it is a 100 kg block and M1 is the Earth.

So we have: M2A = GM1M2/r2
Since M2 appears on both sides of the equation, we can cancel the terms, leaving: A = GM1/r2
Now, I want you to take note of the significance that the mass of the object we are considering has been removed from the equation. This immediately tells us that the acceleration of M2 under the force of gravity does NOT change if you vary the mass of M2. But it doesn't tell us why. Let's continue and see if we can find out.

So, we have: A = GM1/r2
We have solid numbers for both G and M1, leaving us with the only variable, r. Let's say r is the average distance from the center of the Earth to sea level. Our equation becomes: A = (6.67 x 10-11)(5.97 x 1024)/(6.37 x 106)2
Which becomes: A = 9.81 m/s2

Okay! So we have our acceleration of M2! But how does that relate to g? Well... it IS g! (g is defined as the acceleration of an object due to Earth's gravity)

So we've shown above that the mass of the object M2 does NOT affect the acceleration of the object, but let's look further into that and figure out why.


Remember that A = F/M. We have A and we have the mass of the object (100 kg), but let's rearrange the equation real quick.
F = MA
F = 9.81*100
F = 981 newtons
So we know that if an object of 100 kg accelerates at 9.81 m/s2, then it subjected to a force of 981 newtons. Let's go back to the general form of gravitation and plug in our numbers.
F = GM1M2/r2
F = (6.67 x 10-11)(5.97 x 1024)(100)/(6.37 x 106)2
F = 981 newtons!
It's the same as what we found before!

What happens if we double the mass of M2? Per the gravitational equation: F = (6.67 x 10-11)(5.97 x 1024)(200)/(6.37 x 106)2
F = 1962 newtons, or double what it was before.

Now, shouldn't double the force accelerate the object at twice the rate? Let's find out.
A = F/M
A = 1962/200
A = 9.81 m/s2!
No! The increased force does NOT cause an increase in acceleration! This is because in order to double the force, we had to double the mass, which means the object has more inertia and will take twice the amount of force to accelerate at the same rate.

In other words, changing the mass of M2 (whatever object you are dropping) has NO effect on the acceleration of that object under gravity. Note that nowhere did we calculate the acceleration for the Earth. That's the key here. Doubling the mass of M2 WILL change the rate at which the Earth accelerates towards M2. In fact, doubling the mass will double the acceleration. The Earth is just so massive that we don't generally take the acceleration of the Earth into account since it is very, very small compared to the 9.81 m/s2 an object at sea level experiences.

For comparison, the acceleration of the Earth under the influence of a 200 kg object placed 6.67 x 106 meters away from the center of the Earth is:
A = 1962 / (5.97 x 1024).
(Remember that the force from M1 on M2, 1962 newtons, is equal in magnitude to the force from M2 on M1, so our 1962 newtons from the previous equation is also the force exerted on the Earth by the 200 kg object.)
The equation becomes: A = 3.29 x 10-22 m/s2

That's a difference of more than twenty orders of magnitude. Even increasing the mass of the second object a billion-fold would not have a noticeable effect on the Earth. (A then equals 3.29 x 10-13, which is still terribly small) The acceleration of the Earth under any conceivable object short of astronomical bodies is completely and utterly insignificant.
 
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  • #23
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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
No, it is not. Let's work through this problem from the top
No it is not. You're right Drakkith.
But what I mean is not this
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
What I mean is this:
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
g is 9.8m/sec2
g is that, because G is so and so and the earth mass is so and so.
 

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