A question of physics

[The Son of Him]

Nah, despite NASA's incomplete explaination I've followed the proof here. I'd have massive problems constructing it on my own, but laid out like this its screamingly obvious. Good job Aeschylus!

Reading the site NASA's site discribing GALILEO GALILEI's demonstration that all free falling objects fall at the same rate with same acceleration hitting the ground at the same time (provided they fall from same height) takes about a minute to read.

http://quest.arc.nasa.gov/galileo/About/galileobio.html

If you deny that, you are denying GALILEO GALILEI, whos work NEWTON based upon along KEPLER's to give form to the GRAVITATIONAL LAW.
I contend no one in this thread not only does not know enough about science but they do not know about history either.
If you deny NASA's take I suggest you put a man on the moon

 

[The Son of Him]

g = F/m so

g = GM/r^2

So g is also proportional to 1/r^2

PLease see
at :

http://www.lerc.nasa.gov/WWW/K-12/airplane/mofall.html

Acceleration is constant !!!!!!!!!!!!!!

Because you are not taking into account the inertial resistance of the body !!!
g=GM/r^2 is not correct !!

 

[Aeschylus]

No I am taking into account the inertia of the body that is why g is independet of the body's mass. There are sevral cases where accelartion due to the gravity of some body is constant, for example thre case of a flat massive sheet (of any thickness) of infinite area and uniform density (in which case it is depednt on the denisty of the sheet only) or inside a hollow massive spherical shell of uniform denisty where it is zero.

For a point mass or a spherical mass with a density that is the same for any given r it obeys an inverse square law (outside of the object).

The value of 9.8 m/s^2 given by NASA is only an approxiamtion of g at the Earth's surface (even on the Earth's surface g varies, the difremce between g at the poles and at the top of mountains in the equatoor is measurably different).

 

[Aeschylus]

Reading the site NASA's site discribing GALILEO GALILEI's demonstration that all free falling objects fall at the same rate with same acceleration hitting the ground at the same time (provided they fall from same height) takes about a minute to read.

http://quest.arc.nasa.gov/galileo/About/galileobio.html

If you deny that, you are denying GALILEO GALILEI, whos work NEWTON based upon along KEPLER's to give form to the GRAVITATIONAL LAW.
I contend no one in this thread not only does not know enough about science but they do not know about history either.
If you deny NASA's take I suggest you put a man on the moon

Read the thread carefully the independancy of g on the mass of the object (which everyone here certainly with), it's just that when you take into account the universailty of Newton's law of gravity, the result is not the one you might expect.

 

[The Son of Him]

No I am taking into account the inertia of the body that is why g is independet of the body's mass.

The value of 9.8 m/s^2 given by NASA is only an approxiamtion of g at the Earth's surface (even on the Earth's surface g varies, the difremce between g at the poles and at the top of mountains in the equatoor is measurably different).

My friend in F=m.a F is the sum of all forces acting on the body

so in the case of a free falling body F=m.g-m.b in which the term -m.b is the inertial resitance of the body to change its motion so now you can say:

F=GMm/r.r = m.g-m.b

you see F=GMm/r.r=m.g like you propose is wrong !!!

 

[Aeschylus]

My friend in F=m.a F is the sum of all forces acting on the body

so in the case of a free falling body F=m.g-m.b in which the term -m.b is the inertial resitance of the body to change its motion so now you can say:

F=GMm/r.r = m.g-m.b

you see F=GMm/r.r=m.g like you propose is wrong !!!

You have to understand the difference between inertia and inertial forces.

The concept of inertia from Gallileo and Newton and it basically states:

That a body will prove in a straight line with a constant speed, unless acted on by a force.

Further:

When a (constant) force is introduced in some direction it will cause the body to accelerate in this direction with accelartion F/m (where F is the magnitude of the force and m the mass of the body.

Yes F is the sum of all forces acting on the body, but I don't know where you getting mb from as inertia does not cause a real force to act on a body.

Inertial forces are 'fictional' forces that are introduced into accelarted frames of refernces, these forces are directly proportional to the mass of the body they act on and their inclusionm allows us to see things from the point of view of an accelarted observer.

 

[Aeschylus]

So I'll add you should see that in a truly static co-ordinate system your 'm.b' term will always be zero

 

[Dark_Adonis]

But...imagine this. Two cannon balls, exactly the same size (so the same amount of wind resistance). One is hollow, so it weighs 1/5 the amount the other one weighs. Drop them both from a height. They hit the ground at precisely the same time. Or do they? Does not the heavier one fall faster because its greater mass exerts a greater gravitational pull on the earth? Obviously, the difference would be imperceptible (if it's true) - so it's purely a theoretical question, something I've wondered about for a while. Bodies fall toward the earth based on the earth's gravitational attraction...but doesn't the bodies' gravitational attraction (infinitesmal though it is in comparison to that of the earth) play a part, too, making heavier objects fall very slightly faster?

Alright let's see if we can't simplify this, first I would suggest that the problem would be simpler if we neglected wind resistance, I don't think this causes any big change in your problem. Another simplifying assumption is that the Earth can be treated as a point mass as well as the balls, I still don't think that we have left the main point of the problem which is the acceleration of Earth towards the balls. Finally let's clear the rest of the universe for a second, this just makes the whole thing simpler, I don't think the changes the thrust of the argument. Finally I suggest that we test the balls at opposite poles, this will make things easier to see and still keep the main thrust of the argument. Finally let's assume that everything can be explained in the Newtonian way, ie macroscopic objects, low speeds (Beta about equal to zero), low masses (I'm talking in terms of solar masses)...

Alright now the assumption phase is complete. Let's build us some systems. Alright let's define some systems:
1. Universe
1. Earth
2. Solid Ball
3. Hollow Ball

Alright now that we have a system let's see what laws apply:
1. Newton's Third Law- Conservation law
2. Law of Gravity- Defines the force that we will be using

Alright now
Let Fxy be the force between body number x upon body number y
then F11=F22=F33=0. Because there is no external force on our system, and due to Newton's third:
F11+F12+F13+F21+F22+F23+F31+F32+F33+Fexternal = 0 where 0 is the zero vector.Alright Fxy=-Fyx we can see this by breaking things in to subsystems such as the Hollow-Solid system. Now we look at the Law of Gravity and we see that in some unit system Fxy= -(Mx My/|Rxy|^3) Rxy, where Rxy is the vector connecting position of body x to position of body y. Alright so now we know that not only is body x exerting a force but it is also getting exerted upon by the same force and since body 1 has more mass than body 3, body 2 is accelerated towards body 1 slightly more and thus the |R12| will be the radius of the Earth at time t, which precedes the time when body three hits the Earth. Thus it can be observed that it will hit the Earth first. For some definitions of fall we can say that you are correct.

 

[Aeschylus]

I'd avoid having |R12| or |R13| equalling zero as the forces are undefined there. Instaed you can cosnider the Earth to be a rigid ball of uniform density of radius Re and you can still treat it as a point mass.

 

[Dark_Adonis]

I'd avoid having |R12| equalling zero as F12 is undefined there. Instaed you can cosnider the Earth to be a rigid ball of uniform density of radius Re and you can still treat it as a point mass.

Good point, I was starting to get distracted towards the end of writing this.

 

[The Son of Him]

You have to understand the difference between inertia and inertial forces.

The concept of inertia from Gallileo and Newton and it basically states:

That a body will prove in a straight line with a constant speed, unless acted on by a force.

Further:

When a (constant) force is introduced in some direction it will cause the body to accelerate in this direction with accelartion F/m (where F is the magnitude of the force and m the mass of the body.

Yes F is the sum of all forces acting on the body, but I don't know where you getting mb from as inertia does not cause a real force to act on a body.

Inertial forces are 'fictional' forces that are introduced into accelarted frames of refernces, these forces are directly proportional to the mass of the body they act on and their inclusionm allows us to see things from the point of view of an accelarted observer.

You are very respecful in your views, I like that

Lets try this one , please read it:

http://www.physicsclassroom.com/Class/newtlaws/u2l3e.html

Tell me what do you think ?

 

[Aeschylus]

You are very respecful in your views, I like that

Lets try this one , please read it:

http://www.physicsclassroom.com/Class/newtlaws/u2l3e.html

Tell me what do you think ?

I think that website was posted earlier, it is correct in it's own way, but it isn't dealing with exact solutions to Newton's equations only approximations that are used in highschool, so it assumes that dr (the change in the distance between the object and the Earth) is very small and that M >> m which in the cases it describes are true. Therefore it's approximations are going to give the right answer for all practical purposes.

 

[The Son of Him]

I think that website was posted earlier, it is correct in it's own way, but it isn't dealing with exact solutions to Newton's equations only approximations that are used in highschool, so it assumes that dr (the change in the distance between the object and the Earth) is very small and that M >> m which in the cases it describes are true. Therefore it's approximations are going to give the right answer for all practical purposes.

Those are not "approximations" my friend. If that were the case gravitational mass of an object and the inertial mass would not be proportional by g for:

m= m' / g
m=inertial mass
m'=gravitational mass

If g were to depend on distance , you are proposing that the mass of an object is dependant of the distance.

So the gravitational law will be destroyed, and the equivalence principle of EINSTEIN would be wrong !!!!

 

[Dark_Adonis]

Those are not "approximations" my friend. If that were the case gravitational mass of an object and the inertial mass would not be proportional by g for:

m= m' / g
m=inertial mass
m'=gravitational mass

If g were to depend on distance , you are proposing that the mass of an object is dependant of the distance.

So the gravitational law will be destroyed, and the equivalence principle of EINSTEIN would be wrong !!!!

Let's do some unit analysis:
let m be in mass units
let m' be in mass units
let g be in units of acceleration
m=m'/g yields that acceleration is unitless
but acceleration = d^2x/dt^2 which has units of length over time squared.
This is absurd.

 

[The Son of Him]

Let's do some unit analysis:
let m be in mass units
let m' be in mass units
let g be in units of acceleration
m=m'/g yields that acceleration is unitless
but acceleration = d^2x/dt^2 which has units of length over time squared.
This is absurd.

What is your take on the problem?
do both objects hit the ground at the same time or not ?

 

[Chi_Cygni]

I haven't read every post on here but I have seen so many errors it made my head spin.

Aeschylus is correct in principle but hasn't shown a quantitative solution to the general problem.

One thing I didn't see was anyone point out the reason it is obvious Aeschylus is correct.

Think about the mass being dropped and the Earth being a closed system for the experiment in question.

Conservation of linear momentum means that the centre of mass of the system cannot move in space because of this experiment. Thus for the Centre of mass of the Earth/mass system to remain stationary (as it must with no external force acting on the system) then it is obvious that the Earth's position has to shift to account for the mass motion. This shift is also obviously different for masses of two different masses being dropped.



I have just finished deriving a general solution for the time difference between two masses falling.

I will post the full solution to the problem on a new thread in the next hour or so after I have verified the algebra and figured how to display the equations.

 

[Aeschylus]

Chi_Cyngi, it's a trivial matter of putting in the average accelartion derived in post #46 into the equations of constant acceleration. I did consider deriving it from the centre of mass after son of him suggested it, but the results should be the same, because as I said before the time is unchanged under the transformations I applied in order to otain the solution.

 

[Aeschylus]

Those are not "approximations" my friend. If that were the case gravitational mass of an object and the inertial mass would not be proportional by g for:

m= m' / g
m=inertial mass
m'=gravitational mass

If g were to depend on distance , you are proposing that the mass of an object is dependant of the distance.

So the gravitational law will be destroyed, and the equivalence principle of EINSTEIN would be wrong !!!!

You've taken a wrong turning here. At first I wondered where on Earth you got the formulas above from.

The formula's above are from special relativity:

m is the rest mass which is not equivalent to the inertial mass of that object in a given reference frame, m' is the what is sometimes called the relativistic mass and is equiavlent to the inertial mass of the object as measured in a primed reference frame.

The g you have used is something called relativistic gamma and it is given by:

g = 1/sqrt(1 - u^2/c^2) where u is the relative velocity

Now the only times that relativistic gamma is given g, is on physics newsgroups as AFAIK there is no character for small gamma in ASCII. It's a totally different concept to the one we're talking about and I should warn you that the formulas posted on physics newsgroups are not that easy to read as they're written in a *******ized version of what is called 'LATEX'.

 

[The Son of Him]

You've taken a wrong turning here. At first I wondered where on Earth you got the formulas above from.

The formula's above are from special relativity:

m is the rest mass which is not equivalent to the inertial mass of that object in a given reference frame, m' is the what is sometimes called the relativistic mass and is equiavlent to the inertial mass of the object as measured in a primed reference frame.

The g you have used is something called relativistic gamma and it is given by:

g = 1/sqrt(1 - u^2/c^2) where u is the relative velocity

Now the only times that relativistic gamma is given g, is on physics newsgroups as AFAIK there is no character for small gamma in ASCII. It's a totally different concept to the one we're talking about and I should warn you that the formulas posted on physics newsgroups are not that easy to read as they're written in a *******ized version of what is called 'LATEX'.

Please answer me this GALILEO GALILEI states that both objects will hit the ground at the same time.
Are you saying he was wrong ??????????????

 

[Aeschylus]

Please answer me this GALILEO GALILEI states that both objects will hit the ground at the same time.
Are you saying he was wrong ??????????????

Galileo pre-dates Newton's law universal gravitation, so he would of had no way of knowing he was off by a tiny amount, that said his answer gives a good approximation as you're ever going to need.