A question of physics

[Aeschylus]

post edited.

 

[JGMEERT]

post edited.

JM: Ok, I've just e-mailed a physicist friend of mine at NASA the question. If I am wrong, I will cheerfully admit so. Did you ask your physics prof to e-mail me? My e-mail is jmeert@geology.ufl.edu

Cheers

Joe Meert

 

[The Son of Him]

post edited.

My friend your thinking is flawed.

The Weight of a body can be stated as W=m.g (Newtonian law)
Where W is the force exerted by the body on a scale
m is the inertial mass of the body
g is the acceleration of gravity
If g where somehow dependant of the dimension of the body being measured (that is what you are suggesting when you say that different bodies have different accelerations) it will mean that a fat man is accelerated more to earth than a skinny one,. Which is not true, the fat guy is heavier for having more inertial mass , not for being more accelerated !!!!!!!!! as you are suggesting.

 

[JGMEERT]

My friend your thinking is flawed.

The Weight of a body can be stated as W=m.g (Newtonian law)
Where W is the force exerted by the body on a scale
m is the inertial mass of the body
g is the acceleration of gravity
If g where somehow dependant of the dimension of the body being measured (that is what you are suggesting when you say that different bodies have different accelerations) it will mean that a fat man is accelerated more to earth than a skinny one,. Which is not true, the fat guy is heavier for having more inertial mass , not for being more accelerated !!!!!!!!! as you are suggesting.

JM: Correct

 

[Aeschylus]

My friend your thinking is flawed.

The Weight of a body can be stated as W=m.g (Newtonian law)
Where W is the force exerted by the body on a scale
m is the inertial mass of the body
g is the acceleration of gravity
If g where somehow dependant of the dimension of the body being measured (that is what you are suggesting when you say that different bodies have different accelerations) it will mean that a fat man is accelerated more to earth than a skinny one,. Which is not true, the fat guy is heavier for having more inertial mass , not for being more accelerated !!!!!!!!! as you are suggesting.

No you're not quite understanding what I'm saying, I'm not saying that g is different (well I am actually as I'm taking into account the dependancy of g on r^2), what I am saying is that the force on the Earth and hence the acceleration of the Earth which is:

Gm/r^2

is dependent on the mass of the object, this has consequences when considering the inertial forces on the object in the Earth's reference frame.

 

[The Son of Him]

No you're not quite understanding what I'm saying, I'm not saying that g is different (well I am actually as I'm taking into account the dependancy of g on r^2),

That is what I am pointing at g should still be the same , you see there is my problem understanding you .
Are you or are you not saying "g" to be different for different bodies ?

 

[Aeschylus]

That is what I am pointing at g should still be the same , you see there is my problem understanding you .
Are you or are you not saying "g" to be different for different bodies ?

(take M as the mass of Earth and m as the mass of the object)

g = GM/r^2

so as long as r is the same g is the same no matter what the weightof the object, that's not what's being argued.

But there's also a force on the Earth which is exactly the same as the force on the object which is given by:

GMm/r^2

so the accelartion on the Earth is:

Gm/r^2

So this is dependent on the mass of the object, so the Earth accelartes 'more' when the object has a greater mass.

 

[lucaspa]

But according to what others have said (and what I thought) this is wrong. They won't hit the earth at exactly the same time. The heavier will hit first (although the difference in time will be imperceptible).

That's how I see it. You would need a very, very precise determination of the exact moments of impact of each. Otherwise the difference is going to fall within the error bars of your measurement system. Thus the imperceptibility. Of course, it could be for your example that the difference falls within the 10^-43 seconds at which quantum effects take over and time is smeared. In which case they will hit the ground at the same time.

 

[lucaspa]

My friend your thinking is flawed.

The Weight of a body can be stated as W=m.g (Newtonian law)
Where W is the force exerted by the body on a scale
m is the inertial mass of the body
g is the acceleration of gravity
If g where somehow dependant of the dimension of the body being measured (that is what you are suggesting when you say that different bodies have different accelerations) it will mean that a fat man is accelerated more to earth than a skinny one,. Which is not true, the fat guy is heavier for having more inertial mass , not for being more accelerated !!!!!!!!! as you are suggesting.

Son, the post you want is the 3rd or 4th by Logic. Let me post it for you again:

"Fg=G(m1*m2)/r2

G~6.67*10-11 Nm2/Kg2
m1=mass of one of the objects in consideration.
m2=mass of the other object in consideration.
r=distance between the two objects.

So, yes, if one of the object's masses is greater, the force of gravity between the two objects will be greater.

***edit***

ME~5.97*1024Kg

The mass of the other object, on the earth, whether it be a marble or an ocean freighter is negligible."

1. The dimensions of the 2 balls was identical in Bellman's example. They were both spheres of equal diameter. The difference was in mass since one was solid and the other hollow.

2. Now, F= ma. mass times acceleration. Solve for a and you have a = F/m. The F here is the F in the first equation above -- gravitational force. So F is going to be different between the two balls. Thus a is going to be different.

3. The key is that m1 and m2 in the first equation. Since the mass of the earth is so large, 10^24 kg, having a difference of even 10 kilograms between the solid ball and the hollow ball is only going to change the product of m1 * m2 by 1/10^-23. I don't think there is a measuring device that sensitive. We could probably build one -- probably laser -- and then we should be able to detect a difference.

 

[Aeschylus]

Lucas it's the acceleration not the force that's important, as a = F/m the acceleration due to gravity is depednent on r and the mass of the Earth.

 

[The Son of Him]

(So this is dependent on the mass of the object, so the Earth accelartes 'more' when the object has a greater mass.

Let's see;
Acceleration is a=s/t (speed over time, s=d/t ) or a=d/t.t =d/t^2
wich means g=h/t.t (h=height)

If for both balls t were to be different it will mean a different g for each one !!!
Which is not true !!!
We can embelish this with differential equations, but I am trying to keep it as simple as possible

 

[JGMEERT]

Lucas it's the acceleration not the force that's important, as a = F/m the acceleration due to gravity is depednent on r and the mass of the Earth.

JM: lol, so we still don't have a full answer. That's ok. I'll await the reply from my physics friend at NASA. Here's where I have trouble, the argument is somewhat weird in that it seems that if they are dropped at the same time, they will hit at the same time, but if dropped at different times, the time it takes them to hit the ground will be slightly, albeit impercebtibly different. I don't get why at all. The difference you are citing in acceleration is real enough, but it is equally balanced by the fact that it is harder to accelerate the more massive body. Let me do some calcs whilst I am waiting for the definitive answer. As I said, I'm not afraid of being wrong and will cheerfully admit my error, but I don't yet see the error.

Cheers

Joe Meert

 

[The Son of Him]

JM: lol, so we still don't have a full answer. That's ok. I'll await the reply from my physics friend at NASA. Here's where I have trouble, the argument is somewhat weird in that it seems that if they are dropped at the same time, they will hit at the same time, but if dropped at different times, the time it takes them to hit the ground will be slightly, albeit impercebtibly different. I don't get why at all. The difference you are citing in acceleration is real enough, but it is equally balanced by the fact that it is harder to accelerate the more massive body. Let me do some calcs whilst I am waiting for the definitive answer. As I said, I'm not afraid of being wrong and will cheerfully admit my error, but I don't yet see the error.

Cheers

Joe Meert

JOE YOU ARE RIGHT, NO NEED OF PHYSICIST , I am still suprised we are arguing about this simple problem

 

[Aeschylus]

Let's see;
Acceleration is a=s/t (speed over time, s=d/t ) or a=d/t.t =d/t^2
wich means g=h/t.t (h=height)

If for both balls t were to be different it will mean a different g for each one !!!
Which is not true !!!
We can embelish this with differential equations, but I am trying to keep it as simple as possible

No g is (instaneously) the same for both balls, it's the accelartion of the Earth that is different.

The problem isn't as simple as it looks, aslo g is not consatnt it's a function of r

As a minor nitpick a = dv/dt where v is velocity

 

[JGMEERT]

No g is (instaneously) the same for both balls, it's the accelartion of the Earth that is different.

The problem isn't as simple as it looks, aslo g is not consatnt it's a function of r

JM: In this problem, g IS a constant since r is assumed to be the same for both. Unless you correct me, the only difference is that we are timing one more massive body at a slightly different time than the less massive body.

Cheers

Joe Meert

 

[Aeschylus]

JOE YOU ARE RIGHT, NO NEED OF PHYSICIST , I am still suprised we are arguing about this simple problem

I agree we don't need a physicist to solve the problem, the physics discussed here don't reach past whatwas known by 18th century physicists and the physics should stand on it's own without the need to appeal to authority.

All I ask you doo is exmaine my proof careully, I am quite happy to expand on any points.

 

[Aeschylus]

JM: In this problem, g IS a constant since r is assumed to be the same for both. Unless you correct me, the only difference is that we are timing one more massive body at a slightly different time than the less massive body.

Cheers

Joe Meert

g is not constant because the two bodies are moving towards each other which means r is getting less and hence g getting greater, this is why I had to intergrate.

 

[Aeschylus]

I think part of the problem is that I haven't presented my argument very well, especially with all the typos and poor grammar (I should read my posts before I post them), but if you examine the proof (I've just noticed a minor mistake, one of the lmits of intergartion should be h + R not h + r) you should see it is correct.

 

[The Son of Him]

No g is (instaneously) the same for both balls, it's the accelartion of the Earth that is different.

The problem isn't as simple as it looks, aslo g is not consatnt it's a function of r

As a minor nitpick a = dv/dt where v is velocity

It is very simple, you refuse to see it, and it does not change if you want to define a=dv/dt and also v=dh/dt implies a=dh/dt.dt
It does not change a thing
You are mixing two views, and in the process confusing your frame of reference and refuse to see it

 

[JGMEERT]

g is not constant because the two bodies are moving towards each other which means r is getting less and hence g getting greater, this is why I had to intergrate.

JM: This is an error w/respect to the bodies in question. Small g is simply GMe/r^2. Since the bodies are both starting at the same r and ending at the same r, there is no need to worry about the variability in small g as they move toward the surface of the earth (since the variability is the same for both on the way down g slightly changes for both en-route because of the changing r).

Cheers

Joe Meert