"Science vs. Religion: What Scientist...

"Science vs. Religion: What Scientists Really Think"

There are 81867 comments on the Examiner.com story from Jan 22, 2012, titled "Science vs. Religion: What Scientists Really Think". In it, Examiner.com reports that:

It is fascinating to note that atheists boast that most scientists are atheists.

Join the discussion below, or Read more at Examiner.com.

humble brother

Helsinki, Finland

#12278 Aug 8, 2012
Chimney1 wrote:
All HB's computations of the farcical miasma lead to a singularity of theoretical absolute density.
Lets look at your model..

F=ma clearly shows that if you double the force affecting a mass you DOUBLE the acceleration.
F=GMm/r² does NOT even begin to satisfy that.

Above we see that you can double the force by doubling M or m.
You could double the gravitational acceleration on Earth by doubling the mass of Earth.

Apparently gravitational force for a 2 kg lump is twice that of a 1 kg lump.
The gravitational force is doubled, why is not the acceleration doubled?

In reality your model is a miserable failure of gibberish in the language of mathematics.
humble brother

Helsinki, Finland

#12279 Aug 8, 2012
havent forgotten wrote:
that was to am, and this is off topic. just to say hi and check in on our mutual acquaintance! has he apologized yet? are you really in Finland, by the way? and what nationality are you? As I watch Olympics I pay more attention to nationality of folks on topix, but some could be Amricans abroad.
The person behind the alias Skeptic has not uttered much here. Perhaps a couple of times he has had to shout out his opinions when the apparent urge to do that has won.

Yes, I'm in Finland and I am also Finnish. My wife is originally Canadian though :)

Since: Mar 12

Location hidden

#12280 Aug 8, 2012
humble brother wrote:
<quoted text>
Lets look at your model..
F=ma clearly shows that if you double the force affecting a mass you DOUBLE the acceleration.
Correct.
F=GMm/r² does NOT even begin to satisfy that.
Above we see that you can double the force by doubling M or m.
Yes - and there is nothing wrong with that...
You could double the gravitational acceleration on Earth by doubling the mass of Earth.
No, you would double the gravitational FORCE, not the gravitational ACCELERATION. Two different things...
Apparently gravitational force for a 2 kg lump is twice that of a 1 kg lump.
Correct!
The gravitational force is doubled, why is not the acceleration doubled?
Because when you double the mass, you double the inertia (resistance to acceleration). So twice the force acting on twice the mass = the same acceleration.
In reality your model is a miserable failure of gibberish in the language of mathematics.
As the above extremely elementary example shows, its you who needs to understand the basics. You appear to be making the mistake that gravity is a uniform force acting on all objects, i.e. the same force acting on 2kg as on 1kg. But that is not the case. Gravity acts uniformly PER UNIT OF MASS. There IS twice as much total force on a 2 kg object as a 1kg object in the same gravity field, but the same amount of force per unit of mass.

Since: Mar 12

Location hidden

#12281 Aug 8, 2012
humble brother wrote:
<quoted text>
So can you actually explain the logic of the units ending up in G?
Why are those exact units stuffed in G?
Yes. But you will follow only if you read carefully and do not go off half cocked by the second line, as you are prone to do.

Force is a unit derived the fundamental unit MASS and the derived unit ACCELERATION which itself is derived from the fundamental units of DISTANCE and TIME, as follows:

F = ma

F = mass * distance / time / time.
i.e. F = m * d * t^-2

When deriving an expression for the FORCE of a gravitational attraction for two objects, the contributing factors must have a unit expression that is the same as force. BTW, "m" is for mass, not meters, so don't mix them up a second time.

Newton recognised that the force varied as a function of the mass of each object and the distance between the objects squared.(Twice the distance, a quarter the attraction), by a factor of some discoverable constant representing the actual field strength of a G field which was not derivable from pure math but was a discoverable property of the universe. Requiring empirical observation to pin down. However, the units of G were easily derived from the pure equations:

If gravity is an effect of M1 * M2 / r or d (radius, or distance between centres of mass)^2. The units are:

F = m * d * t^-2 but =/= m1 * m2 * d^2

To derive FORCE from the right hand expression, you need to find the ACTUAL expression of field strength applying to gravity in our universe. That is G!!! And G must be expressed in the units that balance the equation as follows:

F = m * d * t^-2 = m1 * m2 * d^-2 * "d^3 * m^-1 * t^-2" (which is G)

i.e. the expression for the actual force generated by a given amount of matter (mass) at a given distance must be expressed in the units to the far right of the equation that I put into "", multiplied by m1 * m2 * d^-2, to give a resultant FORCE that must be expressed as m * d * t^-2

That is the explanation, given from two sides. On one hand, the equation's units must balance. But that is secondary. The primary point is, in physics terms, that to quantify the total force in a given gravity system, you have to know the distance and masses AND know the actual empirically discoverable field strength that any given masses will generate over any given distance i.e. G. Its a real property of our universe (and still valid with Einstein too).

humble brother

Helsinki, Finland

#12282 Aug 8, 2012
Chimney1 wrote:
Correct.
<quoted text>
Yes - and there is nothing wrong with that...
<quoted text>
No, you would double the gravitational FORCE, not the gravitational ACCELERATION. Two different things...
<quoted text>
Correct!
<quoted text>
Because when you double the mass, you double the inertia (resistance to acceleration). So twice the force acting on twice the mass = the same acceleration.
<quoted text>
As the above extremely elementary example shows, its you who needs to understand the basics. You appear to be making the mistake that gravity is a uniform force acting on all objects, i.e. the same force acting on 2kg as on 1kg. But that is not the case. Gravity acts uniformly PER UNIT OF MASS. There IS twice as much total force on a 2 kg object as a 1kg object in the same gravity field, but the same amount of force per unit of mass.
Your whole logic falls apart in this simple scenario:
Suppose a massive spaceship travels to Earth. The mass of this spaceship is Exactly the same as the mass of Earth.

The ship does not go into orbit but moves directly towards Earth and uses its engines to decelerate and enters the atmosphere. When the ship reaches the altitude of 1.0 kilometers it powers its engines so that the movement towards Earth stops. So by using energy the ship hovers in the air.

Suddenly the engine of the ship dies and it begins a free fall.

What is the relative acceleration that an observer standing on Earth will measure for that falling ship that has the same mass as Earth?
humble brother

Helsinki, Finland

#12283 Aug 8, 2012
What is the acceleration if that ship's mass is:

-50 % of Earth's mass
-25 % of Earth's mass
-10 % of Earth's mass
-5 % of Earth's mass
-1 % of Earth's mass

???
humble brother

Helsinki, Finland

#12284 Aug 8, 2012
Chimney1 wrote:
BTW, "m" is for mass, not meters, so don't mix them up a second time.
That's funny. You are the one who confused units with variables of the equations.
Would you like me to quote your post and provide the link to that failure of yours?
humble brother

Helsinki, Finland

#12285 Aug 8, 2012
Chimney1 wrote:
To derive FORCE from the right hand expression, you need to find the ACTUAL expression of field strength applying to gravity in our universe. That is G!!! And G must be expressed in the units that balance the equation as follows:
F = m * d * t^-2 = m1 * m2 * d^-2 * "d^3 * m^-1 * t^-2" (which is G)
"To balance the equation" :D

1. you have a falsifiable equation
2. you need to balance it or it is a fallacy
3. you add units to make the equation balanced

So you have a falsifiable equation which you needed to balance. Do you understand that the balance is as falsifiable as the model?
Chimney1 wrote:
i.e. the expression for the actual force generated by a given amount of matter (mass) at a given distance must be expressed in the units to the far right of the equation that I put into "", multiplied by m1 * m2 * d^-2, to give a resultant FORCE that must be expressed as m * d * t^-2
That is the explanation, given from two sides. On one hand, the equation's units must balance. But that is secondary. The primary point is, in physics terms, that to quantify the total force in a given gravity system, you have to know the distance and masses AND know the actual empirically discoverable field strength that any given masses will generate over any given distance i.e. G. Its a real property of our universe (and still valid with Einstein too).
As you have proven now G is not a property of the universe but falsifiable with your model.
G is your very precious magic constant.

“Think&Care”

Since: Oct 07

Location hidden

#12286 Aug 8, 2012
humble brother wrote:
Haha.. How about this:
<quoted text>
There is no relative to light movement for the observers, only light moves towards them. Do you not understand that?
Huh? No. The light can certainly go away from them (as it does the beacon).
Both ships are 5.0 seconds (emitter time) away from the beacon when first light gets emitted.
OK, good. Now, after 3 seconds where is the light and where are the ships? The light has moved 3 light seconds and the ships have each moved 1.5 light seconds. Now, how far apart, in the beacon's frame, are the light and the ships? Answer: the approaching ship is now 3.5 light seconds away from the beacon and the light is 3 light seconds away. The Receding ship is now 6.5 light seconds from the beacon and the light is 3 light seconds away.

Now, let's look at what things look like after 5 seconds: the approaching ship has moved 2.5 light seconds and is now 2.5 light seconds from the beacon and the light is now 5 light seconds from the beacon. In other words, the ship is now closer to the beacon than the light: the light has passed the ship.

When were they at the same location? Well, the light moves a distance t light seconds in t seconds and the ship moves .5*t light seconds in t seconds. The location of the ship is then 5-.5*t light seconds away from the beacon. and the location of the light is t light seconds away from the beacon. These are equal when t=5-.5*t, so when t=10/3 seconds.

As for the receding ship, after 5 seconds, the light is 5 light seconds away from the beacon and the ship is now 2.5 light seconds farther than it was, so it is now at 7.5 light seconds away from the beacon. In other words, the light is still closer to the beacon than the ship. When are they at the same location? Well, when t=5+.5*t, or when t=10 seconds.

Notice that the speed of light is always 1(=c) at every stage. The speed of the two ships is always .5(=.5*c) at all times.
You are constantly giving that light a different speed for different ship!
No, it do not. I am simply figuring out when the two pass each other in the beacon's frame. Simple algebra.
The ships do not move relative to that light. Once it is emitted it moves at the same observed speed for both ships!
*In their frame*, yes, it does. But that frame differs from the beacon's frame.

In the ship's frame, the beacon starts out 5 light seconds away from both (this is a different scenario than above because of length contraction). After 5 seconds, the *beginning* of the light signal hits the ships.

But, for a 1 second long signal, the *end* of the signal is emitted when the beacon has moved .5 light second cloaser to the approaching ship. So now, the end of the signal is 4.5 light seconds from the approaching ship and 5.5 light seconds from the receding ship. So the end of the signal passes the approaching ship after a total of 1+4.5=5.5 seconds. In other words, the signal looks like it is only .5 seconds long for this ship.

For the receding ship, the end of the signal passes when t=1+5.5=6.5, so the duration of the signal for this ship is 6.5-5=1.5 seconds.

“Think&Care”

Since: Oct 07

Location hidden

#12287 Aug 8, 2012
humble brother wrote:
<quoted text>
Your whole logic falls apart in this simple scenario:
Suppose a massive spaceship travels to Earth. The mass of this spaceship is Exactly the same as the mass of Earth.
The ship does not go into orbit but moves directly towards Earth and uses its engines to decelerate and enters the atmosphere. When the ship reaches the altitude of 1.0 kilometers it powers its engines so that the movement towards Earth stops. So by using energy the ship hovers in the air.
Suddenly the engine of the ship dies and it begins a free fall.
What is the relative acceleration that an observer standing on Earth will measure for that falling ship that has the same mass as Earth?
The *relative* acceleration is 2 g's. The *actual* acceleration is 1 g. The reason is that the earth is also accelerating at 1 g. The Newtonian force of gravity predicts the 1 g correctly.

“Think&Care”

Since: Oct 07

Location hidden

#12288 Aug 8, 2012
humble brother wrote:
What is the acceleration if that ship's mass is:
-50 % of Earth's mass
-25 % of Earth's mass
-10 % of Earth's mass
-5 % of Earth's mass
-1 % of Earth's mass
???
I assume those - signs are not negatives.

In all situations, the acceleration of the ship is 1 g. The acceleration of the earth, however, is .5 g,.25 g,.1 g,.05 g, and .01 g. These absolute accelerations are in opposite directions, so the relative acceleration is obtained by addition.
humble brother

Helsinki, Finland

#12289 Aug 8, 2012
polymath257 wrote:
The *relative* acceleration is 2 g's. The *actual* acceleration is 1 g. The reason is that the earth is also accelerating at 1 g. The Newtonian force of gravity predicts the 1 g correctly.
You're stuck to Earth. So when measuring falling objects on Earth are you measuring *relative* or *actual* acceleration?
humble brother

Helsinki, Finland

#12290 Aug 8, 2012
polymath257 wrote:
I assume those - signs are not negatives.
They're bullet points.
polymath257 wrote:
In all situations, the acceleration of the ship is 1 g. The acceleration of the earth, however, is .5 g,.25 g,.1 g,.05 g, and .01 g. These absolute accelerations are in opposite directions, so the relative acceleration is obtained by addition.
I was asking for the acceleration that can be measured on Earth by aiming an instrument towards the falling ship.

You people seem to apply the equation:
ma = GMm/r²

so that a is the measured *relative* acceleration. That is absolutely false.
humble brother

Helsinki, Finland

#12291 Aug 8, 2012
polymath257 wrote:
Huh? No. The light can certainly go away from them (as it does the beacon).
The theory of relativity deals with observations in different reference frames.
How can you observe light that moves away from you? Answer: you never can.
polymath257 wrote:
But, for a 1 second long signal, the *end* of the signal is emitted when the beacon has moved .5 light second cloaser to the approaching ship. So now, the end of the signal is 4.5 light seconds from the approaching ship and 5.5 light seconds from the receding ship. So the end of the signal passes the approaching ship after a total of 1+4.5=5.5 seconds. In other words, the signal looks like it is only .5 seconds long for this ship.
For the receding ship, the end of the signal passes when t=1+5.5=6.5, so the duration of the signal for this ship is 6.5-5=1.5 seconds.
And if you go back and read what I said, you will realize that you have only just repeated what I said.

The duration of the blink is 1.0 seconds in the rest frame of the beacon. How long is that duration for the observers in the ships towards which the beacon moves at 0.5*c? The answer here is that the ships will observe that 1.0 seconds as 0.87 seconds.

“ad victoriam”

Since: Dec 10

arte et marte

#12292 Aug 8, 2012
humble brother wrote:
What is the acceleration if that ship's mass is:
-50 % of Earth's mass
-25 % of Earth's mass
-10 % of Earth's mass
-5 % of Earth's mass
-1 % of Earth's mass
???


You have to increase the mass in one to get more N.

50% more mass than earth = 50% more N
100% 2x mass of Earth = 100% more N

“Think&Care”

Since: Oct 07

Location hidden

#12293 Aug 8, 2012
humble brother wrote:
<quoted text>
You're stuck to Earth. So when measuring falling objects on Earth are you measuring *relative* or *actual* acceleration?
The Newtonian equation F=GMm/r^2 predicts the actual acceleration. For small objects, the difference between that and the relative acceleration is smaller than the tolerances of all our equipment, so for small objects the difference is irrelevant.

“Think&Care”

Since: Oct 07

Location hidden

#12294 Aug 8, 2012
humble brother wrote:
<quoted text>
They're bullet points.
<quoted text>
I was asking for the acceleration that can be measured on Earth by aiming an instrument towards the falling ship.
You people seem to apply the equation:
ma = GMm/r²
so that a is the measured *relative* acceleration. That is absolutely false.
No, I applied it to find the actual accelerations of both the earth and the ship. Then I add the two to get the relative acceleration. And *that* is correct.
humble brother

Helsinki, Finland

#12295 Aug 8, 2012
Aura Mytha wrote:
You have to increase the mass in one to get more N.
50% more mass than earth = 50% more N
100% 2x mass of Earth = 100% more N
Earth mass = 6 * 10^24 kg
Hammer mass = 6 kg

How can doubling either mass double the gravitational force but only doubling the bigger mass doubles the *relative* acceleration between them?
humble brother

Helsinki, Finland

#12296 Aug 8, 2012
polymath257 wrote:
The Newtonian equation F=GMm/r^2 predicts the actual acceleration. For small objects, the difference between that and the relative acceleration is smaller than the tolerances of all our equipment, so for small objects the difference is irrelevant.
Irrelevant? In the ship example above it is far from irrelevant. Show me how you express mathematically the correct *relative* equation for the previously mentioned ship that travels to Earth?

If you use the laws:
F=ma
F=GMm/r²

How do you express the *relative* acceleration between Earth and the ship using those laws?
humble brother

Helsinki, Finland

#12297 Aug 8, 2012
polymath257 wrote:
No, I applied it to find the actual accelerations of both the earth and the ship. Then I add the two to get the relative acceleration. And *that* is correct.
Well we have an observer in the situation who can only measure the *relative* acceleration. I'd like to see how you model it using the gravity laws.

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