Ps3 is faster than any pc...

I think it's quite a funny conversation. Besides what more can you say about the video? He's a moron. Sorted.

bibalasvegas said:
Mass surely does make a difference when not in a vacuum. There will clearly be a resultant force on the object due to air resistance and this will take into account mg. Hence a water filled balloon will fall faster than an empty one of identical shape.
Click to expand...

That is because of the air inside the balloon and is nothing to do with it's mass.

f=ma

and

f= GMm/R^2 where M is the mass of the earth and R is the radius of the earth

ma=GMm/R^2

a=GM/R^2

a=g as the acceleration is mavity. Showing that the acceloration and thus speed of a falling object is only affected by the mass of the earth not the mass of the object. I do maths at uni btw ;)

If you fired a bullet parallell to the ground and you droped a bullet from the same height, both would hit the ground at the same time as well ;).

:EDIT:

G is the gravitational constant not mavity.

On a side note you can see why a black hole is black. It is becuase the radius is so small and the mass so large that its gravitational field is so strong that it prevents light from escaping.

Also you can see from the equation that the higher up in the sky you go the weaker mavity gets because you are increasing the radius. Thus your terminal velocity will increase the closer to earth you get.
 
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8igdave said:
That is because of the air inside the balloon and is nothing to do with it's mass.

f=ma

and

f= GMm/R^2 where M is the mass of the earth and R is the radius of the earth

ma=GMm/R^2

a=GM/R^2

a=g as the acceleration is mavity. Showing that the acceloration and thus speed of a falling object is only affected by the mass of the earth not the mass of the object. I do maths at uni btw ;)
Click to expand...

Ok, you're still wrong. Let's for simplicity model the resistance from air as a force (R) proportional in some way to the shape of the falling object. Let us have 2 objects of identical shape with different masses (m1 and m2 where m1 < m2). Let the direction the objects are falling in be the positive direction.

The acceleration of shape 1 a1 = (m1*g - R)/m1 = g - R/m1
The acceleration of shape 2 a2 = (m2*g - R)/m2 = g - R/m2

These are different with the more massive object experiencing a greater acceleration. Ok, the resistive force is not that simple but I think you see what I'm getting at.

Edit: apologies for contributing the derailment of this thread but I needed to make this post!
 
bibalasvegas said:
Ok, you're still wrong. Let's for simplicity model the resistance from air as a force (R) proportional in some way to the shape of the falling object. Let us have 2 objects of identical shape with different masses (m1 and m2 where m1 < m2). Let the direction the objects are falling in be the positive direction.

The acceleration of shape 1 a1 = (m1*g - R)/m1 = g - R/m1
The acceleration of shape 2 a2 = (m2*g - R)/m2 = g - R/m2

These are different with the more massive object experiencing a greater acceleration. Ok, the resistive force is not that simple but I think you see what I'm getting at.

Edit: apologies for contributing the derailment of this thread but I needed to make this post!
Click to expand...

I havn't covered this yet, how annoying :P. Well explained though. The thread was derailed long ago, this is just a more interesting direction i think :P. But i am a maths geek...
 
bibalasvegas said:
Ok, you're still wrong. Let's for simplicity model the resistance from air as a force (R) proportional in some way to the shape of the falling object. Let us have 2 objects of identical shape with different masses (m1 and m2 where m1 < m2). Let the direction the objects are falling in be the positive direction.

The acceleration of shape 1 a1 = (m1*g - R)/m1 = g - R/m1
The acceleration of shape 2 a2 = (m2*g - R)/m2 = g - R/m2

These are different with the more massive object experiencing a greater acceleration. Ok, the resistive force is not that simple but I think you see what I'm getting at.

Edit: apologies for contributing the derailment of this thread but I needed to make this post!
Click to expand...

There is something wrong here, perhaps an assumption to far? If R is a force it must also conform to the equation f=ma, as 'a' is the acceleration due to the the air resistance let's call it 'r'. So

R=mr

The equations
The acceleration of shape 1 a1 = (m1*g - R)/m1 = g - R/m1
The acceleration of shape 2 a2 = (m2*g - R)/m2 = g - R/m2

become
The acceleration of shape 1 a1 = (m1*g - m1*r)/m1 = g - r
The acceleration of shape 2 a2 = (m2*g - m2*r)/m2 = g - r

But again I feel we are over simplifying the calculation of air resistance and hence making a mockery of both maths and physics :eek:

Aristotle's theory of mavity states that objects fall at a speed relative to their mass. Galileo's basic reasoning to disprove disprove this:

imagine two objects, one light and one heavier than the other one, are connected to each other by a string. Drop this system of objects from the top of a tower. If we assume heavier objects do indeed fall faster than lighter ones (and conversely, lighter objects fall slower), the string will soon pull taut as the lighter object retards the fall of the heavier object. But the system considered as a whole is heavier than the heavy object alone, and therefore should fall faster. This contradiction leads one to conclude the assumption is false.
Click to expand...

In a vacuum http://en.wikipedia.org/wiki/File:Apollo_15_feather_and_hammer_drop.ogg

On Topic:
The above must have course apply to both PCs and PS3s <chuckle>
 
ecat said:
There is something wrong here, perhaps an assumption to far? If R is a force it must also conform to the equation f=ma, as 'a' is the acceleration due to the the air resistance let's call it 'r'. So

R=mr

The equations
The acceleration of shape 1 a1 = (m1*g - R)/m1 = g - R/m1
The acceleration of shape 2 a2 = (m2*g - R)/m2 = g - R/m2

become
The acceleration of shape 1 a1 = (m1*g - m1*r)/m1 = g - r
The acceleration of shape 2 a2 = (m2*g - m2*r)/m2 = g - r

But again I feel we are over simplifying the calculation of air resistance and hence making a mockery of both maths and physics :eek:

Aristotle's theory of mavity states that objects fall at a speed relative to their mass. Galileo's basic reasoning to disprove disprove this:



In a vacuum http://en.wikipedia.org/wiki/File:Apollo_15_feather_and_hammer_drop.ogg

On Topic:
The above must have course apply to both PCs and PS3s <chuckle>
Click to expand...

i might do some researching online now. Im curious to see if i was right orginaly :P.

Can we at least agree on my orginal point that the PS3 would be faster through the air then the pc? lol.
 
I freely admit I doctored the above so it would fit in with the fact: The acceleration due to mavity is independent of mass.

And this true for probably every object you are likely to come across in real life. The gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2, that's 10e-11 so a very small number. If two objects have sufficiently large mass then I guess the mutual attraction will be the sum of their gravitational forces ie at some point mass does become significant.

bibalasvegas' post still had me puzzling though. The acceleration due to mavity is fixed and independent of mass, but what about the air resistance? If mass is unimportant then you would use the same size of parachute for an egg as you would for a Land Rover - this didn't make any sense.

Think feather, think hammer. A feather will fall more slowly than a hammer, its air resistance, aka drag force, is significant.

Think big hammer, think small hammer. Both fall with equal speed as the air resistances are about equal and are insignificant.

Ok, here's the skinny http://my.execpc.com/~culp/rockets/descent.html

tl;dr

The acceleration due to mavity is independent of mass.
The drag force/wind resistance fights against the gravitational acceleration.
If the drag force is significant then its effect does depend on the mass of the falling object.

Yay! Everyone was right :D

As for the Ps3, I have no dimensional data vs the PC but comparing it to the WII I'd say yes, it would hit the ground first, I hope :eek:... soz, I love that video
 
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