Free Fall vs. Not Free Fall?

  • Thread starter eraemia
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  • #1
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Hello,

If free fall is defined as falling with the absence of friction, then what interactions does an object participate in if it falls with friction? The first obvious answer is that the object participates in a contact friction interaction with the air (or air friction). The second answer is a long-range gravitational interaction, which pulls the falling object toward the center of the earth.

But what other interactions are at play? For example, if air is matter, and if the falling object contacts air, therefore compressing it, could we call this a contact compression interaction, where the air is pushing the object in whatever opposite direction the object is pushing the air? What really is air friction? A contact friction or contact compression interaction? If there is indeed a contact compression interaction with the air, how does this interaction affect the motion of the falling object, and how does it differ from the contact friction's effect on the falling object?

Thanks for the help.
 

Answers and Replies

  • #2
mathman
Science Advisor
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The complete answer can be quite complicated - here are some parts.

1. There is a friction effect (aerodynamicists describe it in terms of boundary layers).
2. There is a general flow effect - the air has to go around the object.
3. Turbulence leads to a drastic lowering of pressure behind the object, so there is slowing resulting from front to back pressure difference.
4. At supersonic speeds, there is a compression effect (sonic boom).

All the above are relevant for any object going through a fluid (air or water), not just falling.
 
  • #3
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I thought that free fall was when the acceleration downward (towards the Earth) was equal to the earths acceleration due to gravity 9.81 m/s. Friction due to drag effects the terminal velocity due to a host of other variables, such as surface area. So I am a bit confused as to what u are asking with contact friction and compression interaction, how do they differ in this case?
 
  • #4
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Thanks mathman, though that sounds beyond my level of physics atm...

I'm just as confused as you, t-money. I'm not even sure if both interactions occur, and if they, what their difference is.
 

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