Terminal velocity is 95%?



  1. Hi forum!

    Why cant falling objects reach terminal velocity in the real world?

    When an object drops (W = mg) it will experience drag force upwards. And since drag is proportional to velocity, both force will eventually be equal therefore, the object will not accelerate anymore, hence reaching terminal velocity.

    So why is it assumed that the object is at terminal velocity when the object has reached 95% of its theoretical terminal velocity?

    Are they assuming this in an ideal world where there is no air resistance and since in object will reach terminal velocity when t = ∞, and since an exponential function will never reach 0, terminal velocity cannot be reached?

    Thank you for reading this!!
     
    Last edited: Jul 18, 2014
  2. jcsd
  3. A.T.

    A.T. 5,621
    Gold Member

    Who is "we"? Reference? Context?

    If the terminal velocity is just an estimate with +/- 5% accuracy then it makes sense to say that.
     
  4. jbriggs444

    jbriggs444 1,576
    Science Advisor

    If you assume a model where drag is proportional to velocity, gravity is constant and the atmosphere does not get thicker as you descend then the difference between actual velocity and terminal velocity will decay exponentially. The two will never be exactly equal.

    If you change the drag formula, the decay may no longer be exponential. But (barring fairly exotic drag formulae) the result will still be that actual velocity approaches terminal velocity asymptotically. You'll hit the ground before they can become equal. Still, you can get close enough for practical purposes. 95% is a reasonable figure for "close enough".

    [If you allow for changes in atmospheric density with altitude, it is possible to have a falling object (momentarily) hit terminal velocity exactly. But that's often a more complex model than the situation warrants -- it's easier to pretend that "terminal velocity" is a constant]
     
  5. Hi jbriggs Could you elaborate more into why it can never be equal? What if you could scale this scenario up, let say you dropped it from the edge of the atmosphere? Is it due to the changing air density at different altitudes which changes the drag? Thank you!
     
  6. jbriggs444

    jbriggs444 1,576
    Science Advisor

    The differential equation of motion would be v'(x) = g - kv(x). If you are able to solve differential equations, this is an easy one and the solution involves an exponential.

    Looking at it from a more simplistic viewpoint, suppose that you jump out of a hot air baloon. You are falling at 0 mph. Your terminal velocity is (let's say) 100 mph.

    You fall for one second against negligible wind resistance, gaining roughly 20 mph. Now wind resistance is equal to about 20% of gravity.

    You fall for another second against this wind resistance. This time you gain only 80% of 20 mph. That's another 16 mph for a total of 36 mph. Now wind resistance is equal to 36% of gravity.

    You fall for another second against this wind resistance. This time you gain only 64% of 20 mph. That's another 12.8 mph for a total of 48.8 mph. Call it 50 mph. Now wind resistance is equal to 50% of gravity.

    The closer your speed gets to terminal velocity the slower your speed increases. It never quite gets there. The difference decays geometrically. It took you three seconds to go from a 100 mph delta to a 50 mph delta. It will take another three seconds to get to a 25 mph delta. Another three to get to a 12.5 mph delta. Another three to get to a 6.25 mph delta.

    So after 12 seconds, you're roughly at 95% of terminal velocity.

    [The difference between the simplistic picture and the differential equation is that the differential equation is the ideal limit as one does the simplistic calculation using smaller and smaller time intervals]

    If you factor in the changes in air density, then "terminal velocity" starts out quite high at the edge of the atmosphere and reduces to more normal values as one descends. So a falling astronaut will fall at a high rate of speed initially (greater than the speed of sound). As he falls into denser air the corresponding terminal velocity will be decreasing. He will eventually be going faster than the [reduced] terminal velocity and will be slowing down rather than speeding up.
     
    Last edited: Jul 18, 2014
  7. A.T.

    A.T. 5,621
    Gold Member

    The closer you get to terminal velocity, the smaller the net force that accelerates you.
     
  8. Related problem...

    Stand 20ft from a wall. Every 5 mins walk forwards half the remaining distance to the wall. How long will it take to reach the wall? In theory never. In practice at some point you will decide you are close enough.
     
  9. russ_watters

    Staff: Mentor

    Since the equation is exponential and terminal velocity is the asymptotic, it is easy to see mathematically that you never reach it. 95% is just a common, convenient "close enough".
     
Know someone interested in this topic? Share a link to this question via email, Google+, Twitter, or Facebook

Have something to add?

0
Draft saved Draft deleted
Similar discussions for: Terminal velocity is 95%?
Loading...