Doubt about approximation and limiting case

[Bestfrog]

Homework Statement

A ball is dropped from rest at height ##h##. We can assume that the drag force from the air is in the form ##F_d=-m \alpha v##.
I know then the position in function of the height $$y(t)=h-\frac{g}{\alpha} (t-\frac{1}{\alpha} (1 - e^{-\alpha t}))$$
If I take ##\alpha t<<1##, the equation above (expanding ##e^{-\alpha t}## with Taylor's series) becomes $$y(t)=h-\frac{g}{\alpha} (t-\frac{1}{\alpha} (1 -(1 - \alpha t + \frac{(\alpha t)^2}{2}...)))$$
If I would neglect the terms of higher order in ##\alpha t## I have ##y(t) \sim h##, but Morin's Introduction to classical mechanics says ##y(t) \sim h - \frac{(gt)^2}{2}## for ##\alpha t<<1##. How is it possible?

 

[DrClaude]

If I would neglect the terms of higher order in ##\alpha t## I have ##y(t) \sim h##

Indeed, to first-order, the ball doesn't move! You need to keep higher-order terms to get a meaningful approximation.

 

[Ray Vickson]

Homework Statement

A ball is dropped from rest at height ##h##. We can assume that the drag force from the air is in the form ##F_d=-m \alpha v##.
I know then the position in function of the height $$y(t)=h-\frac{g}{\alpha} (t-\frac{1}{\alpha} (1 - e^{-\alpha t}))$$
If I take ##\alpha t<<1##, the equation above (expanding ##e^{-\alpha t}## with Taylor's series) becomes $$y(t)=h-\frac{g}{\alpha} (t-\frac{1}{\alpha} (1 -(1 - \alpha t + \frac{(\alpha t)^2}{2}...)))$$
If I would neglect the terms of higher order in ##\alpha t## I have ##y(t) \sim h##, but Morin's Introduction to classical mechanics says ##y(t) \sim h - \frac{(gt)^2}{2}## for ##\alpha t<<1##. How is it possible?

Why not continue by re-writing what you have, including the quadratic terms ##(\alpha t)^2/2##? If you are going to include them in line 1 you should also include them in line 2.

 

[Bestfrog]

Indeed, to first-order, the ball doesn't move! You need to keep higher-order terms to get a meaningful approximation.

Why not continue by re-writing what you have, including the quadratic terms ##(\alpha t)^2/2##? If you are going to include them in line 1 you should also include them in line 2.

Thanks for your replying, I understood what is wrong.
@DrClaude Can you explain me in depth why your affirmation is true? (Because for very very very little time, the ball seems not to fall)

 

[DrClaude]

@DrClaude Can you explain me in depth why your affirmation is true? (Because for very very very little time, the ball seems not to fall)

The entire point of the procedure is to get a good first approximation to how things are changing. The first term will always be the constant term, which doesn't describe a change, so it is uninteresting. You therefore need to consider the lowest order term which is not constant.

That said, the fact that there is no first-order term and that you have to go to second order to get a function of time is interesting in itself. It means that even to first order, nothing moves, which tells you there is less motion than in the absence of air resistance.

 

[Bestfrog]

That said, the fact that there is no first-order term and that you have to go to second order to get a function of time is interesting in itself. It means that even to first order, nothing moves, which tells you there is less motion than in the absence of air resistance.

Wow, that's interesting, thank you!