Re: Two gravity questions driving me nuts! Please!

We all have been taught that objects fall the same when air resistance is not involved.  However, in the real world there is always some air resistance.  Of course, we could exaggerate the difference of air resistance on the flight, or in this drop of an object, by picking an object that is not very dense and has a relatively low Reynolds number.   Reynolds number is a useful quantity when talking about things moving through a fluid.  

Reynolds Number = Inertial Forces / Viscous Force

Or in other words, the Reynolds Number is the ratio of the movement to the fluid friction that resists motion.   There are a few responses on the effects of air resistance, or drag on the flight of objects keywords like air drag, air resistance, terminal velocity, here are a few that I found relevant to this discussion:

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I also refer you to a book I often cite: The Physics of Baseball by Robert K. Adair.  I got to imaging it is available through a few public libraries or certainly online bookstores.

The drag force on a ball is pretty close to proportional to the velocity of the ball.  I say velocity because the drag force is always opposed the direction of motion.   However, as the speed increases the air separates from the ball more quickly and the drag force drops, i.e. the basic physical mechanism changes.  Later the drag rises again.  A physicist or engineer would say that there are different regimes of effects that depend on Reynolds number.

The really cool part of drag forces on a ball is that it depends on how rough the surface is on average.   A smooth ball will have more drag than a rough ball, but only at speeds greater than around 50 miles per hour according to Adair’s figure 2.1 on page 8.

If I drop a ball from 50 feet and ignore air resistance, how fast is the ball going when it hits the ground.

Speed = Speed_initial + g * time

Height_now -   Height_initial =  Speed_initial * time + ˝ g * time²

where Speed is the speed measured in appropriate units of length/time, Speed_initial is the start speed, for a dropped ball this is zero, g is the local acceleration of gravity, and time is the current running time since the ball was dropped (time = 0 when dropped).  I have chosen down to be the positive direction.  A bigger positive speed means a faster falling ball.   Since I am using down as positive, let’s set the Height_initial at zero.

Our equation reduce to:

Height_now = ˝ g * time²

For the fun of it, let’s work in American/British units of feet and seconds and convert to miles per hour later.  Using the second equation we can solve for the time it takes to fall.

time_fall = SQRT (2 Heigh_fall / g)

              = SQRT (2 * 50 feet / 32 feet/sec²)

              = 1.77 seconds

Plugging this value into the first equation:

Speed_atground = 32 feet/second² * 1.77 seconds

                     = 56.57 feet /second

                      = 56.57 feet/second * 1 mile/5280 feet * 3600 seconds/hour

                      = 38.57 miles per hour

The Drag force is going to be given by the simple formula (that I will not derive here, but a fluid mechanic text like Fluid Mechanics by Pijush K. Kundu in combination with Adair’s book and a general physics text will do)

Force_drag = C_d * density * area * speed²

where C_d is the coefficient of drag, density is density of the fluid (air), area is exposed cross section (pi * radius2 for a sphere), and speed is the relative speed of the ball in the fluid. 

The book Fluid Mechanics by Pijush K. Kundu, has a plot of the coefficient of drag as function of Reynolds number which shows that for Reynolds numbers of 10³ to 10⁵ the C_d is about constant value of ˝ and Adair states that the C_d = ˝ for a baseball of speeds in the range of 50 mph to 200 mph which.  Incidentally, the value of ˝ is value that yields the force required to push the column of air out of the way.

For much lower Reynolds number the coefficient of drag is proportional to 24/(Reynolds number), which means the drag is proportional to speed.

For low Reynolds Number:  Force_drag = K * r * viscosity * speed (opposed to direction of motion)

where K is a constant the value of which doesn’t matter really here), r is the radius, viscosity is the fluid friction of the air, and speed is the speed of the ball.

For intermediate to high Reynolds Number : Force_drag = ˝* density * pi * radius² * speed² (opposed to direction of motion)

So which is correct?  That depends on the speed of the ball or really the Reynolds number.

The dynamic viscosity of air is tiny with units in (newtons *seconds) /meter² or (pounds * seconds)/ft².  The kinematic viscosity is given in ft²/sec.

From the site http://www.lmnoeng.com/fluids. htm

The kinematic viscosity of air is 1.64 x 10^{- 4} ft²/sec and the density is 0.00233 slug/ft³.  The dynamic viscosity is given by kinematic viscosity*density = 1.64 x 10^-4 ft²/sec * (0.00233 slug /ft³) = 3.8 10^-7 slug /(ft * s) = 3.8 x10^-7 pounds * seconds/ft².

For a  bowling ball ball of radius 0.1 to 0.5 feet going about 40 feet /second: we get

Re = 40 (ft/sec) * 0.5 (ft)  / 1.64 x 10-4 (ft²/sec)

Re = 2.4 x 10⁴ to 120 x 10⁴

which is well into the high Reynolds number range.  We should use the drag force relationship proportional to speed squared.  During the fall the system transfers from the low Reynolds number regime to the high Re regime.  I expect that since we are in the high Re regime most of the time that we can deal with just that equation.

I will compute the ratio of drag force to gravity force.  This will tell us how important the opposing force of air resistance is compared to gravity.

We can take the ratio of the two forces:

Force_drag / Force_gravity = ˝* density * pi * radius² * speed² / mass * gravity

Note that mass * gravity is the weight.

For a bowling ball, radius about 0.5 foot, weight of 12 pounds, speed on the order of 30 feet/second

Force_drag / Force_gravity = 0.5 * 0.00233 (slug/ft³) * pi * 0.25 ft² * 900 (ft²/sec²)/ 12 pounds

                                   = 0.069

For a golf ball, radius about 0.1 foot, weight about 0.3 pounds, speed of 30 feet /second

Force_drag / Force_gravity = 0.5 * 0.00233 (slug/ft³) * pi * 0.01 ft² * 900 (ft²/sec²)/ 0.3 pounds

                                   = 0.110

For both of these, the air drag is significant, however the drag on the bowling ball is slightly less.  Therefore, I expect the bowling ball to hit the ground first by a small fraction of a second.  I just picked a representative speed during the fall of each ball.

You could take this argument farther to other speeds, but remember the drag changes as speed goes up.   But you could assume it works like this all the time.  You could figure out the range of speeds need for drag to matter.  And you could figure out the terminal velocity too.  It is the point at which the ratio of forces is 1.  According to figure 2.2 in The Physics of Baseball, the terminal ratio of drag to weight is equal to 1.0 around 90 mph for a baseball (or 132 feet/sec).  Which seems to be pretty close to my values of 114 ft/sec for a bowling ball, and 90 ft/sec for a golf ball.

Note that I didn’t try to solve the exact equation here.  It is difficult to do.  It is best solved numerically using small time intervals and methods like the Runga-Kutta solution.

On the second question,  yes a coin could do some serious hurt on a person from a height of more than 50 feet.

Sincerely,

Tom “Terminal Free Fall” Cull