Falling Bodies

The Physics Hypertextbook™
© 1998-2008 by Glenn Elert -- A Work in Progress
All Rights Reserved -- Fair Use Encouraged

prev | up | next

Discussion

Want to see an object accelerate?

In each of these examples the acceleration was the result of gravity. Your object was accelerating because gravity was pulling it down. Even the object tossed straight up is falling -- and it begins falling the minute it leaves your hand. If it wasn't, it would have continued moving away from you in a straight line. This is the acceleration due to gravity.

What are the factors that affect this acceleration due to gravity? If you were to ask this of a typical person, they would most likely say "weight" by which the actually mean "mass" (more on this later). That is, heavy objects fall fast and light objects fall slow. Although this may seem true on first inspection, it doesn't answer my original question. "What are the factors that affect the acceleration due to gravity?" Mass does not affect the acceleration due to gravity in any measurable way. The two quantities are independent of one another. Light objects accelerate more slowly than heavy objects only when forces other than gravity are also at work. When this happens, an object may be falling, but it is not in free fall. Free fall occurs whenever an object is acted upon by gravity alone.

Try this experiment.

Something else is getting in the way here -- and that thing is air resistance (also known as aerodynamic drag). If we could somehow reduce this drag we'd have a real experiment. No problem.

We're getting closer to the essence of this problem. If only somehow we could eliminate air resistance altogether. The only way to do that is to drop the objects in a vacuum. It is possible to do this in the classroom with a vacuum pump and a sealed column of air. Under such conditions, a coin and a feather can be shown to accelerate at the same rate. (In the olden days in Great Britain, a guinea coin was used and so this demonstration is sometimes still called the "guinea and feather".) A more dramatic demonstration was done on the surface of the moon -- which is as close to a true vacuum as humans are likely to experience any time soon. Astronaut David Scott released a rock hammer and a falcon feather at the same time during the Apollo 15 lunar mission in 1971. In accordance with the theory I am about to present, the two objects landed on the lunar surface simultaneously (or very nearly so). Only an object in free fall will experience a pure acceleration due to gravity.

Let's jump back in time for a bit. In the Western world prior to the Sixteenth Century, it was generally assumed that the acceleration of a falling body would be proportional to its mass -- that is, a 10 kg object was expected to accelerate ten times faster than a 1 kg object. The ancient Greek philosopher Aristotle (384-322 BCE), included this rule in what was perhaps the first book on mechanics. It was an immensely popular work among academicians and over the centuries it had acquired a certain devotion verging on the religious. It wasn't until the Italian scientist Galileo Galilei (1564-1642) came along that anyone put Aristotle's theories to the test. Unlike everyone else up to that point, Galileo actually tried to verify his own theories through experimentation and careful observation. He then combined the results of these experiments with mathematical analysis in a method that was totally new at the time, but is now generally recognized as the way science gets done. For the invention of this method, Galileo is generally regarded as the world's first scientist.

In a tale that may be apocryphal, Galileo (or an assistant, more likely) dropped two objects of unequal mass from the Leaning Tower of Pisa. Quite contrary to the teachings of Aristotle, the two objects struck the ground simultaneously (or very nearly so). Given the speed at which such a fall would occur, it is doubtful that Galileo could have extracted much information from this experiment. Most of his observations of falling bodies were really of bodies rolling down ramps. This slowed things down enough to the point where he was able to measure the time intervals with water clocks and his own pulse (stopwatches and photogates having not yet been invented). This he repeated "a full hundred times" until he had achieved "an accuracy such that the deviation between two observations never exceeded one-tenth of a pulse beat."

With results like that, you'd think the universities of Europe would have conferred upon Galileo their highest honor, but such was not the case. Professors at the time were appalled by Galileo's comparatively vulgar methods even going so far as to refuse to acknowledge that which anyone could see with their own eyes. In a move that any thinking person would now find ridiculous, Galileo's method of controlled observation was considered inferior to pure reason. Imagine that! I could say the sky was green and as long as I presented a better argument than anyone else, it would be accepted as fact contrary to the observation of nearly every sighted person on the planet.

Galileo called his method "new" and wrote a book called Discourses on Two New Sciences wherein he used the combination of experimental observation and mathematical reasoning to explain such things as one dimensional motion with constant acceleration, the acceleration due to gravity, the behavior of projectiles, the speed of light, the nature of infinity, the physics of music, and the strength of materials. His conclusions on the acceleration due to gravity were that …

the variation of speed in air between balls of gold, lead, copper, porphyry, and other heavy materials is so slight that in a fall of 100 cubits a ball of gold would surely not outstrip one of copper by as much as four fingers. Having observed this I came to the conclusion that in a medium totally devoid of resistance all bodies would fall with the same speed.

For I think no one believes that swimming or flying can be accomplished in a manner simpler or easier than that instinctively employed by fishes and birds. When, therefore, I observe a stone initially at rest falling from an elevated position and continually acquiring new increments of speed, why should I not believe that such increases take place in a manner which is exceedingly simple and rather obvious to everybody?

I greatly doubt that Aristotle ever tested by experiment.

Despite that last quote, Galileo was not immune to using reason as a means to validate his hypothesis. In essence, his argument ran as follows. Imagine two rocks, one large and one small. Since they are of unequal mass they will accelerate at different rates -- the large rock will accelerate faster than the small rock. Now place the small rock on top of the large rock. What will happen? According to Aristotle, the large rock will rush away from the small rock. What if we reverse the order and place the small rock below the large rock? It seems we should reason that two objects together should have a lower acceleration. The small rock would get in the way and slow the large rock down. But two objects together are heavier than either by itself and so we should also reason that they will have a greater acceleration. This is a contradiction.

Here's another thought problem. Take two objects of equal mass. According to Aristotle, they should accelerate at the same rate. Now tie them together with a light piece of string. Together, they should have twice their original acceleration. But how do they know to do this? How do inanimate objects know that they are connected? Let's extend the problem. Isn't every heavy object merely an assembly of lighter parts stuck together? How can a collection of light parts, each moving with a small acceleration, suddenly accelerate rapidly once joined? We've argued Aristotle into a corner. The acceleration due to gravity is independent of mass.

Galileo made plenty of measurements related to the acceleration due to gravity but never once calculated its value (or if he did, I have never seen it reported anywhere). Instead he stated his findings as a set of proportions and geometric relationships -- lots of them. His description of constant speed required one definition, four axioms, and six theorems. All of these relationships can now be written as the single equation v = Δs / Δt in modern notation.

Algebraic symbols can contain as much information as several sentences of text, which is why they are used. Contrary to the common wisdom, mathematics makes life easier.

The generally accepted value is

g = 9.8 m/s2

or in non-SI units …

g = 32 feet/s2 = 35 kph/s = 22 mph/s

It is useful to memorize this number (as millions of people around the globe already have), however, it should also be pointed out that this number is not a constant. Although mass has no effect on the acceleration due to gravity, there are other factors that do.

Everyone reading this should be familiar with the images of the astronauts hopping about on the moon and should know that the gravity there is weaker than it is on the earth -- about one sixth as strong or approximately 1.6 m/s2. That is why the astronauts were able to hop around on the surface easily despite the weight of their space suits. In contrast, gravity on Jupiter is stronger than it is on the earth -- about two and a half times stronger or 25 m/s2. Astronauts cruising through the top of Jupiter's thick atmosphere would find themselves struggling to stand up inside their space ship. The acceleration due to gravity varies with location.

Furthermore, even on the earth, this value varies with latitude and altitude (to be discussed in later chapters). The acceleration due to gravity is greater at the poles than at the equator and greater at sea level than atop Mount Everest. There are also local variations that depend upon geology. The value of 9.8 m/s2 is thus merely a convenient average over the entire surface of the earth. This value is accurate to two significant digits up to the altitude at which commercial jets fly (18 km, 29,000 feet, or 5.5 miles). The acceleration due to gravity is effectively 9.8 m/s2 over the entire surface of the earth.

How crazy are you for accuracy?

As was discussed earlier, don't confuse the phenomena of acceleration due to gravity with the unit of the same name. While the quantity g has a value that depends on location and is approximately 9.8 m/s2 on earth, the unit gravity has the defined value of …

g = 9.80665 m/s2

You may also have noticed they use slightly different symbols. The unit uses the roman or upright g while the natural phenomena uses the italic or oblique g. Don't confuse g with g.

The unit g is often used to measure the acceleration of a reference frame. "Say what?" This is technical language that will be elaborated upon later in another section of this book, but I will explain it with examples for now. As I write this, I'm sitting in front of my computer in my home office. Gravity is drawing my body down into my office chair, my arms toward the desk, and my fingers toward the keyboard. This is the normal 1 g (one gee) world we're all accustomed to. I could take a laptop computer with me to an amusement park, get on a roller coaster, and try to get some writing done there. Gravity works on a roller coaster just as it does at home, but since the roller coaster is accelerating up and down (not to mention side to side) the sensation of normal earth gravity is lost. There will be times when I feel heavier than normal and times when I fell lighter than normal. These correspond to periods of more than one g and less than one g. I could also take my laptop with me on a trip to outer space. After a brief period of 2 or 3 g (two or three gees) accelerating away from the surface of the earth, most space journeys are spent in conditions of apparent weightlessness or 0 g (zero gee). This happens not because gravity stops working (gravity has infinite range and is never repulsive), but because a spacecraft is an accelerating reference frame. As I said earlier, this concept will be discussed more thoroughly in a later section of this book.

Summary

Problems

practice

  1. Life Magazine Cover
    [magnify]
    The following passages are excerpts from "The Long, Lonely Leap" by Captain Joseph W. Kittinger, Jr. USAF as they appeared in National Geographic magazine. It is the story of his record-setting, high altitude parachute jump from a helium balloon over New Mexico on 16 August 1960.

    An hour and thirty-one minutes after launch, my pressure altimeter halts at 103,300 feet. At ground control the radar altimeters also have stopped on readings of 102,800 feet, the figure that we later agree upon as the more reliable. It is 7 o'clock in the morning, and I have reached float altitude.

    At zero count I step into space. No wind whistles or billows my clothing. I have absolutely no sensation of the increasing speed with which I fall.

    Though my stabilization chute opens at 96,000 feet, I accelerate for 6,000 feet more before hitting a peak of 614 miles an hour, nine-tenths the speed of sound at my altitude. An Air Force camera on the gondola took this photograph when the cotton clouds still lay 80,000 feet below. At 21,000 feet they rushed up so chillingly that I had to remind myself they were vapor and not solid.

    Verify the speed claim of the author. (At this altitude g = 9.72 m/s2.)

    [magnify]

    Solution …

    For most skydivers, the acceleration experienced during free fall is not constant. As a skydiver's speed increases, so too does the aerodynamic drag until their speed levels out at a typical terminal velocity of 55 m/s (120 mph). Air resistance is not negligible in such circumstances. The story of Captain Kittinger is an exceptional one, however. At the float altitude where his dive began, the earth's atmosphere has only 1.5% of its density at sea level. It is effectively a vacuum and offers no resistance to a person falling from rest.

    The acceleration due to gravity is often said to be constant, with a value of 9.8 m/s2. Over the entire surface of the earth up to an altitude of 18 km, this is the value accurate to two significant digits. In actuality, this "constant" varies from 9.81 m/s2 at sea level to 9.75 m/s2 at 18 km. At the altitude of Captain Kittinger's dive, the acceleration due to gravity was closer to 9.72 m/s2.

    Given this data it is possible to calculate the maximum speed of Captain Kittinger during his descent. First we will need to convert the altitude measurements. To save calculation time we will only convert the change in altitude and not each altitude. Given that he stepped out of the gondola at 102,800 feet, fell freely until 96,000 feet, and then continued to accelerate for another 6,000 feet; the distance over which he accelerated uniformly was …

    102,800 − 96,00 + 6,000  = 12,800 feet
      12,800 feet   1609 m  = 3900 m  
    1 5280 feet

    It's now just a matter of choosing the correct formula and plugging in the numbers.

    v =  ?? v2 =  v02 + 2aΔs
    v0 =  0 m/s v =  √(2aΔs)
    a =  9.72 m/s2 v =  √(2(9.72 m/s2)(3900 m))
    Δs =  3900 m v =  275 m/s

    This result is amazingly close to the value recorded in Kittinger's report.

    614 mile   1609 m   1 hour  = 274 m/s
    1 hour 1 mile 3600 s

    As one would expect the actual value is slightly less than the theoretical value. This agrees with the notion of a small but still non-zero amount of drag. This number is another world record -- the fastest speed attained by a human without the use of an engine.

  2. A basketball dropped from rest 1.00 m above the floor rebounds to a height of 0.67 m. Calculate the basketball's velocity …
    1. before it hit the floor on the way down and
    2. just after it left the floor on the way up.

    Solutions …

    1. The first half of this problem is much like the the skydiving problem we just solved.
             
      v0 =  0 m/s v2 =  v02 + 2aΔs
      Δs =  1.00 m v =  √(2aΔs)
      a =  9.8 m/s2 v =  √(2(9.8 m/s2)(1.00 m))
      v =  ?? v =  4.4 m/s down
             
    2. The second half of this problem is also like the first only the velocity reduces to zero instead of starting at zero.
             
      v =  0 m/s v2 =  v02 + 2aΔs
      Δs =  0.67 m v0 =  √(2aΔs)
      a =  9.8 m/s2 v0 =  √(2(9.8 m/s2)(0.67 m))
      v0 =  ?? v0 =  3.6 m/s up
             
    3. There is another way to solve the second half of this question using the notion of proportionality. 0.67 m is approximately 2/3 of 1.00 m. Velocity is proportional to the square root of distance when acceleration is constant (and the initial velocity is zero). Thus …
                                     
      v ∝ √Δs ⇒ v2  =  ⎛
      ⎜
      ⎝      		 Δs2 ⎞½
      ⎟
      ⎠      		 ⇒ v2  =  ⎛
      ⎜
      ⎝      		 0.67 m ⎞½
      ⎟
      ⎠      		  ≈ √⅔ ⇒ v2 = 3.6 m/s up
      v1 Δs1 4.4 m/s 1.00 m
                                     
      This method is no easier, but it serves as a check on our first calculation. Since both results are identical, we must have done it right.
  3. A diver jumps from a 3 m board with an initial upward velocity of 5.5 m/s. Determine …
    1. the time the diver was in the air,
    2. the maximum height to which she ascended, and
    3. her velocity on impact with the water.

    Solutions …

    1. This is a problem in which close attention to signs is a must. Let's assume arbitrarily that up is positive.
               
      v0 =  +5.5 m/s Δs =  v0Δt + ½ aΔt2
      Δs =  −3 m (−3 m) =  (+5.5 m/s)Δt + ½ (−9.8 m/s2)Δt2
      a =  −9.8 m/s2 0 =  4.9Δt2 − 5.5Δt − 3
      v =  ?? Δt =  +5.5 ± √[(5.5)2 + 4(4.9)(3)]  
          2(4.9)  
          Δt =  +1.5 s or −0.40 s
             
      Quadratics have two solutions. What do they mean? Surely the negative solution is nonsense. How could a diver dive and yet strike the water before she left the board? The answer is 1.5 s.
    2. At the point of maximum height, the diver's velocity would be zero. Thus …
                   
      v0 =  +5.5 m/s v2 =  v02 + 2aΔs   s = s0 + Δs
      v =  0 m/s Δs =  v2 − v02     s = +3 m + 1.5 m
      a =  −9.8 m/s2 2a s = +4.5 m up
      s0 =  +3 m Δs =  (0 m/s)2 − (+5.5 m/s)2    
      Δs =  ?? 2(−9.8 m/s2)  
          Δs =  +1.5 m      
             
    3. Solving for the impact velocity is perhaps the easiest problem. Just get the direction right.
             
      Δs =  −3 m v2 =  v02 + 2aΔs
      v0 =  +5.5 m/s v =  √(v02 + 2aΔs)
      a =  −9.8 m/s2 v =  √[(+55 m/s)2 + 2(−9.8 m/s2)(−3 m)]
      v =  ?? v =  8.0 m/s down
             
  4. Sketch the following graphs of motion for an object thrown straight upward.
    1. displacement-time,
    2. velocity-time, and
    3. acceleration-time

    Solutions …

    1. The displacement-time graph is the easiest for most people to think about. The object goes up for a while and then comes down. Since the velocity is changing, the graph is curved. Looking at the slope of the tangent to the curve, we can see that the object starts with a positive (upward) velocity that decreases to zero at the top where the graph turns around. The slope then turns negative and gets steeper as the velocity increases downward.
    		 
         
    1. The velocity-time graph is trickier since many people can't separate it mentally from the previous graph. Using the slope of the tangent of the displacement-time graph can help us. The object's velocity starts out large and positive, decreases at a uniform rate until it reaches zero, then keeps decreasing (or gets more negative, whichever you prefer). At the end, the object is moving downward just as fast as it was moving upward at the beginning.
    		 
         
    1. The acceleration due to gravity is constant and directed downward. It neither increases, nor decreases, nor changes in any significant way. Looking at the previous graph, this should be apparent. The velocity of an object grows in the negative (downward) direction at a constant rate. When a value is constant, its graph with respect to time is a horizontal line.
    		 

conceptual

  1. A ball is thrown straight up over level ground. State the direction of the velocity and acceleration vectors …
    1. on the way up,
    2. at the highest point, and
    3. on the way down.

numerical

  1. Whole body reaction time is about three-tenths of a second. Would you have enough time to throw yourself clear of a brick falling from a point 3 m directly overhead if you saw it the moment it came loose?
  2. A bullet leaves the muzzle of a 1.000 m long rifle with a velocity of 400.0 m/s when fired straight up. Determine the muzzle velocity when the rifle is fired straight down instead.
  3. The terminal velocity of a skydiver is 55 m/s (120 mph) at typical jump altitudes.
    1. Determine the minimum time and displacement needed to reach to reach this velocity for a skydiver starting from rest.
    2. Why are these values minimums?
  4. In the early 2000s, three skydivers proposed attempts to break Joseph Kittinger's 1960 world record parachute jump: Michel Fournier of France, Cheryl Stearns of the United States, and Rodd Millner of Australia. All planned to jump from an altitude of 40 km (130,000 feet) -- 8 km (5 miles) higher than Captain Kittinger. With this additional distance, it is quite possible that one of them would have exceeded the speed of sound. Due to a series of technical and financial troubles, however, none of them have managed to get off the ground.
    1. At what altitude might one of these skydivers break through the sound barrier? Assume that the acceleration due to gravity is 9.70 m/s2, the speed of sound is 300 m/s, and that air resistance is negligible.
    2. Captain Kittinger believed that air resistance was negligible down to about 27.5 km (90,000 feet). Assuming a continued acceleration of 9.72 m/s2 after exceeding the speed of sound, determine a possible maximum speed during such a jump.
  5. A baseball is thrown upward at 20 m/s. At what time is the ball …
    1. 10 m above the point at which it was released?
    2. 20 m above the point at which it was released?
    3. 30 m above the point at which it was released?
  6. A stone is thrown upward from a point 72 m above the ground and is airborne for 6 s.
    1. Determine the initial velocity of the stone.
    2. At some later time the stone is moving downward at 12 m/s.
      1. When does this occur?
      2. Where does this occur?
  7. Two acrobats are about to perform a stunt; one on a trampoline and another 5.0 m above on a platform. At the instant that the acrobat on the platform steps off, the acrobat on the trampoline is moving upward at 7.5 m/s.
    1. When do the two acrobats pass each other?
    2. At what height above the trampoline are they?
    3. What are their respective velocities?

investigative

  1. You can determine a person's reaction time using a centimeter ruler. Find several volunteers and have them hold their open hand out so that you can drop a ruler between their thumb and fingers. Suspend the ruler vertically with the zero mark of the ruler at the same level as the top of your volunteer's open hand. Tell your volunteers to grab the ruler the instant you drop it. Release it without warning and record the position on the ruler where they grabbed it. (Take the reading from the top of their fingers.) Repeat this test a few times for each volunteer to obtain an average value. Record your results along with your volunteer's ages (or another demographic variable that you think may be relevant). Determine the relation, if any, between age and reaction time in your sample. (A list of demographic variables to test might include such things as gender, place of birth, hours spent playing video games, hours spent watching TV, number of siblings, whatever.)
  2. How high should a domed baseball stadium be built if it is to accommodate even the highest pop fly? There are two ways to solve this problem …
    1. using the time a pop fly is in the air or
    2. using the speed of a batted ball.
    Use whichever method you find easiest.

Resources

prev | up | next

Another quality webpage by

Glenn Elert		 eglobe logo home | contact

bent | chaos | eworld | facts | physics