Articles

1.2: Newton Mechanics - Free Fall


Dimensions are useful not just to debunk incorrect arguments but also to generate correct ones. As a contrary example showing what not to do, here is how many calculus textbooks introduce a classic problem in motion:

A ball initially at rest falls from a height of h feet and hits the ground at a speed of v feet per second. Find v assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

The units such as feet or feet per second are highlighted in boldface because their inclusion is so frequent as to otherwise escape notice, and their inclusion creates a significant problem. Because the height is h feet, the variable h does not contain the units of height: h is therefore dimensionless. (For h to have dimensions, the problem would instead state simply that the ball falls from a height h; then the dimension of length would belong to h.) A similar explicit specification of units means that the variables g and v are also dimensionless. Because g, h, and v are dimensionless, any comparison of v with quantities derived from g and h is a comparison between dimensionless quantities. It is therefore always dimensionally valid, so dimensional analysis cannot help us guess the impact speed.

Giving up the valuable tool of dimensions is like fighting with one hand tied behind our back. Thereby constrained, we must instead solve the following differential equation with initial conditions:

[frac{d^{2}y}{dt^{2}} = -g, ext{ with } y(0) = h ext{ and } dy/dt = 0 ext{ at } t = 0, label{1.1}]

where y(t) is the ball’s height, dy/dt is the ball’s velocity, and g is the gravitational acceleration.

Problem 1.3 Calculus solution

Use calculus to show that the free-fall differential equation (d^{2}y/dt^{2}) = −g with initial conditions y(0) = h and dy/dt = 0 at t = 0 has the following solution:

[frac{dy}{dt} = -gt ext{ and } y = -frac{1}{2}gt^{2} + h. label{1.2}]

Question

Using the solutions for the ball’s position and velocity in Problem 1.3, what is the impact speed?

When y(t) = 0, the ball meets the ground. Thus the impact time t is (sqrt{2h/g}). The impact velocity is −gt(_{0}) or − (sqrt{2gh}). Therefore the impact speed (the unsigned velocity) is (sqrt{2gh}).

This analysis invites several algebra mistakes: forgetting to take a square root when solving for (t_{0}), or dividing rather than multiplying by g when finding the impact velocity. Practice in other words, making and correcting many mistakes reduces their prevalence in simple problems, but complex problems with many steps remain minefields. We would like less error-prone methods.

One robust alternative is the method of dimensional analysis. But this tool requires that at least one quantity among v, g, and h have dimensions. Otherwise, every candidate impact speed, no matter how absurd, equates dimensionless quantities and therefore has valid dimensions.

Therefore, let’s restate the free-fall problem so that the quantities retain their dimensions:

  • A ball initially at rest falls from a height h and hits the ground at speed v. Find v assuming a gravitational acceleration g and neglecting air resistance.

The restatement is, first, shorter and crisper than the original phrasing:

  • A ball initially at rest falls from a height of h feet and hits the ground at a speed of v feet per second. Find v assuming a gravitational acceleration of g feet per second squared and neglecting air resistance.

Second, the restatement is more general. It makes no assumption about the system of units, so it is useful even if meters, cubits, or furlongs are the unit of length. Most importantly, the restatement gives dimensions to h, g, and v. Their dimensions will almost uniquely determine the impact speed—without our needing to solve a differential equation.

The dimensions of height h are simply length or, for short, L. The dimensions of gravitational acceleration g are length per time squared or (LT^{−2}), where T represents the dimension of time. A speed has dimensions of (LT^{-1}), so v is a function of g and h with dimensions of (LT^{-1}).

Problem 1.4 Dimensions of familiar quantities

In terms of the basic dimensions length L, mass M, and time T, what are the dimensions of energy, power, and torque?

Question

What combination of g and h has dimensions of speed?

The combination (sqrt{gh}) has dimensions of speed.

((underbrace{mathrm{LT}^{-2}}_{mathrm{g}} imes underbrace{mathrm{L}}_{mathrm{h}})^{1 / 2}=sqrt{mathrm{L}^{2} mathrm{~T}^{-2}}=underbrace{mathrm{LT}^{-1}}_{ ext {speed }} .) [label{1.3}]

Question

Is (sqrt{gh}) the only combination of g and h with dimensions of speed?

In order to decide whether (sqrt{gh}) is the only possibility, use constraint propagation [43]. The strongest constraint is that the combination of g and h, being a speed, should have dimensions of inverse time ((T^{−1})). Because h contains no dimensions of time, it cannot help construct (T^{-1}).

Because g contains (T^{-2}), the (T^{-1}) must come from (sqrt{g}). The second constraint is that the combination contain (L^{1}). The (sqrt{g}) already contributes (L^{1/2}), so the missing (L^{1/2}) must come from (sqrt{h}). The two constraints thereby determine uniquely how g and h appear in the impact speed v.

The exact expression for v is, however, not unique. It could be (sqrt{gh}), (sqrt{2gh}), or, in general, (sqrt{gh}) × dimensionless constant. The idiom of multiplication by a dimensionless constant occurs frequently and deserves a compact notation akin to the equals sign:

[v∼ sqrt{gh} label{1.4}]

Including this ∼ notation, we have several species of equality:

∝ equality except perhaps for a factor with dimensions,

∼ equality except perhaps for a factor without dimensions,

≈ equality except perhaps for a factor close to 1.

The exact impact speed is (sqrt{2gh}), so the dimensions result (sqrt{gh}) contains the entire functional dependence! It lacks only the dimensionless factor (sqrt{2}), and these factors are often unimportant. In this example, the height might vary from a few centimeters (a flea hopping) to a few meters (a cat jumping from a ledge). The factor-of-100 variation in height contributes a factor-of-10 variation in impact speed. Similarly, the gravitational acceleration might vary from 0.27 m(s^{−2})(on the asteroid Ceres) to 25 m(s^{−2})(on Jupiter). The factor-of-100 variation in g contributes another factor-of-10 variation in impact speed. Much variation in the impact speed, therefore, comes not from the dimensionless factor (sqrt{2}) but rather from the symbolic factors which are computed exactly by dimensional analysis. Furthermore, not calculating the exact answer can be an advantage. Exact answers have all factors and terms, permitting less important information, such as the dimensionless factor such as (sqrt{gh}). As William James advised, “The art of being wise is the art of knowing what to overlook” [19, Chapter 22].

Problem 1.5 Vertical throw

You throw a ball directly upward with speed v0. Use dimensional analysis to estimate how long the ball takes to return to your hand (neglecting air resistance). Then find the exact time by solving the free-fall differential equation. What dimensionless factor was missing from the dimensional-analysis result?


2.1: Introduction to Newtonian Mechanics

  • Contributed by Douglas Cline
  • Professor (Physics) at University of Rochester

It is assumed that the reader has been introduced to Newtonian mechanics applied to one or two point objects. This chapter reviews Newtonian mechanics for motion of many-body systems as well as for macroscopic sized bodies. Newton&rsquos Law of Gravitation also is reviewed. The purpose of this review is to ensure that the reader has a solid foundation of elementary Newtonian mechanics upon which to build the powerful analytic Lagrangian and Hamiltonian approaches to classical dynamics.

Newtonian mechanics is based on application of Newton&rsquos Laws of motion which assume that the concepts of distance, time, and mass, are absolute, that is, motion is in an inertial frame. The Newtonian idea of the complete separation of space and time, and the concept of the absoluteness of time, are violated by the Theory of Relativity as discussed in chapter (17). However, for most practical applications, relativistic effects are negligible and Newtonian mechanics is an adequate description at low velocities. Therefore chapters (2-16) will assume velocities for which Newton&rsquos laws of motion are applicable.


Free Fall Definition

The everyday use of the term "free fall" is not the same as the scientific definition. In common usage, a skydiver is considered to be in free fall upon achieving terminal velocity without a parachute. In actuality, the weight of the skydiver is supported by a cushion of air.

Freefall is defined either according to Newtonian (classical) physics or in terms of general relativity. In classical mechanics, free fall describes the motion of a body when the only force acting upon it is gravity. The direction of the movement (up, down, etc.) is unimportant. If the gravitational field is uniform, it acts equally on all parts of the body, making it "weightless" or experiencing "0 g". Although it might seem strange, an object can be in free fall even when moving upward or at the top of its motion. A skydiver jumping from outside the atmosphere (like a HALO jump) very nearly achieves true terminal velocity and free fall.

In general, as long as air resistance is negligible with respect to an object's weight, it can achieve free fall. Examples include:

  • A spacecraft in space without a propulsion system engaged
  • An object thrown upward
  • An object dropped from a drop tower or into a drop tube
  • A person jumping up

In contrast, objects not in free fall include:

  • A flying bird
  • A flying aircraft (because the wings provide lift)
  • Using a parachute (because it counters gravity with drag and in some cases may provide lift)
  • A skydiver not using a parachute (because the drag force equals his weight at terminal velocity)

In general relativity, free fall is defined as the movement of a body along a geodesic, with gravity described as space-time curvature.


Linear Restoring Force

An important class of problems involve a linear restoring force, that is, they obey Hooke&rsquos law. The equation of motion for this case is

Then the equation of motion then can be written as

[ label ddot + omega_0^2 x = 0 ]

which is the equation of the harmonic oscillator. Examples are small oscillations of a mass on a spring, vibrations of a stretched piano string, etc.

The solution of this second order equation is

[ label x(t) = A sin ( omega _0 t - delta ) ]

This is the well known sinusoidal behavior of the displacement for the simple harmonic oscillator. The angular frequency ( omega_0 )

Note that for this linear system with no dissipative forces, the total energy is a constant of motion as discussed previously. That is, it is a conservative system with a total energy (E) given by

The first term is the kinetic energy and the second term is the potential energy. The Virial theorem gives that for the linear restoring force the average kinetic energy equals the average potential energy.


1.2: Newton Mechanics - Free Fall

According to the National Institute of Standards and Technology (NIST) — which is the U.S. government agency that investigated the World Trade Center’s destruction — the Twin Towers came down “essentially in free fall.” 1

NIST’s theory of the collapses hinges on the idea that the upper section of each tower could continuously accelerate through the lower stories at nearly the rate of gravity, while in the process completely dismembering the steel frames and pulverizing nearly all of the concrete to a fine powder.

Yet NIST provided no modeling or calculations to demonstrate that such behavior was possible. Instead, NIST arbitrarily stopped its analysis at the moment of “collapse initiation,” asserting that total collapse was “inevitable” once the collapses initiated. 2

Astonishingly, NIST’s entire explanation for why the lower sections failed to stop or even slow the descent of the upper sections is limited to half a page of its 10,000-page report, in a section titled “Events Following Collapse Initiation,” 3 which asserts:

“The structure below the level of collapse initiation provided minimal resistance to the falling building mass at and above the impact zone. The potential energy released by the downward movement of the large building mass far exceeded the capacity of the intact structure below to absorb that through energy of deformation.

“Since the stories below the level of collapse initiation provided little resistance to the tremendous energy released by the falling building mass, the building section above came down essentially in free fall, as seen in videos.” — p. 146, NIST NCSTAR 1

In 2007, a group of scientists, an architect, and two 9/11 family members filed a “Request for Correction” to the NIST report under the Information Quality Act. They argued that, among other things, NIST failed to establish the likely technical cause of the building failures because it did not explain why, after collapse initiation, total collapse had ensued. 4 They wrote:

“Here, NIST has not offered any explanation as to why (i.e. the technical cause of) the story below the collapse zone was not able to arrest the downward movement of the upper floors. The statement “as evidenced by the videos from several vantage points” is only an explanation of what occurred, but gives the reader absolutely no idea why it occurred. Basic principles of engineering (for example, the conservation of momentum principle) would dictate that the undamaged steel structure below the collapse zone would, at the very least, resist and slow the downward movement of the stories above…. The families of the firefighters and WTC employees that were trapped in the stairwells when the entirety of the WTC Towers collapsed on top of them would surely appreciate an adequate explanation of why the lower structure failed to arrest or even resist the collapse of the upper floors.” — p. 20, Request for Correction

NIST responded to the Request for Correction with the remarkable admission that it was not able to provide a full explanation of the total collapse: 5

“NIST carried its analysis to the point where the buildings reached global instability. At this point, because of the magnitude of deflections and the number of failures occurring, the computer models are not able to converge on a solution…. [W]e are unable to provide a full explanation of the total collapse.” — p. 3-4, NIST Response to Request for Correction

Total Collapse Explained

While NIST failed to provide an explanation for the total collapse of the Twin Towers, several independent researchers have taken on that challenge.

The upper section of the North Tower.

Central to their analysis has been to measure the downward motion of the upper section of WTC 1 (the North Tower). Two papers in particular have found that, in the four seconds before the upper section disappeared from view, the rate of acceleration remained constant, at approximately 64 percent of free fall, 6 and there was never an observable deceleration. 7

Based on Newton’s Third Law of Motion, which states that for every action there is an equal and opposite reaction, we know there would have been a deceleration of WTC 1’s upper section if it had impacted and crushed the intact structure below it. The absence of deceleration is incontrovertible proof that another force (i.e., explosives) must have been responsible for destroying the lower structure before the upper section reached it.

Figure 1: This graph from David Chandler’s “Destruction of the World Trade Center North Tower and Fundamental Physics” (Journal of 9/11 Studies, February 2010) shows that the North Tower’s upper section traveled at nearly uniform downward acceleration of -6.31 m/s 2 (with an R 2 value of 0.997), or 64% of free fall.

In 2011, the ASCE’s Journal of Engineering Mechanics published a paper by Dr. Zdeněk Bažant and Jia-Liang Le titled “Why the Observed Motion History of the World Trade Center Towers Is Smooth,” 8 in which the authors attempted to argue that the upper section’s deceleration was “far too small to be perceptible,” thus accounting for why the observed motion is “smooth.” Specifically, they calculated, the deceleration was “three orders of magnitudes smaller than the error of an amateur video, and thus undetectable.”

In response, researchers Tony Szamboti and Richard Johns submitted a Discussion paper to the Journal of Engineering Mechanics in May 2011. 9 Their paper argued that Bažant and Le had used incorrect values for (1) the resistance of the columns, (2) the lower structure’s floor mass, and (3) the upper section’s total mass. Szamboti and Johns showed that when the correct values are applied, Bažant and Le’s analysis actually proves that the deceleration of the upper section would have been significant and detectable (if it were a true fire-induced progressive collapse), and that the collapse would have arrested within three seconds.

Unfortunately, the Journal of Engineering Mechanics inexplicably rejected Szamboti and Johns’ Discussion paper as “out of scope” after holding it in review for 27 months. So Szamboti and Johns, along with Dr. Gregory Szuladziński, a world-renowned expert in structural mechanics, wrote another paper refuting Bažant and Le’s analysis and submitted it to the International Journal of Protective Structures. That paper, titled “Some Misunderstandings Related to the WTC Collapse Analysis,” 10 was published in June 2013.

So little research has been published on why the Twin Towers underwent total collapse that Bažant and Le’s 2011 paper, and Bažant’s three earlier papers on the subject, are the only analysis that exists to support the official explanation of a fire-induced progressive collapse. That analysis has now been indisputably debunked by Szamboti, Johns, Szuladziński, and others.

Endnotes

[1] NIST: Final Report of the National Construction Safety Team on the Collapses of the World Trade Center Towers (December 1, 2005), p.146. (NIST NCSTAR 1)

[2] NIST NCSTAR 1, p.xxxvii, p. 82.

[8] Bažant, Zdeněk and Le, Jia-Liang: “Why the Observed Motion History of the World Trade Center Towers is Smooth,” Journal of Engineering Mechanics (January 2011).

[10] Szuladziński, Gregory and Szamboti, Tony and Johns, Richard: “Some Misunderstandings Related to WTC Collapse Analysis,” International Journal of Protective Structures (June 2013).


ANSHS Physics Classroom-xanmechanics

How Fast?
Things fall because of the force of gravity. When a falling object is free of all restrains- no friction, air or otherwise-and falls under the influence of gravity alone, it is in free fall.
Velocity acquired= acceleration x time
For an object falling from rest, the instantaneous velocity v at anytime t can be expressed in shorthand notation as v= gt.
The letter v symbolizes both speed and velocity.
How far?
The distance traveled by the uniformly accelerating object starting from rest is
Distance = 1/2gt2. Where d is the distance the object falls, g is the acceleration and t is the time of fall in seconds.
Sample Problems
1.
a. How long does it take a ball to fall from a roof to the ground 7.0 m below?
b. With what speed does it strike the ground?
ANSWER:
In kinematics problems, start off with a davfvit table. Use this format to list the information given and identify the quantity being solved for. Then identify the relationship between the given and the unknown quantities, substitute the values into the relationship, and solve fro the unknown.

1.
d 7.0 m [down]
a 9.8 m/s2 [down]
vf
vi 0
t ?
THROW-UP PROBLEMS
Throw up problems refer to situations where an object’s initial velocity is opposite to its acceleration. The key is to choose a frame of reference. For example, if “up” is +, then “down” is -. The frame of reference must be used consistently throughout the solving process.
ANSWER:
2. frame of reference: down = +
d 7.0 m
a 9.8 m/s2
vf
vi – 2.0 m/s *
t ?
Now we can see we need a relationship between d, a, vi, and t

(7.0 m) = (-2.00)t + (0.5)(9.8)t2
t = <1.42, -1.01>
Since t < 0 has no meaning,
t = 1.42 s

In catch-up problems, two objects with different motions end up at the same place at the same time. Sometimes, these problems have the appearance of not having enough information to be solved. However, in physics we trust.


These problems are complex because they describe two different motions. The approach used is to simplify the problem by breaking it down into simple problems. This is down by using two columns in the davfvit table: one column for each motion.


EXAMPLE
3. A ball is dropped from a roof to the ground 8.0 m below. A rock is thrown down from the roof 0.600 s later. If they both hit the ground at the same time, what was the initial speed of the rock?

ANSWER:
3.
ball rock
d 8.0 m [down] 8.0 m [down]
a 9.8 m/s2 [down] 9.8 m/s2 [down]
vf
vi 0 ?
t ? ?

We need time to find speed. Since we have more information about the ball we start off solving for time for the ball.
Now we can see we need a relationship between d, a, vi, and t

d = vit + (0.5)at2
8.0 = (0)t + (0.5)(9.8 m/s2)t2

t=1.28s
The time of travel for the rock is 0.600 s less than for the ball. Now our table looks like this:
ball rock
d 8.0 m [down] 8.0 m [down]
a 9.8 m/s2 [down] 9.8 m/s2 [down]
vf
vi 0 ?
t 1.28 s 0.68 s

For the rock we need a relationship between d, a, vi, and t

d = vit + (0.5)at2
8.0 = vi(0.68 s) + (0.5)(9.8 m/s2)(0.68 s)2
vi = 8.43 m/s


6 Answers 6

it is because the Force at work here (gravity) is also dependent on the mass

gravity acts on a body with mass m with

you will plug this in to $F=ma$ and you get

and this is true for all bodies no matter what the mass is. Since they are accelerated the same and start with the same initial conditions (at rest and dropped from a height h) they will hit the floor at the same time.

This is a peculiar aspect of gravity and underlying this is the equality of inertial mass and gravitational mass (here only the ratio must be the same for this to be true but Einstein later showed that they're really the same, i.e. the ratio is 1)

Newton's gravitational force is proportional to the mass of a body, $F=frac imes m$, where in the case you're thinking about $M$ is the mass of the earth, $R$ is the radius of the earth, and $G$ is Newton's gravitational constant.

Consequently, the acceleration is $a=frac=frac$, which is independent of the mass of the object. Hence any two objects that are subject only to the force of gravity will fall with the same acceleration and hence they will hit the ground at the same time.

What I think you were missing is that the force $F$ on the two bodies is not the same, but the accelerations are the same.

There are two ways that mass could effect the time of impact:

(1) An object which is very massive has a stronger attraction to the earth. Logically, this might make the object fall faster and so reach the ground sooner.

(2) An object which is very massive is difficult to get moving. (I.e. it has very high inertia.) Thus one might logically expect the very massive object to be more difficult to get moving and so to lose the race.

The miracle is that in the world we live in, these two effects exactly balance and so the heavier mass reaches the ground at the same time.

Now let me give a simple explanation for why it's natural that this comes about. Suppose we have two very heavy masses. If we drop them separately they take some time T to fall. On the other hand, if we attach them together, will they take the same length of time? Think about a sphere split into two halves:

Two two halves of the sphere would fall at the same speed as each other. So if you dropped them next to each other, they'd fall together. And dropping them next to each other isn't going to be any different from screwing them together and dropping them together. That is, there won't be any force on the screws. So the combined (or screwed together) sphere has to fall at the same rate as the split sphere.


Free Fall with Examples

Free fall is a kind of motion that everybody can observe in daily life. We drop something accidentally or purposely and see its motion. At the beginning it has low speed and until the end it gains speed and before the crash it reaches its maximum speed. Which factors affect the speed of the object while it is in free fall? How can we calculate the distance it takes, time it takes during the free fall? We deal with these subjects in this section. First, let me begin with the source of increasing in the amount of speed during the fall. As you can guess, things fall because of the gravity. Thus, our objects gain speed approximately10m/s in a second while falling because of the gravitation. We call this acceleration in physics gravitational acceleration and show with &ldquog&rdquo. The value of g is 9,8m/s² however, in our examples we assume it 10 m/ s² for simple calculations. Now it&rsquos time to formulize what we said above. We talked about the increase in speed which is equal to the amount of g in a second. Thus our velocity can be found by the formula

V=g.t where g is gravitational acceleration and t is the time.

Look at the given example below and try to understand what I tried to explain above.

Example The boy drops the ball from a roof of the house which takes 3 seconds to hit the ground. Calculate the velocity before the ball crashes to the ground. (g=10m/s²)


V=10m/ s².3s=30m/s

We have learned how to find the velocity of the object at a given time. Now we will learn how to find the distance taken during the motion. I give some equations to calculate distance and other quantities. Galileo found an equation for distance from his experiments.

Using this equation we can find the height of the house in given example above. Let&rsquos found how height the ball has been dropped? We use 10 m/s ² for g.

I think the formula now a little bit clearer in your mind. We will solve more problems related to this topic. Now, think that if I throw the ball straight upward with an initial velocity. When it stops and falls back to the ground? We answer these questions now.


Picture shows the magnitudes of velocity at the bottom and at the top. As you can see the ball is thrown upward with an initial v velocity, at the top it&rsquos velocity becomes zero and it changes it&rsquos direction and starts to fall down which is free fall. Finally at the bottom before the crash it reaches its maximum speed which shown as V&rsquo. We have talked about the amount of increase in the velocity in free fall. It increases 9,8m/s in each second due to the gravitational acceleration. In this case, there is also g but the ball&rsquos direction is upward so the sign of g is negative. Thus, our velocity decreases in 9,8m/s in each second until the velocity becomes zero. At the top, because of the zero velocity, the ball changes its direction and starts to free fall. Before solving problems I want to give the graphs of free fall motion.

As you see in the graphs our velocity is linearly increases with an acceleration &ldquog&rdquo, second graphs tells us that acceleration is constant at 9,8m/s², and finally third graphic is the representation of change in our position. At the beginning we have a positive displacement and as the time passes it decreases and finally becomes zero. Now we can solve problems using these graphs and explanations.

Example John throws the ball straight upward and after 1 second it reaches its maximum height then it does free fall motion which takes 2 seconds. Calculate the maximum height and velocity of the ball before it crashes the ground. (g=10m/s²)

Example An object does free fall motion. It hits the ground after 4 seconds. Calculate the velocity of the object after 3 seconds and before it hits the ground. What can be the height it is thrown?

Two examples given above try to show how to use free fall equations. We can find the velocity, distance and time from the given data. Now, I will give three more equations and finishes 1D Kinematics subject. The equations are


First equation is used for finding the velocity of the object having initial velocity and acceleration. Second one is used for calculating the distance of the object having initial velocity and acceleration. Third and last equation is timeless velocity equation. If distance, initial velocity and acceleration of the object is known then you can find the final velocity of the object. Now let&rsquos solve some problems using these equations to comprehend the subject in detail.

Example Calculate the velocity of the car which has initial velocity 24m/s and acceleration 3m/s² after 15 second.

We use the first equation to solve this question.

Example The car which is initially at rest has an acceleration 7m/s² and travels 20 seconds. Find the distance it covers during this period.


Free Fall

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 .

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 they 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. occurs whenever an object is acted upon by gravity alone.

  • Obtain a piece of paper and a pencil. Hold them at the same height above a level surface and drop them simultaneously. The acceleration of the pencil is noticeably greater than the acceleration of the piece of paper, which flutters and drifts about on its way down.

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.

  • Repeat the experiment, but before you begin, wad the piece of paper up into the tightest ball possible. Now when the paper and pencil are released, it should be obvious that their accelerations are identical (or at least more similar than before).

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 coin called a guinea was used and so this demonstration is sometimes 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 nearly so). Only an object in free fall will experience a pure acceleration due to gravity.

The leaning tower of Pisa

Let's jump back in time for a bit. In the Western world prior to the 16th 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 of Stagira (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 round objects 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.

Galileo Galilei, 1638

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 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.

Location, location, location

The generally accepted value for the acceleration due to gravity on and near the surface of the Earth is…

g = 35 kph/s = 22 mph/s = 32 feet/s 2

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 three factors that do. They are location, location, location.

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 1.6 m/s 2 . That's 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 Earth — about two and a half times stronger or 25 m/s 2 . Astronauts cruising through the top of Jupiter's thick atmosphere would find themselves struggling to stand up inside their space ship.

On the Earth, gravity varies with latitude and altitude (to be discussed in a later chapter). 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/s 2 — with only two significant digits — is true for all places on the surface of the Earth and holds for altitudes up to +10 km (the altitude of commercial jet airplanes) and depths down to 󔼜 km (far below the deepest mines).

How crazy are you for accuracy? For most applications, the value of 9.8 m/s 2 is more than sufficient. If you're in a hurry, or don't have access to a calculator, or just don't need to be that accurate rounding g on Earth to 10 m/s 2 is often acceptable. During a multiple choice exam where calculators aren't allowed, this is often the way to go. If you need greater accuracy, consult a comprehensive reference work to find the accepted value for your latitude and altitude.

If that's not good enough, then obtain the required instruments and measure the local value to as many significant digits as you can. You may learn something interesting about your location. I once met a geologist whose job it was to measure g across a portion of West Africa. When I asked him who he worked for and why he was doing this, he basically refused to answer other than to say that one could infer the interior structure of the Earth from a prepared from his findings. Knowing this, one might then be able to identify structures where valuable minerals or petroleum might be found.

Like all professions, those in the gravity measuring business () have their own special jargon. The SI unit of acceleration is the meter per second squared [m/s 2 ]. Split that into a hundred parts and you get the centimeter per second squared [cm/s 2 ] also known as the [Gal] in honor of Galileo. Note that the word for the unit is all lowercase, but the symbol is capitalized. The gal is an example of a Gaussian unit.

00 1 Gal = 1 cm/s 2 = 0.01 m/s 2
100 Gal = 100 cm/s 2 = 1 m/s 2 .

Split a gal into a thousand parts and you get a [mGal].

1 mGal = 0.001 Gal = 10 𕒹 m/s 2

Since Earth's gravity produces a surface acceleration of about 10 m/s 2 , a milligal is about 1 millionth of the value we're all used to.

1 g ≈ 10 m/s 2 = 1,000 Gal = 1,000,000 mGal

Measurements with this precision can be used to study changes in the Earth's crust, sea levels, ocean currents, polar ice, and groundwater. Push it a little bit further and it's even possible to measure changes in the distribution of mass in the atmosphere. Gravity is a weighty subject that will be discussed in more detail later in this book.

Gee, Wally

Don't confuse the phenomenon of acceleration due to gravity with the unit of a similar name. The quantity g has a value that depends on location and is approximately

almost everywhere on the surface of the Earth. The unit g has the exact value of…

They also use slightly different symbols. The defined unit uses the roman or upright g while the natural phenomenon that varies with location uses the italic or oblique g. Don't confuse g with g.

As mentioned earlier, the value of 9.8 m/s 2 with only two significant digits is valid for most of the surface of the Earth up to the altitude of commercial jet airliners, which is why it is used throughout this book. The value of 9.80665 m/s 2 with six significant digits is the so called or . It's a value that works for latitudes around 45° and altitudes not too far above sea level. It's approximately the value for the acceleration due to gravity in Paris, France — the hometown of the International Bureau of Weights and Measures. The original idea was to establish a standard value for gravity so that units of mass, weight, and pressure could be related — a set of definitions that are now obsolete. The Bureau chose to make this definition work for where their laboratory was located. The old unit definitions died out, but the value of standard gravity lives on. Now it's just an agreed upon value for making comparisons. It's a value close to what we experience in our everyday lives — just with way too much precision.

Some books recommend a compromise precision of 9.81 m/s 2 with three significant digits for calculations, but this book does not. At my location in New York City, the acceleration due to gravity is 9.80 m/s 2 . Rounding standard gravity to 9.81 m/s 2 is wrong for my location. The same is true all the way south to the equator where gravity is 9.780 m/s 2 at sea level — 9.81 m/s 2 is just too big. Head north of NYC and gravity gets closer and closer to 9.81 m/s 2 until eventually it is. This is great for Canadians in southern Quebec, but gravity keeps keeps increasing as you head further north. At the North Pole (and the South Pole too) gravity is a whopping 9.832 m/s 2 . The value 9.806 m/s 2 is midway between these two extremes, so it's sort of true to say that…

This is not the same thing as an average, however. For that, use this value that someone else derived…

Here are my suggestions. Use the value of 9.8 m/s 2 with two significant digits for calculations on the surface of the Earth unless a value of gravity is otherwise specified. That seems reasonable. Use the value of 9.80665 m/s 2 with six significant digits only when you want to convert m/s 2 to g. That is the law.

The unit g is often used to measure the acceleration of a reference frame. 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.


A motion is said to be uniformly accelerated when, starting from rest, it acquires during equal time intervals, equal increments of speed.

Let us first look more closely at Galileo's proposed definition.

Is this the only possible way of defining uniform acceleration? Not at all! Galileo says that at one time he thought a more useful definition would be to use the term uniform acceleration for motion in which speed increased in proportion to the distance traveled, D d, rather than to the time fit. Notice that both definitions met Galileo's requirement of simplicity. (In fact, both definitions had been discussed since early in the fourteenth century.)
Furthermore, both definitions seem to match our common sense idea of acceleration about equally well. When we say that a body is "accelerating," we seem to imply "the farther it goes, the faster it goes," and also "the longer time it goes, the faster it goes." How should we choose between these two ways of putting it? Which definition will be more useful in the description of nature? This is where experimentation becomes important. Galileo chose to define uniform acceleration as the motion in which the change of speed v is proportional to elapsed time D t, and then demonstrate that this matches the behavior of real moving bodies, in laboratory situations as well as in ordinary, "un-arranged," experience. As you will see later, he made the right choice. But he was not able to prove his case by direct or obvious means, as you shall also see.

Describe uniform speed without referring to dry-ice pucks and strobe photography or to arty particular object or technique of measurement.

Express Galileo's definition of uniformly accelerated motion in words and in the form of an equation.

What two conditions did Galileo want his definition of uniform acceleration to meet?

Galileo cannot test his hypothesis directly

After Galileo defined uniform acceleration so that it would match the way he believed freely falling objects behaved, his next task was to devise a way of showing that the definition for uniform acceleration was useful for describing observed motions.

Suppose we drop a heavy object from several different heights say, from windows on different floors of a building. We want to check whether the final speed increases in proportion to the time it takes to fall-that is, whether D v is "proportional to" D t, or what amounts to the same thing, whether D v/ D t is constant. In each trial we must observe the time of fall and the speed just before the object strikes the ground.

But there's the rub. Practically, even today, it would be very difficult to make a direct measurement of the speed reached by an object just before striking the ground. Furthermore, the entire time intervals of fall (less than 3 seconds even from the top of a 10-story building) are shorter than Galileo could have measured accurately with the clocks available to him. So a direct test of whether D v/ D t is constant was not possible for Galileo.

Which of these are valid reasons why Galileo could not test directly whether the final speed reached by a freely falling object is proportional to the time of fall?
(a) His definition was wrong.
(b) He could not measure the speed attained by an object just before it hit the ground.
(c) There existed no instruments for measuring time.
(d) He could not measure ordinary distances accurately enough.
(e) Experimentation was not permitted in Italy.

Looking for logical consequences of Galileo's hypothesis

Large distances of fall and large time intervals for fall are, of course, easier to measure than the small values of D d and D t that would be necessary to find the final speed just before the falling body hits. So Galileo tried to find, by reasoning, how total fall distance ought to increase with total fall time if objects did fall with uniform acceleration. You already know how to find total distance from total time for motion at constant speed. Now we will derive a new equation that relates total fall distance to total time of fall for motion at constant acceleration. In this we shall not be following Galileo's own derivation exactly, but the results will be the same. First, we recall the definition of average speed as the distance traversed D d divided by the elapsed time D t:

vav = D d/ D t
This is a general definition and can be used to compute the average speed from measurement of D d and D t, no matter whether D d and D t are small or large. We can rewrite the equation as
D d = vav x D t
This equation, still being really a definition of vav is always true. For the special case of motion at a constant speed v, then vav = v and therefore, D d = v x D t. When the value of v is known (as, for example, when a car is driven with a steady reading of 60 mph on the speedometer), this equation can be used to figure out how far ( D d) the car would go in any given time interval ( D t). But in uniformly accelerated motion the speed is continually changing-so what value can we use for vav?

The answer involves just a bit of algebra and some plausible assumptions. Galileo reasoned (as others had before) that for any quantity that changes uniformly, the average value is just halfway between the beginning value and the final value. For uniformly accelerated motion starting from rest (where vinitial = 0 and ending at a speed vfinal this rule tells us that the average speed is halfway. More generally the average speed would be between 0 and vfinal - that is,


vav= 1/2 vfinal.
(More generally, the average velocity would be
vav=(vinitial + vfinal)/2.
If this reasoning is correct, it follows that
D d = 1/2 vfinal x D t for uniformly accelerated motion starting from rest. This relation could not be directly tested either, because the last equation still contains a speed factor. What we are trying to arrive at is an equation relating total distance and total time, without any need to measure speed.
Now we look at Galileo's definition of uniform acceleration: a = D v/ D t. We can rewrite this relationship in the form
D v = a x D t. The value of D v is just vfinal - vinitialr and vinitial = 0 for motion that begins from rest. Therefore we can write
D v=a x D t
vfinal - vinitial = a x D t
vfinal = a x D t
Now we can substitute this expression for vfinal into the equation for D d above. Thus if the motion starts from rest, and if it is uniformly accelerated (and if the average rule is correct, as we have assumed) we can write
D d = 1/2 vfinal x D t
= 1/2 (a x D t) x D t
Or, regrouping terms.
D d = 1/2 a( D t) 2

This is the kind of relation Galileo was seeking-it relates total distance D d to total time D t, without involving any speed term.

Before finishing, though, we will simplify the symbols in the equation to make it easier to use. If we measure distance and time from the position and the instant that the motion starts (dinitial= 0 and tinitial = 0), then the intervals D d and D t have the values given by dfinal and tfinal. Because we will use the expression dfinal/ t 2 final , many times, it is simpler to write it as d/t 2

-it is understood that d and t mean total distance and time interval of motion, starting from rest. The equation above can therefore be written more simply as

dfinal = 1/2 a x t 2 final
Remember that this is a very specialized equation-it gives the total distance fallen as a function of total time of fall but only if the motion starts from rest (vinitial = 0), if the acceleration is uniform (a = constant), and if time and distance are measured from the start (tinitial = 0 and dinitial = 0).

Galileo reached the same conclusion, though he did not use algebraic forms to express it. Since we are dealing only with the special situation in which acceleration a is constant, the quantity 2a is constant also, and we can cast the conclusion in the form of a proportion: in uniform acceleration from rest, the distance traveled is proportional to the square of the time elapsed, or

dfinal / t 2 final
For example, if a uniformly accelerating car starting from rest moves 10 m in the first second, in twice the time it would move four times as far, or 40 m in the first two seconds. In the first 3 seconds it would move 9 times as far-or 90 m. Another way to express this relation is to say that the ratio dfinal to t 2 final has a constant value, that is, dfinal / t 2 final = constant . Thus a logical result of Galileo's original proposal for defining uniform acceleration can be expressed as follows: if an object accelerates uniformly from rest, the ratio d/t 2 should be constant. Conversely, any motion for which this ratio of d and t 2 is found to be constant for different distances and their corresponding times, we may well suppose to be a case of motion with uniform acceleration as defined by Galileo. Of course, we still must test the hypothesis that freely falling bodies actually do exhibit just such motion. Recall that earlier we confessed we were unable to test directly whether D v/ D t has a constant value. Galileo showed that a logical consequence of a constant value of v/ D t would be a constant ratio of dfinal to t 2 final. The values for total time and distance of fall would be easier to measure than the values of short intervals D d and D t needed to find D v. However, measuring the time of fall still remained a difficult task in Galileo's time. So, instead of a direct test of his hypothesis, Galileo went one step further and deduced an ingenious, indirect test.

Why was the equation d = 1/2at 2 more promising for Galileo than a = D v/ D t in testing his hypothesis?

If you simply combined the two equations D d = v x D t and D v = a x D t it looks as if one might get the result D d = a x D t 2 . What is wrong with doing this?

Realizing that a direct quantitative test with a rapidly and freely falling body would not be accurate, Galileo proposed to make the test on an object that was moving less rapidly. He proposed a new hypothesis:

if a freely falling body has an acceleration that is constant, then a perfectly round ball rolling down a perfectly smooth inclined plane will also have a constant, though smaller, acceleration.

Thus Galileo claimed that if d/t 2 is constant for a body falling freely from rest, this ratio will also be constant, although smaller, for a ball released from rest and rolling different distances down a straight inclined plane.

Here is how Salviati described Galileo's own experimental test in Two New Sciences:


This picture painted in 1841 by G. Bezzuoli, attempts to reconstruct an experiment Galileo is alleged to have made during his time as lecturer at Pisa. Off to the left and right are men of ill will: the blasé Prince Giovanni de Medici (Galileo had shown a dredging-machine invented by the prince to be unusable) and Galileo's scientific opponents. These were leading men of the universities they are shown here bending over a book of Aristotle, where it is written in black and white that bodies of unequal weight fall with different speeds. Galileo, the tallest figure left of center in the picture, is surrounded by a group of students and followers.

Angle of Incline
For each angle, the acceleration is found to be a constant. Spheres rolling down planes of increasingly steep inclination. At 90° the inclined plane situation matches free fall. (Actually, the ball will start slipping instead of rolling long before the angle has become that large.)

Free Fall-Galileo Describes Motion

In general, for each angle of incline, the value of d / t1 2 was constant. Galileo did not present his experimental data in the full detail which has become the custom since. However, his experiment has been repeated by others, and they have obtained results which parallel his. This is an experiment which you can perform yourself with the help of one or two other students.
(b) Galileo's second experimental finding relates to what happens when the angle of inclination of the plane is changed. He found that whenever the angle changed, the ratio d / t 2 took on a new value, although for any one angle it remained constant regardless of distance of roll. Galileo confirmed this by repeating the experiment "a full hundred times" for each of many different angles. After finding that the ratio d / t 2 was constant for each angle of inclination for which measurements of t could be carried out conveniently, Galileo was willing to extrapolate. He concluded that the ratio dl / t 2 is a constant even for larger angles, where the motion of the ball is too fast for accurate measurements of t to be made. Finally, Galileo reasoned that in the particular case when the angle of inclination became 90°, the ball would move straight down-and so becomes the case of a falling object. By his reasoning, d/t 2 would still be some constant in that extreme case (even though he couldn't say what the numerical value was.)

Because Galileo had deduced that a constant value of d/t 2 was characteristic of uniform acceleration, he could conclude at last that free fall was uniformly accelerated motion.

Now that you are familiar with the historical concepts of free fall, move on to the experiment. Or you can look at a spreadsheet of some actual student data. The data were collected and manipulated according to the described experiment. It definitely demonstrates that there is a constant acceleration. A good question to ask the students is why there is so much error. Then have your students modify the experiment. When it is all said and done, take the quiz.


Watch the video: ΕΟΔΥ ΔΩΡΕΑΝ RAPID TEST ΣΤΗ ΝΕΑΠΟΛΗ 05 01 2021 (October 2021).