Derivate the forumla for the acceleration due to gravity

[JohnDubYa]

Yes, I earlier dropped the term "constant" in my definition.

While students are learning kinematics, I don't give g a name. I tell them that near the surface of the Earth a body in freefall accelerates at 9.8 m/s/s, and we give that value the label g. Later, after discussing gravity, I define it in terms of the gravitational force and call it "the gravitational field."

Overall, a good discussion.

 

[robphy]

One last post...
Here are some comments on "acceleration due to gravity" by some physics teachers [from "The Physics Teacher"].

http://scitation.aip.org/dbt/dbt.jsp?KEY=PHTEAH&Volume=37&Issue=5
SURVEY OF HIGH-SCHOOL PHYSICS TEXTS: Quibbles, misunderstandings, and egregious mistakes
Physics Textbook Review Committee
The Physics Teacher--May 1999 Volume 37, Issue 5, pp. 297-308

7. The texts differentiate between mass and weight in various chapters. All of them refer to g as the “acceleration due to gravity.” This is surely confusing, if not misleading, to students who wonder why acceleration should have anything to do with the weight of an object sitting still on a pan balance. A more useful definition of g is that it is the gravitational field strength. The analogy with the electric field strength can be made early as well as late. Thus F_grav = mg is analogous to F_electric = qE. To two significant figures, g =9.8 N/kg; thus the weight of 1.0 kg is 9.8 N. Otherwise the weight of 1.0 kg is 9.8 kg m/s2, which of course is valid, but confusing for a student.

and

http://scitation.aip.org/dbt/dbt.jsp?KEY=PHTEAH&Volume=38&Issue=3
NOTES: g-whizz
John A. McClelland
The Physics Teacher--March 2000 Volume 38, Issue 3, p. 150

In the review of current textbooks¹, it is pointed out that g should be described as the strength of the gravitational field rather than as an acceleration. The distinction may benefit from a little elaboration.

A book of mass m rests on a table. What is its weight w? A typical textbook answer is that w = mg where g is “the acceleration due to gravity.”

Given that the book is in equilibrium² it has no acceleration due to gravity, nor has anything else in the figure, so the statement is self-contradictory. A worse case arises where a body is falling but is not in free fall. Its weight is still described as the “mass times the acceleration due to gravity,” but its acceleration is less than this—potentially very confusing.

Even if g is called the acceleration of a body in free fall, it is not at all obvious to a beginner that the product of a mass m kilogram and g m/s2 gives w Newton. It is more logical and much more transparent if g is introduced as the gravitational field at the surface of Earth in Newton/kilogram. This may or may not be related to GM/R² according to the level of the course. If the concept of a field is not required, g can be called the strength or intensity of gravity. The gravitational field (or the strength of gravity) at a point can be defined as the force in Newtons experienced by each kilogram of mass placed there. It is not hard to see that m kg x g N/kg gives w N. Because this is analogous to how we deal with forces on electric charges in an electric field it also makes pedagogical sense.

Applying Newton’s second law to a body at Earth’s surface shows that, when w = mg is the net force, the acceleration of the body, in m/s2 is numerically equal to the strength of the field in Newton/kilogram. This is free fall. In all other cases, the net force is different from w and the acceleration is different from the acceleration in free fall. In other words, the acceleration in free fall is both a consequence of the strength of the gravitational field and a special case.

Away from Earth’s surface, there really is no satisfactory way to arrive at the acceleration of a body in free fall except through the strength of the gravitational field there. For example, at the surface of the Moon, g=1.6 N/kg, so the acceleration of a body in free fall is 1.6 m/s2.

There are probably two reasons why it has become conventional to define g as an acceleration rather than as a field strength. One is that most introductory textbooks cover kinematics before dynamics and use examples of bodies taken to be falling freely, often including projectiles. Having introduced a value for the acceleration in free fall and attributed it to gravity, it then becomes “the acceleration due to gravity” in all that follows. The other is that it is possible to obtain a quite reasonable value for this acceleration experimentally. To be logically and pedagogically sound, we should quote g in Newton/kilogram, refer to “the acceleration in free fall,” and mothball the expression “acceleration due to gravity.”

References

1. Editorial comment, “Quibbles, misunderstandings and egregious mistakes,” Phys. Teach. 37, 297 (1999).

2. If the table is on a rotating Earth it is not in equilibrium.

 

[JohnDubYa]

RE: "This is surely confusing, if not misleading, to students who wonder why acceleration should have anything to do with the weight of an object sitting still on a pan balance. A more useful definition of g is that it is the gravitational field strength."

Oooh, that's wierd. It is as if I wrote it. (But I didn't. They explained their views better than I did.)

Thanks for the references.

 

[woodysooner]

If accuracy is what you desire then we should drop Newton's second law altogether and use general relativity to solve classical physics problems.

Johndubya, what GR equation do you use to do this?

 

[pmb_phy]

By the way, the term "acceleration due to gravity" is a misnomer, because it assumes g is an acceleration. It is not. The correct term, in my opinion, should be "gravitational field constant."

Since this is in the Classical Physics forum then I'm assuming that this is a discussion on Newton's Law of gravity.

In that sense = What is your justification for your statement it assumes g is an acceleration. It is not.

g(r) is a vector function of a vector variabe and as such varies from place to place and therefore can't be considered to be a constant. g(r) is an acceleration vector as a function of position since

\bold g(\bold r) = \frac{d^2\bold r}{dt^2}

The only constant is G in

\bold F_{12} = \frac{Gm_{1}m_{2}}{r^2} \bold e_{12}

Pete

 

[pmb_phy]

Given that the book is in equilibrium2 it has no acceleration due to gravity, nor has anything else in the figure, so the statement is self-contradictory.

That seems to be a common error in thinkng in physics. Its a result of a misuse/misunderstanding of the principle of superposition. F = mg refers to that acceleration what would occur if this were the only force acting on the body in the absense of all other forces. This same thing carries over to EM. Griffitth quite nicely explains all this in a most beautiful fashion.

Pete

 

[JohnDubYa]

What is your justification for your statement it assumes g is an acceleration. It is not.

The fact that the constant g can factor into a non-accelerating system. When students are told that g is an acceleration, then they assume that a body with a weight mg must accelerate at g.

Sure, WE know better.

 

[JohnDubYa]

woody, I'm not sure what you are asking.

Are you a Sooner?

 

[woodysooner]

A Sooner I am, can't wait to get back to Norman only few more days.

If accuracy is what you desire then we should drop Newton's second law altogether and use general relativity to solve classical physics problems.

I just meant how in the instance do we use GR, guess I need to get with it and learn the math behind GR, just thought if you gave me the equation for that instance in GR i could run with it.

Thanx

 

[JohnDubYa]

Gawd, I miss Oklahoma. Are you a physics major at the University?

My allusion to GR was a jab at those that conjure GR to explain the most elementary physics.

 

[woodysooner]

good question, at the present I'll be a junior in chemical eng. and eng physics, but bout to give all that up and start over lol in Physics and Math.

if dad let's me, he runs the show sadly, but he shouldn't cause he don't pay for it the state and a lot of other do.

 

[JohnDubYa]

Take all the classes that Bruce Mason teaches. I had him as an undergraduate and he may be the best Prof I have ever had at any level. Doc Watson and Mike Morrison too.

 

[cepheid]

I'm a student and I'd just like to offer my two cents on this debate over terminology:

When we began our study of physics in high school, we began with kinemetics. Certain problems (involving freefall and later projectile motion) involved use of the physical quantity g, which I learned was the acceleration of objects in freefall. (Our physical model for falling objects always neglected air resistance, so I did not trouble myself over it). My teacher referred to this as the acceleration due to gravity, which made perfect sense to me, because it meant that gravity caused objects to accelerate at 9.81m/s^2. More precisely, when gravity was the sole force on an object, it's resulting acceleration would be g. I accepted this to be true, even though I had no idea how or why it was so, i.e. was was the nature of the graviational force and why did all objects under it's influence accelerate like so?

Later, when we studied dynamics, we attempted to determine how such 'action at a distance' was possible, and we were introduced to the notion of fields. From Newton's Universal Law of Gravitation, we derived the 'graviational field strength' at the Earth's surface to be g. I must admit to being mometarily confused. Why should these two quantities with totally different names be represented by the same symbol? I decided that the quanitites must actually be one and the same, and set about explaining it to myself. I asked myself the question: "Why is the acceleration of objects under the sole influence of Earth's gravity equal to the strength of the graviational field at the Earth's surface? The answer is simple: From the definition of a field, we see that the Earth's graviational field exerts a force of 9.81 Newtons on every kilogram of an object's mass. It therefore exerts a force of 9.81 Newtons on a 1kg object in freefall. Since the definition of a Newton is the force required to accelerate a 1kg object at 1 m/s^2, it stands to reason that our 1kg object in freefall will accelerate at 9.81 m/s^2, since 9.81 N of force are being exerted on every kilogram of mass. Therefore, the two quantities are one and the same: the acceleration of an object in freefall is equal to the strength of the gravitational field. I was sure this was true when I observed that

\frac{N}{kg} = \frac{m}{s^2}

In short, I sorted it out on my own and I really don't care whether you call g 'acceleration due to gravity' or 'gravitational field strength' since the fact that the two must have the same magnitude (and direction) at the Earth's surface is direct consequence of the Law of Gravitation and Newton's Second Law.