Derivate the forumla for the acceleration due to gravity

[cutesoqq]

Hello,
Does anyone can tell me how to derivate the forumla for the acceleration due to gravity, i mean how to find the g of other planets like the moon?...

 

[Gokul43201]

Look at https://www.physicsforums.com/showthread.php?t=30396

The last post tells you how to derive g = MG/R^2

M : planet's mass,
G : Universal Grav. Const
R : planet's radius

 

[JohnDubYa]

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."

 

[cutesoqq]

Sorry, do u mean that using the equation above, the g of the moon or other planets can be solve?!...Actually, I still don't really understand...

 

[Doc Al]

Sorry, do u mean that using the equation above, the g of the moon or other planets can be solve?!...Actually, I still don't really understand...

Short answer: yes.

Start with Newton's law of gravity. Find the gravitational force on a mass (m) on the surface of a planet/moon (radius = R) due to the mass of that planet/moon (M). F = GMm/R^2. Does that make sense?

Once you have the force due to gravity, the "acceleration" due to gravity is gotten via Newton's 2nd law: a = F/m, so:

a = g = GM/R^2

Make sense?

 

[TeV]

so:

a = g = GM/R^2

Make sense?

Yup,if the planet is a perfect sphere with radially symmetric mass distribution (density function in spherical coordinate system)

 

[dedaNoe]

a = g = GM/R^2
Make sense?

Not quite.
GM/R^2=V/R is gravitational potential (V) over that distance(R).
I don't se why it would be accel.
By the way, is every body in some gravity field if accelerating?

 

[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."

I disagree. The term "acceleration due to gravity" refers to the fact that if you place a point test mass at a particular location then a free particle will accelerate at a rate g.

Why don't you think that g is an acceleration? The term "field" as applied to both EM and Newtonian gravity refers to what happens to a free particle when a test particle is at a particular location. For gravity and EM its defined as F/m and F/q respectively.

 

[Doc Al]

Not quite.
GM/R^2=V/R is gravitational potential (V) over that distance(R).

What does that statement mean?

I don't se why it would be accel.
By the way, is every body in some gravity field if accelerating?

The term "acceleration due to gravity" means the acceleration a test mass would experience just due to the gravitational force. (See pmb_phy's post above.) It is a measure of the strength of the field.

 

[cutesoqq]

Short answer: yes.

Start with Newton's law of gravity. Find the gravitational force on a mass (m) on the surface of a planet/moon (radius = R) due to the mass of that planet/moon (M). F = GMm/R^2. Does that make sense?

Once you have the force due to gravity, the "acceleration" due to gravity is gotten via Newton's 2nd law: a = F/m, so:

a = g = GM/R^2

Make sense?

Do u mean that the equation is available for any planets?! Then, I must put G be the universal gravitational constant, M be mass of the moon as well as R be the radius of the moon, right?!

 

[Doc Al]

Do u mean that the equation is available for any planets?! Then, I must put G be the universal gravitational constant, M be mass of the moon as well as R be the radius of the moon, right?!

Yes. This of course assumes a simplified model of the moon (or planet) of being spherically symmetric. (As TeV explained.) But close enough!

 

[Gokul43201]

If you did not actually look at that post, here it is :

r=R+h

So,

U =\frac {-GMm} {r} = \frac {-GMm} {R+h} = \frac {-GMm} {R} (1 + h/R)^{-1}

For h<<R, you can expand the last term binomially, and neglect terms of second order and up. So,

U = \frac {-GMm} {R} (1 - h/R) =\frac {-GMm} {R} + \frac {GMmh} {R^2}
= Constant + (m)*(\frac {GM} {R^2})*h = Constant + mgh

Since we are interested only in changes in potential, we can throw away the constant term.

 

[HallsofIvy]

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."

Are you talking about "g" or "G"? The "g" in f= mg is definitely an acceleration: it is the acceleration of any object on the surface of the earth: approximately 9.81 m/s².

On the other hand "G", in F= \frac{GmM}{r^2} is the "universal gravitational constant".

Gokul43201's original response told how to calculate "g" for other planets.

 

[JohnDubYa]

RE: "I disagree. The term "acceleration due to gravity" refers to the fact that if you place a point test mass at a particular location then a free particle will accelerate at a rate g."

Not if there are other forces acting on the particle, hence the problem. So defining g as the acceleration due to gravity causes confusion unless you stick the qualifier onto the definition, which is awkward.

If I had a dime for every student who had an object accelerating at g when it wasn't supposed to, I would be a rich man. And students make this mistake because we use confusing terminology.

Defining g "the acceleration due to gravity" is not really wrong, but unfortunate in my opinion. But I think some definitions are more practical than others.

The constants g and E are directly analagous. One indicates the force that would act on a mass if the mass was placed at that point in space. The other indicates the force on a charge if placed at that point in space. One we call the electric field. The other then is obviously the gravitational field. So it makes sense to call g "the gravitational field constant." When an object is placed at a point in space where g is defined, it may or may not accelerate at a value coinciding with g depending on the value of the other forces acting on the object.

 

[Janus]

Not if there are other forces acting on the particle, hence the problem. So defining g as the acceleration due to gravity causes confusion unless you stick the qualifier onto the definition, which is awkward.

The qualifier is already there in "due to gravity. That's what it means; that it is that part of any acceleration the object undergoes (if any), that is attributed to gravity.

If there are other forces involved then there will of course be accelerations due to them and the net acceleration would be the result of all the accelerations combined.

 

[JohnDubYa]

RE: "The qualifier is already there in "due to gravity. That's what it means; that it is that part of any acceleration the object undergoes (if any), that is attributed to gravity."

A ball dropped in molasses does not accelerate at g. To say that it accelerates downward at g, and that this motion is partially canceled by an acceleration upward, makes no intuitive sense. Sure it would work mathematically, but that doesn't make it sensible.

However, the Earth apply a gravitational force down on the ball and the molasses can apply a separate force upwards. But there is only one resulting motion.

 

[Nenad]

calling it 'acceleration due to gravity' is a correct interpretation, if you've ever studied general relativity, there is no difference between a gravitational field force and acceleration. An observer cannot distinguish between the two. Therefore making gravity and acceleration the same thing.

 

[JohnDubYa]

RE:"calling it 'acceleration due to gravity' is a correct interpretation, if you've ever studied general relativity, there is no difference between a gravitational field force and acceleration."

Sure there is. When I have a box sitting on a desk, it is located inside a gravitational field. But it is not accelerating.

Besides, explanations that resort to general relativity are of little practical use to a person studying purely classical physics.

RE: "If you have trouble getting the planet's weight then you could still use another not-so-practical method to measure "g". You can get on a rocket on that planet or celestial body you want to measure "g" at. Start the engines and accelerate until your velocity (Vel) is such that you manage to scape from the gravitational field of that planet or celestial body. You might have to do this several times until you get it right... =( Then: g = (Vel^2)/(2*R)"

This equation assumes that the spaceship does not develop any thrust after time t = 0, right? A more apt example would be to use a cannon ball instead of a rocket ship.

 

[Nenad]

yes, general relativity is of little use to clessical physics, but it is a lot more accurate.

 

[JohnDubYa]

RE: "yes, general relativity is of little use to clessical physics, but it is a lot more accurate."

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

Using the term "acceleration due to gravity" is misleading and causes students to miss problems they normally wouldn't. Calling g the gravitational field constant does not, in my experience, lead students to assign accelerations of 9.8 m/s/s to bodies that are not in free fall.

 

[zoobyshoe]

Using the term "acceleration due to gravity" is misleading and causes students to miss problems they normally wouldn't. Calling g the gravitational field constant does not, in my experience, lead students to assign accelerations of 9.8 m/s/s to bodies that are not in free fall.

So, if you don't use the acceleration due to gravity to define and describe the force that keeps everything stuck to the planet's surface, what do you use?

 

[JohnDubYa]

I am not sure about your question, but here goes:

We define a gravitational field at every point in space surrounding a planet, such as Earth. This field is a vector and points towards the center of the planet. We denote the magnitude of this field as g. The purpose of this value is to indicate the magnitude of the gravitational force that would exert on any mass placed at that point,

F_g = mg.

If there are no other forces acting on the mass, then

\sum F = F_g = mg = ma,

g = a,

and so in this situation (free fall) the mass accelerates at a value that corresponds numerically to g.

Now there is no confusion. Since you are not calling g an acceleration, students are less likely to assign an acceleration of 9.8 m/s/s to objects that are not in free fall. They consider g merely a constant. Objects in free fall happen to accelerate at a value that agrees with that constant.

I have used this approach and found it very effective. In fact, in those situations where a student assigns g to non-freely objects, I ask them "Where did you get the idea that the object was falling at 9.8 m/s/s?" Their answer is invariably "Because that is the acceleration due to gravity, as it says in the book (or in their last class)."

 

[zoobyshoe]

I guess I'm not clear about the situations where the students misapply 9.8 m/sec².

 

[robphy]

In my experience, I see lots of students apply "g" as the acceleration down an incline, the acceleration in an Atwood machine, etc... possibly because these masses involved do move "due to gravity" [although the value of acceleration is not only due to the force of gravity].

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."

I agree with JohnDubYa that "acceleration due to gravity" is a poor term.
A better term might be "freefall acceleration".

As he points out above in a recent post in this thread, the g appearing in the definition of weight is identified as an "acceleration" only after using Newton's Second Law.

The best term is probably "gravitational field [of the earth]" in complete [classical] analogy to "electric field [of a point charge]".

F_{Grav}=m\frac{GM}{r^2}=mg
F_{Elec}=q\frac{kQ}{r^2}=qE

Like \vec E, the "gravitational field" \vec g is a radial vector field, which is approximately constant in a small enough region of space. As has been discussed above, the magnitude g=9.8 m/s^2 when r is equal to the radius of the earth.

One can continue to find analogies in defining a "potential" and a "potential energy". Indeed, one can apply Gauss' Law to both vector fields. Wouldn't the use of the "gravitational field" \vec g be a good introduction to next semester's electric field?

By the way, this does not mean that you have to introduce Newton's Law of Gravitation right away. Simply use the term "gravitational field" for "g". If you need to describe the "freefall acceleration" when first discussing kinematics, maybe a_{freefall}=9.8\ m/s^2 is a better notation. Some bright student might recognize that a_{freefall} has the same units and value as g (unlike E). You can use that as a hook to mention relativity, if you wish.

 

[JohnDubYa]

Freefall acceleration would certainly be better than "acceleration due to gravity" and light-years better than the God-awful "acceleration of gravity" (as if gravity accelerates).

As with robphy, I find calling g the gravitational field constant a nice way to introduce students to terminology they will find in the second semester.

By the way, this is not merely a matter of semantics. Words mean things. Poor terminology can turn a good student into a confused student, so instructors should take great care in phrasing physics concepts.

 

[Doc Al]

As with robphy, I find calling g the gravitational field constant a nice way to introduce students to terminology they will find in the second semester.

By the way, this is not merely a matter of semantics. Words mean things. Poor terminology can turn a good student into a confused student, so instructors should take great care in phrasing physics concepts.

I like it and will begin using it immediately! Thanks, JohnDubYa. (Yes, terminology does matter.)

I'll probably start calling g the "gravitional field strength/constant at the Earth's surface".

 

[zoobyshoe]

By the way, this is not merely a matter of semantics. Words mean things.

This is a disturbing couplet of sentences to read from someone concerned about avoiding confusion. "Words mean things." suggest that semantics has nothing to do with the meanings of words. But of course it does.

Poor terminology can turn a good student into a confused student, so instructors should take great care in phrasing physics concepts.

I agree, absolutely. Physics has specific and rigorous terminology in many cases, which helps to keep things clear. Once in a while it comes up that people aren't all on the same page about terms, though. To the extent that textbooks have perpetuated confusion, it should be corrected, but I'm not sure to whom you would appeal for that.

 

[chris_tams]

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."

Surely g is an accelleration due to gravity! How can it not be, when gravity causes something to accelerate at g. If it isn't an acceleration, how is it a field constant? It isn't, becuase g isn't constant throughout a field?

 

[ArmoSkater87]

cutesoqq, its this simple, g = Gm/r^2, u can think of g as the acceleration due to gravity of any planet. See for urself by pluging in the values for Earth, and u will get 9.8 m/s^2

 

[JohnDubYa]

RE: "Surely g is an accelleration due to gravity!"

This would imply that if an object is placed inside the Earth's gravitational field, it will accelerate at g. Not necessariy so, but the confused student will do this everytime.

 

[zoobyshoe]

John,

Do you have any objection to the way "acceleration" is used in this example from a physics text:

"Suppose next that a body is thrown vertically upward with an initial velocity of 96 ft/sec. How far will it rise above the starting point, and how long will it be in motion? In order to solve this problem it is important to realize that as soon as it leaves the hand the moving body experiences a constant downward acceleration of 32 ft/sec², at every instant. This causes the upward velocity to decrease to zero at the point of maximun height and then produces an increasing downward velocity as the body falls back down to its starting point. It may help to understand this behavior if we anticipate somewhat our disscussion of dynamics in order to point out that accelerations are produced by forces. After the body leaves the hand the only force (air resistance neglected) acting on it is the pull of the earth. Furthermore, since this is a constant force (near the surface of the earth), the acceleration is therefore constant and is downward in the direction of the force."

Fundamentals of University Physics
W. Wallace McCormick 1969

 

[russ_watters]

RE: "Surely g is an accelleration due to gravity!"

This would imply that if an object is placed inside the Earth's gravitational field, it will accelerate at g. Not necessariy so, but the confused student will do this everytime.

The confused student just needs to have explained to him/her exactly what it means, not leaving out things like "...at sea level" and "...neglecting air resistance."

I really think that "gravitational field constant" will just create more confusion.

 

[robphy]

I really think that "gravitational field constant" will just create more confusion.

Given the unfortunate use of "acceleration due to gravity" used in current textbooks, I agree. However, as I argued above, I would advocate replacing the term in textbooks with "gravitational field" in analogy to the "electric field".

The confused student just needs to have explained to him/her exactly what it means, not leaving out things like "...at sea level" and "...neglecting air resistance."

Indeed, "knowing when a formula applies" is a stumbling block for many students.

Hence, if one wants to emphasize the notion of acceleration (as one does in kinematics), I suggested above the more descriptive term "freefall acceleration" for "g". In my opinion, "freefall acceleration [at sea level]" is better than "acceleration due to gravity [neglecting air resistance] [neglecting any other forces] [at sea level]".



Consider now the definition of weight, W=F_grav=mg. In the context of a book at rest on a table, "g" is probably better thought of as the "gravitational field" [at the book's location] (in analogy to the "electric field" E in F_elec=qE) rather than "the acceleration the book would experience if it were in freefall", an interpretation which must make use of Newton's Second Law.

 

[krab]

JohnDubYa: I think you make too big a deal of this. I don't hold to the notion that ambiguous terminology always impedes learning. Sure, "acceleration due to gravity" can be misleading, but even worse are words like "force" and "work". The latter 2 words existed in English before there were physics definitions, so students have to do a lot of thinking to e.g. get over the fact that pushing something that does not move results in no (physics) work done on that object. In my own experience, conflicts like these force a person to think. I am of the opinion that all real learning comes from thinking, not from remembering definitions. So e.g. when I learned g=acc. due to gravity, I pondered the fact that the reason I am not accelerating is because the floor knows exactly the force needed to cancel gravity. I learned a lot from this thought. Calling g by some generic name would also rob the students of making a direct connection between g and one of the (non-obvious) wonders of nature: that things freefall at the same rate. Contemplating this fact has taught me a lot.

 

[jcsd]

g is an accelartion and guess what causes it! gravity! Now I'm sure every student is aware that objects in gravitaional fields don't necessarily accelerate, but it's easy to remember that F = ma and that 'g' just slots into a to create F = mg. IMO describing it as a "universal constant" is X1000 more confusing.

 

[JohnDubYa]

Zooby, the description sounds okay to me.

JohnDubYa: I think you make too big a deal of this. I don't hold to the notion that ambiguous terminology always impedes learning."

It doesn't ALWAYS impede learning, but it sometimes does. I have seen the effects of calling g an acceleration, and they are very real.

RE: " Sure, "acceleration due to gravity" can be misleading, but even worse are words like "force" and "work".

Sure, but that is a battle that cannot be won because these definitions are so firmly entrenched throughout all of science. And I know of no better words to use as replacements.

RE: "In my own experience, conflicts like these force a person to think."

So you are advocating that we inject confusion into the curriculum to force students to sort it out on their own?

I think we should make physics as approachable as possible. This means we should use clear, precise language that minimizes confusion. Not everyone is an A student. Some students can work their way through the muck, but many others can't.

I don't disagree that calling g an acceleration and having students figure out the inconsistency can have a beneficial educational effect. But is the payoff worth it? Have you ever read a confusing instruction manual? When you finally figured it out, were you glad that the instruction manual was confusing? Sure, you may have learned something interesting while trying to sort out what to do, but the end result was just anger, wasted time, and a negative attitude towards the product. That is what I want to avoid.

Clarity and accuracy are the two most important features of quality writing.

 

[jcsd]

g is an accelration, it may not necessarily refer to the actual acceleration of an object, but nevertheless it is an acceleration.

 

[Nenad]

the equivalence principle states tha there is no difference between gravity and acceleration.

 

[JohnDubYa]

RE: "g is an accelration, it may not necessarily refer to the actual acceleration of an object, but nevertheless it is an acceleration."

g is a number. Acceleration is a physical event. What you mean to say is that objects, in certain situations, can accelerate at values numerically equal to g.

Sure, the distinction is subtle, but in my opinion important. Because defining g as an acceleration causes students problems.

RE: "the equivalence principle states tha there is no difference between gravity and acceleration."

Hmmm... I have a book sitting on my desk. Gravity acts on it, but it is not accelerating. Huh?

The units of acceleration are m/s^2. If gravity and acceleration are the same thing, do we assign m/s^2 as the units of gravity? If so, I have never seen gravity defined in such a manner. In fact, how exactly are you defining "gravity"?

A student in an introductory physics course is going to ask these questions. How would you answer them without leaving him scratching his head?

 

[zoobyshoe]

Zooby, the description sounds okay to me.

OK. So do you object to him saying that on the way up the ball undergoes a constant acceleration downward at 32ft/sec²? I would anticipate that you would since the ball is not even traveling downward, toward the earth, on it's way up.

 

[krab]

g is a number. Acceleration is a physical event. What you mean to say is that objects, in certain situations, can accelerate at values numerically equal to g.

Just a nit here, but an important one: g is not a number. It is a physical quantity. If you omit units, g can be any number.

 

[jcsd]

RE: "g is an accelration, it may not necessarily refer to the actual acceleration of an object, but nevertheless it is an acceleration."

g is a number. Acceleration is a physical event. What you mean to say is that objects, in certain situations, can accelerate at values numerically equal to g.

Sure, the distinction is subtle, but in my opinion important. Because defining g as an acceleration causes students problems.

As krab points out g has units of m/s^2, so it certianly is not just a number, it is also a vector and it is most defintely an acceleration. If it were not an acceleration we shouldn't be allowed to substiute it into F = ma. Even in non-relativistic physics acceleration can be 'transformed away' by using d'Alembert's principle, so whether an accelartion is an 'actual acceleration' (i.e. a co-ordinate acceleration) is subject to frame of reference. In fact when you transform an inertial frame to a non-inertial reference frame you find that there is a very real simlairty between the acceleration in the inertial reference frame and g.

 

[quarkman]

I am currently reviewing a freshman text and am in the middle of the section on gravity. Maybe I am complicating the subject but this is what the text states:

"The gravitational acceleration a computed with [a = GM/r^2] is not the same as the free-fall acceleration g that we would measure for the falling particle ... The two accelerations differ for three reasons: (1) Earth is not uniform, (2) it is no0t a perfect sphere, and (3) it rotates. Moreover, because g differs from a, the weight mg of the particle differs from the gravitational force on the particle as given by [F = GMm/r^2] for the same three reasons." Halliday, Resnick, Walker "Fundamentals of Physics Extended: Fifth Ed." 1997, pg. 326.

I originally found this confusing, but I feel it helps distinguish between the approximation g = 9.8 m/s/s and the actual force the mass of the Earth would exert on a particle at a distance r from its center of mass.

 

[LURCH]

Hmmm... I have a book sitting on my desk. Gravity acts on it, but it is not accelerating. Huh?

I think the basic principles of relativity would respond, "not accelerating relative to you". But according to the equivalence principle earlier cited, could not beset with equal accuracy that both you and the book are accelerating parallel to one another?

 

[VantagePoint72]

Yes, and also don't forget that Einstein defines freely falling reference frames as the benchmark for accelerated motion (spacetime itself and its curvature being the benchmark, freely falling reference frames being the way to recognize said curvature.) According to a reference point that is accelerating relative to you, an object at rest to you would be accelerating. Thus, according to the benchmark defined by Einstein (it's not a preferred frame by which to define motion, this would violate SR, it's only GR's embodiment of the equivalence principle), the desk your book is sitting on is accelerating upward as is the book, and much like when you feel pressed into your seat when your car accelerates, the book is pressed into the desk. It's the equivalence principle in action.

 

[JohnDubYa]

RE: "So do you object to him saying that on the way up the ball undergoes a constant acceleration downward at 32ft/sec2? I would anticipate that you would since the ball is not even traveling downward, toward the earth, on it's way up."

Huh? I don't see why you would think I would object. His explanation is perfectly reasonable.

RE: "I think the basic principles of relativity would respond, "not accelerating relative to you". But according to the equivalence principle earlier cited, could not beset with equal accuracy that both you and the book are accelerating parallel to one another?"

So what? How does that possibly answer the question. This appears to be a non-sequitur.

In retrospect (having seen all of your arguments), calling g the free-fall acceleration wouldn't bother me too much. (I still like "gravitational field" better.) Calling it the acceleration due to gravity, however, does. And according to quarkman's argument, it isn't even accurate.

I think the allusions to relativity are pointless in this debate.

 

[VantagePoint72]

Why the issue with calling it the acceleration due to gravity? It doesn't say that is how fast a given object will accelerate towards the earth, it just says that's the magnitude of gravity's contribution to its acceleration: approx. 9.81 m/s^2 towards the centre of mass of the earth. It's perfectly accurate. My previous references to relativity weren't pointless, relativity is the refinement of classical physics. While it's effects are hardly noticable at velocities we experience, it's explanations of the reason things happen they way do is always applicable. If you ignore it then you aren't looking at the full picture. The point is, an object in a gravitational field is accelerating. Always. That's what gravity IS. Calling g the acceleration due to gravity is absolutely accurate, it's the general, all encompassing way of saying it. Calling it "free-fall acceleration" is more specific. It excludes the situation of the book on the desk in one of your previous posts. The book is still accelerating! Acceleration due to gravity is a broader definition, and hence more applicable.

 

[JohnDubYa]

RE: "Why the issue with calling it the acceleration due to gravity? It doesn't say that is how fast a given object will accelerate towards the earth, it just says that's the magnitude of gravity's contribution to its acceleration: approx. 9.81 m/s^2 towards the centre of mass of the earth. It's perfectly accurate."

But very confusing to weaker students. They will think that a block sliding down a plane must accelerate at g. After all, gravity is making the block slide down the plane, and the acceleration due to gravity is g. So the block must accelerate at g.

Sure, WE know better. But they don't. And you can throw in as many caveats as you wish, the bad terminology will ultimately dominate.

This is one of the most common mistakes made in introductory physics in my experience, and is caused by a misleading description.

RE: "My previous references to relativity weren't pointless, relativity is the refinement of classical physics."

This argument centers around defining g in introductory physics. Definitions that depend on an understanding of special relativity are not going to help students.

RE: "The point is, an object in a gravitational field is accelerating. Always."

Sure, tell that to a student in an introductory physics class and watch the dropout rate.

The first thing that a student will ask is, "Are you saying that the forces acting on an object can never cancel?" What do you say in response?

RE: "That's what gravity IS. Calling g the acceleration due to gravity is absolutely accurate, it's the general, all encompassing way of saying it. Calling it "free-fall acceleration" is more specific. It excludes the situation of the book on the desk in one of your previous posts. The book is still accelerating!"

The student is going to ask, in which direction? What do you say in response?

 

[ArmoSkater87]

This argument about the terminology has dragged on this long probably because there hasnt been an actual students opinion...well I am here to solve that problem, I am a high school student that took physics this past year. Acceleration due to gravity is most certainly the easiest way to get across to anyone, what g actually is. Its very clear, and obviously self explanatory. Saying anything else would confuse the crap out of a student. It might not seem that way to some, but it will because not every student that wants to know physics is necessarily a very bright one. Anyways, I, as a student agree that with everyone that prefers "acceleration due to gravity" instead of "universal constant". In fact if i didnt know any physics, i would think that g is the same for all planets and object in the universe from that terminology (universal constant).

 

[VantagePoint72]

OK, JohnDubya, I think I understand your point of view now. You're questioning the simplicity of calling it "acceleration due to gravity" not the accuracy of it. Fair enough, I can see where there would be confusion among some students about it. The point is though, scientifically it's the most correct name. What would answer if the student asked me which way the book is accelerating? Upward, as the equivalence principle predicts and is confirmed by the freely falling observer. And reading that again, yes that would confuse the heck out of most students. Even so, the fact that the book even IS accelerating isn't introduced till GR. So, leaving some of the more complicated details until later in their studies, by calling g the "acceleration due to gravity" the students who do choose to go on to more complicated subjects won't have to completely switch around their thinking. I know, like ArmoSkater I'm a student too who happened to be so taken in with physics in my first course, that I've been teaching myself relativity, quantum theory, etc. Earlier, I had been a little confused about why everything accelerates at g in the Earth's gravity, even if we view them as being at rest, but I didn't exactly lose sleep over it. And it made understanding the equivalence principle that much easier.