Why is gravitational acceleration regarded as a constant?

[Ronald Wieck]

I am hoping to recover some of the knowledge of physics I gained in high school that has vanished into the mists of time. Currently, I am being abused on Facebook by a self-proclaimed authority who ridicules my contention that gravitational acceleration is a constant. I get the idea that gravity varies slightly with altitude, but the sources I have consulted refer to gravitational acceleration as a constant. I suspect that the twit is playing word games, splitting hairs, but I simply don't know enough.

Can a genuine authority help me?

 

[f95toli]

Do you understand the difference between g and G? They are both sometimes referred to as "constants" but the former -which is about 9.8 m/s^2- depends on where you are on the Earth's surface and is actually a "compound" constant which includes capital G and the mass or the earth.
Capital G is an actual constant, is independent of where you are, and is what appears in the expression for gravitational force in Newtonian mechanics.

 

[A.T.]

my contention that gravitational acceleration is a constant.

Does this look like a constant to you?

 

[Ronald Wieck]

Many thanks for your prompt response.

Yes, I have been doing some reading and the distinction between g and G is becoming clearer. The guy on FB tried to embarrass me by asking me where the acceleration is greatest when a ball is tossed up in the air. I replied that the acceleration is constant.
He immediately started ranting about air resistance, but I believe that if I had given a different answer, he would have reproached me for not knowing that the acceleration is constant.

 

[Ronald Wieck]

A.T., I confess that your chart is not clear to me. Could you elucidate?

 

[paisiello2]

For all practical non-esoteric purposes g can be treated as a constant.

 

[A.T.]

A.T., I confess that your chart is not clear to me. Could you elucidate?

http://en.wikipedia.org/wiki/Gravity_of_Earth

 

[A.T.]

where the acceleration is greatest when a ball is tossed up in the air. I replied that the acceleration is constant.

Ignoring air resistance that is a good approximation for a such small range of altitudes. With air resistance it is initially maximal, then decreases.

 

[Ronald Wieck]

In your experience, has a physics text ever asked such a question concerning acceleration and expected an answer that took air resistance into consideration?

 

[Bandersnatch]

Hi Ronald Wieck.

The strenght of a gravitational field has the general form of:
$$g=\frac{GM}{R^2}$$
This is just the Newtonian gravitational force divided by mass of the object being accelerated, which incidentally means that it doesn't matter how massive an object is - all are accelerated at the same rate (but read on). It's the same as acceleration caused by the mass in the absence of any other forces.

The only actual constant in that equation is the gravitational constant G. The mass M and distance from its centre R depend on what is pulling you in, and how far from it you are standing.

Now, since most of the people that have ever lived are interested in acceleration on Earth's surface, we can just plug in Earth's mass and the distance from Earth's centre to the surface.

Earth's mass is the same for everybody, although denser mass concentrations below some areas of the surface have some tiny, yet measurable effect.
The radius varies depending on where you're standing - on the summit of Mt Everest or on the shores of Dead Sea, on one of the poles or on the equator.
However, the difference is not that great when compared to the average radius of the Earth. Looking at the equation above, it doesn't matter a whole lot if you divide GM by say, 6 370 000 m, squared, or by 6 378 000 m, squared (roughly, the height of Mt Everest compared to sea level). If you're thinking of differences in the range of a few dozen metres it becomes even more negligible.

That's why, for most intents and purposes, when one is talking about the acceleration from Earth's gravity on its surface, he can say it's everywhere the same and equals g (9.81m/s).

It's an approximate statement, and one needs to remember that it works only for the special case of the vicinity of the surface of the earth. In this sense it is constant. If you're a scientist whose work is to map the variation of g with very precise instruments, then of course, by definition, you won't treat it as constant. If you're plotting a trajectory of a space ship, or modelling its strength below the surface (what AT's graph shows), then you won't treat it as constant. If you're a regular shmoe, then it's as good as constant, and treating it otherwise without any good reason is pretty much pointless.

By the way, air resistance has no bearing on the strength of the gravitational field. It does have a bearing on the NET acceleration (like a parachutist at some point reaches 0 acceleration).
If you're asked to compute the acceleration due to gravity alone, you use the equation provided, or just use 9.81 if you're on Earth and not concerned with super-high accuracy. If you're asked to compute net acceleration (like with a projectile motion with air resistance - a standard exercise for college freshmen studying physics), then the acceleration will vary significantly. You'd still use the gravitational acceleration (treated as constant) in there, but that's a whole different kettle of fish.

 

[Doc Al]

In your experience, has a physics text ever asked such a question concerning acceleration and expected an answer that took air resistance into consideration?

Acceleration of a falling body certainly is affected by air resistance. But free-fall acceleration is the acceleration in the absence of forces other than gravity. Think of it as a measure of the strength of gravity.

That said, can you state the context of your disagreement? The acceleration due to gravity certainly depends on altitude (see A.T.'s chart), but for most practical purposes can be considered constant over short distances from the earth.

 

[Ronald Wieck]

Doc Al, I was asked by my opponent, a 9/11 "truther," to state the point of greatest acceleration when a ball is tossed up in the air. I replied that the acceleration is a constant. He has been ridiculing me ever since without managing to express his precise position in a few coherent sentences.

 

[Doc Al]

Doc Al, I was asked by my opponent, a 9/11 "truther," to state the point of greatest acceleration when a ball is tossed up in the air. I replied that the acceleration is a constant. He has been ridiculing me ever since without managing to express his precise position in a few coherent sentences.

Note that "acceleration" and "acceleration due to gravity" are not quite the same, since the first presumably considers other factors such as air resistance.

What is his claim? (And why do you waste time arguing with "truthers"?)

 

[Bandersnatch]

a 9/11 "truther,"

Run away, man. There's plenty less wasteful ways to spend one's day.

What is his claim?

I'm quite sure it's the 'the building could only collapse this fast if its supporing structure were blown up' thing.
Arglebargle.

 

[Nugatory]

The guy on FB tried to embarrass me by asking me where the acceleration is greatest when a ball is tossed up in the air. I replied that the acceleration is constant.
He immediately started ranting about air resistance, but I believe that if I had given a different answer, he would have reproached me for not knowing that the acceleration is constant...
Doc Al, I was asked by my opponent, a 9/11 "truther," to state the point of greatest acceleration when a ball is tossed up in the air. I replied that the acceleration is a constant. He has been ridiculing me ever since without managing to express his precise position in a few coherent sentences.

If you find someone doing that here, report them - please!. Any day in which I get to use the banhammer is a good day.
And kidding aside, PF strives to be different from a lot of the rest of the internet - check out "The Physics Forums Way" at the bottom of the right-hand sidebar on the main page.

In your experience, has a physics text ever asked such a question concerning acceleration and expected an answer that took air resistance into consideration?

Only if that was clearly the point of the exercise and it was mentioned in the exercise. In more advanced physics classes, it will be expected that you can figure out for yourself which effects are small enough to ignore and how to model the ones that are not, but that's not what we're talking about here.

Bottom line: If you only care about one or two significant digits of accuracy, which is the case for just about all everyday problems involving objects on ballastic trajectories near the surface of the earth, you can ignore air resistance and tidal effects, and use 9.8 m/sec^2 (or 10, if you're in a hurry) and you'll be fine. If you're talking about building collapses, then take comfort in something that a wise structural engineering professor once said: "If you're looking at the second significant digit of the tensile strength, you've already screwed up".

 

[Ronald Wieck]

I am greatly impressed by the knowledge and understanding of the members of these forums. Facebook is a good place to study people in the grip of irrational beliefs, twoofers being particularly grotesque examples.

Extracting an infected molar from an unanesthetized Siberian tiger with one's bare hands would be easier than pinning my opponent to a coherent position. As close as I can discern one, he rejects the statement found here:

http://www.engineeringtoolbox.com/accelaration-gravity-d_340.html?hc_location=ufi

"Acceleration of gravity is one of the most used physical constants - known from
Newton's Second Law"

 

[Nugatory]

Extracting an infected molar from an unanesthetized Siberian tiger with one's bare hands would be easier than pinning my opponent to a coherent position. As close as I can discern one, he rejects the statement found here:

Like Bandersnatch said... Let it go. You cannot win this one.

As your question has been answered, this thread is closed. You have an oasis of sanity here whenever you need it.