The gravitational acceleration of a freely falling object, expressed in terms of the rate of increase of velocity per second. On earth, it is 9.80665 meters per second per second.

Abbreviation: 'g'
The acceleration of an object does not change regardless of the mass of that object. However the force exerted does. Also, the acceleration of the Earth towards the object does change with mass. So a large rock would pull more on the earth than a small feather. In both cases, the pull is so small, it's considered non-existant. Also, the acceleration of gravity changes with your distance from the surface. All this can be proven quite easily.

       G m1 m2
F = ----------
This is the gravitational law equation, which gives the force two objects exert on each other due to gravity. G is the Universal gravitational constant, and equals 6.67259E-11. Next, we have Newton's Second Law, which is:

F = ma

or force is mass times acceleration. Now, let m be the mass of one of the objects, say m1. Substitute the second for the first equation.

       G m1 m2
m1a = ----------

and cancel. units to get:
       G m2
a = ----------

Something to think about is that weight and mass are often mislabled. This mistake occurs just as much in metric as imperial units. In imperial (english) units, pounds is often incorrectly referred to as a measure of mass. The correct unit for mass is the slug. In metric, the kilogram is often miscorrectly referred to as a unit of weight. The unit of weight for metric is the newton. Mass is a collection of particles, Weight is the force exerted on a body by gravity.

Another common misinterpretation of the gravity law is, if your stand on a chair, x distance from the surface of the earth, then stand on a chair twice as tall, or 2x, you should weight a fourth as much. (since r is increased by two, then the term r2 should decrease the force by four). While this is mathmatically correct, the distance r is measured not to the surface of an object, but to the distance between the two center of gravities for the objects object. For spheres, such as the earth, the center of gravity is the same as the center of the sphere. So in reality, r has only increased by an very small amount.

An elementary proof that acceleration due to gravity is independent of mass

Well, "everybody knows" that acceleration of a falling body (absent friction) is independent of the body's mass. How come?

We've all heard this from friends, physics teachers, science museum demonstrators, and the rest of the popular science mob. But it's not very convincing, is it? (Well, I think the Smithsonian Institute has a device that drops lead and feathers down 2 tubes in vacuum which shows you this; that's also pretty convincing...)

So here's a simple argument. It also works to show the period of a pendulum is independent of its mass.

Take your falling object and drop it. It falls at some acceleration. OK, take it and drop it again. It falls with the same acceleration as it did last time, right? OK, so take an identical copy of your falling object; it still takes the same time to fall.

Right. Take the original and the copy, hold them together in one hand, and drop both. Each takes the same time to fall. Now do the same, but instead of calling them "two identical objects", pretend you've joined them and they're now a single object. Of course, if they accelerate at different speeds the "single" object would fall apart, but we know that if we drop this "single" object both halves fall at the same rate.

We've just shown that the acceleration of a falling object is the same as the acceleration of a falling object with twice the mass!

To generalise, take any two objects (the ratio of whose masses is a rational number) and pretend each one is made up of some number of very small objects with identical mass. The above argument shows that the small objects all fall together, so the two objects must fall at the same rate.

For objects with irrational mass ratios, it doesn't work. But any continuity argument will help you out.

I discovered this last week, but I doubt I can claim priority for it. I'd guess somebody rediscovers it every week...
ariels' argument is very old indeed. I believe I have heard it attributed to the venerable Galileo myself. While attempting to find a citation for that argument (which, I agree, has probably been discovered and re-discovered repeatedly in the last half-millenium) I found a great little write up by Yaskar Safkan, a physics doctoral candidate at some backwater hick school called MIT. Safkan's not sure who came up with ariel's argument, either, but he thinks it may have been Galileo, and so I am therefore (by the principle of repeated assertion) correct.

Safkan then goes on to explain why ariel's argument isn't entirely satisfactory, from a modern physics point of view, and then ties it in to a very accurate and easy to read variation of the material StormHunter presented in the writeup above. I think it's good enough that I am going to run the risk of rampant downvoting for redundancy and node it. The text will be my own, however, and the original was found at in the "Ask the Experts" column.

First: the problem with ariel's "elementary proof". It makes no reference to acceleration due to gravity at all, only acceleration due to an unknown force. Is that a problem? Well, consider a situation where we are releasing objects which accelerate towards their destination via the electromagnetic force. Each object has a static (unchanging) charge q and a mass m. The rate at which out objects fall now will be proportional to q/m - an object with no charge won't fall at all, an object with a large charge and little mass will fall very quickly, and an object will a large charge and huge mass will fall quite slowly. And tying two objects together will cause them to fall at a rate proportional to (q1 + q2) / (m1 + m2), which is an averaged rate for the two objects.

Safkan then suggests an alternate, more accurate conclusion for ariel's argument: "If all objects which have equal weight fall at the same rate, then all objects will fall at the same rate, regardless of their weight."

Now we'll jump back to our force equations, staring with gravitational force:

F = GMm1/r2

where G is the constant of gravitation, M is the mass of the attracting body (i.e. Earth), r is the distance between the objects, and m1 is the gravitational mass of the object. How does this releate to the acceleration of our objects?

F = m2a

where a is acceleration and m2 is the inertial mass of the object.

If we do a bit of algebra we quickly come up with

a = G(m1/m2)M/r2

And now we ask ourselves the key question: for any object, is the ratio of inertial mass and the gravitational mass of the object the same? Note that it isn't necessary for the ratio to be one; if inertial mass was always double that of gravitational mass, we could just adjust G by a factor of half and we'd have the same effect: the removal of the mass of our object from the calculation of gravitational acceleration on the object. It so happens, nicely enough, that general relativity requires that inertial mass and gravitational mass be equivalent.

Thus: the answer to the question "Why do objects with different masses fall at the same rate in a vacuum?" is simply "Because the ratio of gravitational and inertial mass is constant."*

* The author of this writeup assumes no responsibility if your 4-year old finds this explanation unsatisfying.

Log in or register to write something here or to contact authors.