What does the gravitational force of attraction mean?

myink

I learned the formulas [tex]F = G \frac{m_1 m_2}{r^2}[/tex] and [tex]F = m \cdot a[/tex] and also saw mathematically that objects have the same gravitational acceleration and will fall down at the same rate, also demonstrated in a video by people that were on our moon, even though one object clearly has more force with the body of mass that it is falling on.

So if the force does not affect how strongly the object is pulled towards the mass, what is the point of it, since all masses are moving towards each other at the same rate?

CWatters

The force does affect how "strongly the object is pulled towards the mass"...

On Earth write F=m1g.

One way to look at it is to say that different masses DO have a different force acting on them BUT they also need a bigger force to accelerate them. Net result is constant acceleration.

Lets say m2 is the mass of a moon and m1 the hammer or feather dropped on the moon..

The force pulling m1 down is the force causing it to accelerate so..

Gm₁m₂/r² = m₁a

Oh look m1 cancels so "a" is independant of m1..

a = Gm₂/r²


As for the last part of the question....

I think that's a question for the philosophy forum.

myink

I meant what purpose does it have in mathematics and physics, since at first glance it doesn't seem to have any usefulness. I don't see how it can be applied to anything useful.

So comparing the hammer and the feather:

[tex]F_h = [G \frac{m_1}{r^2}] \cdot m_h[/tex]
and
[tex]F_f = [G \frac{m_1}{r^2}] \cdot m_f[/tex]

[tex]F_h > F_f[/tex]

So the force on the hammer is clearly greater

But as you showed, the force on either of them doesn't matter at all, since they will both move towards gravity at the same acceleration. I don't see where the force fits into all of this

tiny-tim

hi myink!
on the moon there's no point …

you can perfectly well go straight to the acceleration

but in air, or in a magnetic field, or under any other force (in addition to gravity), it does matter …

you have to subtract (or add) the forces (and then divide by the mass)

Villyer

As for determining how long it takes an object to fall to the surface of earth or the moon, there is no need to use force.

But the gravitational force is useful for other problems, such as the tension in a wire that supports a mass. In that case, the tension has to oppose the gravitational force, so a larger mass would need a larger tension.

CWatters

Likewise in problems involving friction. Boxes sliding down slopes, and cars on banked tracks etc Solving these types of problems involves working out the force on an object due to gravity and sometimes the component of that force acting in directions other than vertically downwards.

PeterO

Try reading about Einstein's warped space.
One summary of the effect is that space is warped about a mass, and that warping will "tell" an object how to move.
If you put two masses in essentially the same place [held side by side above the moon's surface for example] the warping will be the same, so the instructions will be the same, so not surprisingly the two objects move in the same way - they both accelerate to the surface, with the rate of acceleration determined by the warping. Einstein has no need for the "Force due to gravity" in his model.

If instead, we assume there is a force due to gravity, as suggested by Newton, and follow the laws of motion presented by Newton, we will have equations predicting the sort of motion we observe.

myink

So now I am wondering, why the force of gravity on an object has no effect on accleration, even though the force of gravity is pulling the hammer down much more strongly than it's pulling down on the feather? If I pull on a light object, I'd move it much faster than if I tried pulling a very heavy object, wouldn't I?

My textbook is not really helping, and I'm afraid this might turn into a typed lecture, so to prevent that does anyone have any books or textbooks that shed some light on this, for an absolute beginner that has never taken Physics before?

Muphrid

Your arm produces the same force on both the light and heavy objects, which is why the light object accelerates much more quickly.

Not everything in nature works like gravity, where the force is proportional to mass. EM is a good example, where the force is proportional to charge instead.

We need to distinguish between force and acceleration precisely because there are some forces that don't get stronger because the object is heavier.

Villyer

Maybe it will help to look at relationships between the forces and accelerations.
From [itex]F = G\frac{m_1m_2}{r^2}[/itex], [itex]F[/itex] is directly proportional to [itex]m_1[/itex]
From [itex]F = m_1a[/itex], [itex]F[/itex] is directly proportional to [itex]a[/itex], and [itex]m_1[/itex] is inversely proportional to [itex]a[/itex].

Therefore, as an object gets more massive, the force increases by the same factor. This normally causes acceleration to increase by the same factor as well, but since the object is more massive, acceleration wants to decrease by that very same factor.
So the net result is that the acceleration doesn't change.


To use the formulas a bit more, lets consider two masses, [itex]m_1[/itex] and [itex]km_1[/itex] (so the second is a scalar k more massive than the first).

The gravitational force on [itex]m_1[/itex] is [itex]F = G\frac{m_1m_2}{r^2}[/itex]
[itex]F = m_1a[/itex]
[itex]G\frac{m_1m_2}{r^2} = m_1a[/itex]
[itex]a = G\frac{m_2}{r^2}[/itex]

The gravitational force on [itex]km_1[/itex] is [itex]F = G\frac{km_1m_2}{r^2}[/itex]
[itex]F = km_1a[/itex]
[itex]G\frac{km_1m_2}{r^2} = km_1a[/itex]
[itex]a = G\frac{m_2}{r^2}[/itex]

The accelerations on each are equal.



I'm not sure if this is any more clear than other explainations offered, but I hope it helps.

Nano-Passion

One object has more force due to its mass, but that is irrelevant in this case and you need differentiate the formulas from each other. You have a formula for the force of gravity, and that just says that there is a field of force at every point in space, and in every point in space an object is attracted to the other inversely proportional to the distance between them.

The other force that your referring to when a body has more mass, is irrelevant in this case, it is just the potential force something can have when it hits another body. But it does not relate to gravity.

Force does affect how strongly an object is pulled towards the mass. The higher the gravitational force, the harder the object is pulled.

All masses move toward the same rate because of something we have dubbed "inertia." We noticed that the higher the mass of an object, the higher its resistance to change, so we called it inertia. So what actually happens is that it acts as a balance, making all objects accelerate at the same speed even though they have different masses -- because higher masses also have a higher resistance to change.

Nano-Passion

Because of inertia. And this is essentially what Villyer was saying.

This is how it works in physics, we make guesses, observations, and if it doesn't work we go back to the drawing board. Isaac newton came up with a powerful new concept, he wanted to describe everything by what he calls "force." You can think of force as what causes an imbalance, if you don't accelerate something, that is if you don't change its speed-- then you have applied no force.

[tex] F = ma[/tex]
[tex] F = \frac{Gm_em}{r^2}[/tex]
[tex] ma = \frac {G m_e m}{r^2}[/tex]
[tex] a = \frac {G m_e}{r^2}[/tex]

So this shows that acceleration does not depend on the mass of an object.

Now they stop and wonder, why is that? We invent a concept called inertia which says that things are more resistant to change when they have bigger masses. You can think about this intuitively.

Now as PeterO said, Einstein does away with it completely when it comes to gravity. But we still use the concept of inertia to describe why it is that heavier things are harder to push.

CWatters

At the risk of confusing you...

When we write..

Gm₁m₂/r₂ = m₁a

we are assuming that "gravitational mass" (m₁ on the left hand side) is the same as "Inertial mass" (m₁ on the right hand side) but this is not necessarily true. It's possible that they aren't exactly the same, in which case the force of gravity would have an effect on acceleration.

However so far no experiment has shown them to be different but the effect could be too small for us to measure.

PeterO

Take a friend out into the street and have them sit on a bicycle. Now push that bike as hard as you can and see what you achieve in the first 3 seconds.

Now have the friend sit in a car in the street. Push that as hard as you can and see what you achieve in the first 3 seconds.

Note: Analysis of those two situations will probably show that you pushed the car with a greater force than the bike, despite the effects of the forces.

NoPoke

Its a model. The math is there to describe what is observed. The meaning is still in the experimental observation and not in mathematical constructs like "force of gravity".

The math associated with force of gravity, the inverse square law, works so well that there is a risk that we start to think that force of gravity has some special meaning. we lose sight that it is a model of the world, a model of how nature behaves.

You can get a different model if you assume that objects fall at the same rate because there is NO force on them at all. But now straight lines have to become curved paths: the force of gravity has been replaced by changes in the geometry of space.

That too would be a model, a way of visualising nature. Idealy models have predictive power and simplicity. Sometimes you can't have both sometimes the simpler model is good enough.

Why the left-field answer? I suspect that what you really want to ask for is a description of what is meant by "force". Which is the name given to the effect that arises when momentum transfer takes place between two particles, in particular to the rate at which those momentum transfer events occur.

Or maybe I have that completely wrong. In which case I'll learn something when the flaws in what I've said are pointed out.

Nano-Passion

Very good point!

Unless someone disputes the logic of the above, there seems to be something inconsistent about the framework. Why is why it is best to think of a gravitational mass as setting up a field of acceleration. Where the field is a = Gm_1/r^2. Force only has a part to play with respect to the object being accelerated. But this also starts a loop of consistency, because by definition, for anything to be accelerated there must be a force!

A question to those who have studied general relativity: does this inconsistency carry over to general relativity since things are also accelerated?

myink

Mathematically, I think I sort of understand it...

[tex]F_h = [G \frac{m_1}{r^2}] \cdot m_h[/tex]
and
[tex]F_f = [G \frac{m_1}{r^2}] \cdot m_f[/tex]

[tex]F_h > F_f[/tex]

but

[tex]F_h = m_h \cdot a_h[/tex]
[tex]\frac{F_h}{m_h} = a_h[/tex]
and
[tex]F_f = m_f \cdot a_f[/tex]
[tex]\frac{F_f}{m_f} = a_f[/tex]

It seems that it actually turns out to be that:

[tex]\frac{F_h}{m_h} = \frac{F_f}{m_f}[/tex]

So this has already been proven mathematically earlier in this thread, but I just didn't intuitively understand it. The way I did it is just algebraically slightly different, but I think about it better in the manner that I calculated it (by plugging in actual numbers).

I am guessing that the previously mentioned "inertia" is what I am missing, because I haven't learned about that yet, which might be why it turns out the gravitational acceleration is the same for every object on Earth (and other planets too?).

So I am going to study inertia right now, but as an initial question, does inertia basically make any given mass fall slower towards gravity if they have more mass than an object with less mass, which will fall faster, so the inertia balances out the uneven gravitational forces which results in making both objects fall at the same acceleration?