# If an object with more mass experiences a greater gravitational force, why don't more massive objects fall faster? [duplicate]

According to Sir Isaac Newton, the gravity equation runs like this:

$$F = \frac{Gm_1m_2}{r^2}$$

where $$F$$ is the gravitational force, $$G$$ the gravitational constant, $$m_1$$ and $$m_2$$ are the respective masses of two bodies being pulled together by said force, and $$r$$ is the distance between them.

Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared.

Question:

Is this actually true?

It is clear from the above equation that the larger the object's mass (either of the $$m$$'s), the greater the force ($$F$$).

In other words, should we leave the distance ($$r$$) the same but increase the mass of either object, the force ($$F$$) would increase proportionately, resulting in a greater acceleration rate. Which was what Aristotle and everyone else after him thought until Galileo decided Aristotle was wrong and proved it by throwing two different objects from the tower of Pisa. They hit the ground at the same time: yes. So far as the naked eye could judge, anyway.

To clarify:

Compared to the Earth's mass, the masses of the two objects would have been so small that any difference in their respective acceleration rates would have been too tiny to detect, i.e. NEGLIGIBLE.

In other words, to all, or most, PRACTICAL (i.e. earth-based) purposes, Galileo was right.

But was he TECHNICALLY right?

Einstein's theory states that gravity and inertia are the same exact force. This strikes me as perfectly reasonable. However, it is asserted that this somehow confirms Galileo's conclusion. I don't see how.

• What is your argument against Galileo's conclusion? – my2cts Mar 24 '19 at 9:38
• $F=ma$, so the gravitational acceleration is independent of the mass of the falling body. However, see physics.stackexchange.com/q/3534 – PM 2Ring Mar 24 '19 at 9:49
• Inertia isn't a force – BioPhysicist Mar 24 '19 at 10:24
• @Ricky The difference is due to how much the Earth accelerates upwards to meet the falling object, in the centre of mass frame of (the Earth + object), the object's acceleration towards the centre of mass is still independent of its mass. – PM 2Ring Mar 24 '19 at 10:32
• – knzhou Mar 24 '19 at 13:13

To put more words into @Dale answer, the greater the gravitational mass,the greater the force due to gravity per Newton's law of gravity. But the greater the mass the greater its inertial mass as well, i.e., its resistance to a change in velocity, and therefore a greater force is required to accelerate the mass per Newton's second law, $$F=ma$$. Gravitational mass equals inertial mass, thus the acceleration is the same for all masses in the gravitational field.

Hope this helps.

Now Galileo asserted that regardless of the difference in mass of two objects falling onto Earth, their acceleration rate will always be the same, i.e. 9.8 meters per second squared. Question: Is this actually true?

Yes, and it is fairly straightforward to derive. For an object in free fall near the earth’s surface.

$$F=\frac{GMm}{r^2}$$ $$ma=\frac{GMm}{r^2}$$ $$a=\frac{GM}{r^2}$$ $$a=g$$

Where $$F$$ is the gravitational force, $$G$$ is the universal gravitational constant, $$M$$ is the mass of the earth, $$m$$ is the mass of the object, $$r$$ is the radius of the earth, and $$a$$ is the object’s acceleration.

Others have addressed the mathematics; here is an intuitive explanation. (I'm not sure if this will help the asker, given that their question shows a good grasp of the maths, but it may help others.)

Let's say my identical twin and I are pushing two different boulders along the ground. Because we're identical, we provide the same force. But my boulder is much heavier than his. The result is that my boulder moves more slowly.

So I find someone much more muscled to push my boulder for me. Muscly McMuscles applies a greater force, enough to get the boulder moving exactly as fast as my twin's.

Yes, bigger objects have a greater force acting on them.

No, this doesn't make them fall faster.

The greater force and the greater mass exactly balance each other out, so the big and small objects move just as fast (more precisely, they accelerate the same).

• Again an answer which ignores the problem of observing the phenomenon from a non-inertial frame. It's a pity that the OP was not able to grasp the point. – GiorgioP Mar 24 '19 at 21:13
• @If thinking that makes you happy, then ... etc ... – Ricky Mar 25 '19 at 0:13
• @GiorgioP: I answered the question as summarised in the title: doesn't greater force produce greater acceleration? I'd love to see an answer that delves into the maths of the non-inertial frame (it's been far too long since I did any of that stuff myself!), but even your own answer and comments stop at saying "there's a difference but it's unobservably small". – Tim Pederick Mar 25 '19 at 10:12

The correct replies to this question have already been written elsewhere in SE.Physics (better if ignoring the negative score answers).

However, it is probably useful to stress the basic issue underlying the answer to this question: if we describe motions in non-inertial frames, accelerations depend on the frame. Therefore, the fact that different masses are acclerated in the same way in inertial frames, does not imply that the same is true in non-inertial frames. And observing free fall of bodies on the Earth, means that we are using a non-inertial frame whose acceleration depends on the force between Earth and falling body. It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations.

• ". It is only the huge ratio between the mass of Earth and that of (reasonable size) falling bodies which makes practically unobservable the difference of accelerations." That's what I said. "Practically unobservable" does not mean that it doesn't exist. – Ricky Mar 24 '19 at 20:36
• @Ricky yeah, but I would not insist on claiming that , in absence of air drag, a iron ball would fall faster than a feather, if observed from Earth. From the experimental point of view unobservable differences are equivalent to no difference at all. – GiorgioP Mar 24 '19 at 21:04
• @Ricky Unless the falling body has a mass that's a significant fraction of Earth's, the acceleration it induces on the Earth is so tiny that not only is it immeasurable, it will also be swamped by various other effects, mainly air resistance, but also including the asymmetric distribution of mass in the Earth, seismic disturbances, tidal forces from the Sun, Moon, and the other planets (primarily Jupiter), other people dropping things at other locations, et cetera. If you insist on including the Earth's acceleration in your model you should also include those other things. – PM 2Ring Mar 25 '19 at 8:22
• @PM2Ring: I absolutely agree with you on all points here. – Ricky Mar 25 '19 at 16:11

This might sound confusing i agree. A body of greater mass does experience a greater force compared to a lighter object , but then the acceleration is the ratio of Force experienced over the mass which equals gravitational constant times Mass of earth( in this case) over distance squared , which is independent of the mas of the object that is falling . You can also understand it this way : a greater force is greater is required to maintain a given acceleration for a more massive body than the force that is required to maintain the same accelaration for a less massive body.In both cases force varies but accelaration remains same

The gravitational force is proportional to the masses of the two interacting objects. Since force is mass times acceleration, instantaneous gravitational acceleration is independent of mass. This is a approximation. The solution to the full two body problem does depend on both masses. See: https://en.m.wikipedia.org/wiki/Gravitational_two-body_problem.