Why do two bodies of different masses fall at the same rate (in the absence of air resistance)?

I'm far from being a physics expert and figured this would be a good place to ask a beginner question that has been confusing me for some time.

According to Galileo, two bodies of different masses, dropped from the same height, will touch the floor at the same time in the absence of air resistance.

BUT Newton's second law states that $a = F/m$, with $a$ the acceleration of a particle, $m$ its mass and $F$ the sum of forces applied to it.

I understand that acceleration represents a variation of velocity and velocity represents a variation of position. I don't comprehend why the mass, which is seemingly affecting the acceleration, does not affect the "time of impact".

Can someone explain this to me? I feel pretty dumb right now :)

• Minor caveat for VERY heavy masses: physics.stackexchange.com/q/3534/2451 – Qmechanic Jul 5 '11 at 19:52
• You are right to think of neglecting air resistance, but you also have to neglect air buoyancy due to Archimedes' principle. This is also an is easily observed effect by setting the right conditions. – babou Jun 12 '13 at 6:06

it is because the Force at work here (gravity) is also dependent on the mass

gravity acts on a body with mass m with

$$F = mg$$

you will plug this in to $$F=ma$$ and you get

$$ma = mg$$ $$a = g$$

and this is true for all bodies no matter what the mass is. Since they are accelerated the same and start with the same initial conditions (at rest and dropped from a height h) they will hit the floor at the same time.

This is a peculiar aspect of gravity and underlying this is the equality of inertial mass and gravitational mass (here only the ratio must be the same for this to be true but Einstein later showed that they're really the same, i.e. the ratio is 1)

• This is not a good answer does not explain things from first principles unlike the answers which start with the $F = \frac{GM}{R^2} \times m$ equation. – Autodidact May 25 at 22:32

Newton's gravitational force is proportional to the mass of a body, $F=\frac{GM}{R^2}\times m$, where in the case you're thinking about $M$ is the mass of the earth, $R$ is the radius of the earth, and $G$ is Newton's gravitational constant.

Consequently, the acceleration is $a=\frac{F}{m}=\frac{GM}{R^2}$, which is independent of the mass of the object. Hence any two objects that are subject only to the force of gravity will fall with the same acceleration and hence they will hit the ground at the same time.

What I think you were missing is that the force $F$ on the two bodies is not the same, but the accelerations are the same.

There are two ways that mass could effect the time of impact:

(1) An object which is very massive has a stronger attraction to the earth. Logically, this might make the object fall faster and so reach the ground sooner.

(2) An object which is very massive is difficult to get moving. (I.e. it has very high inertia.) Thus one might logically expect the very massive object to be more difficult to get moving and so to lose the race.

The miracle is that in the world we live in, these two effects exactly balance and so the heavier mass reaches the ground at the same time.

Now let me give a simple explanation for why it's natural that this comes about. Suppose we have two very heavy masses. If we drop them separately they take some time T to fall. On the other hand, if we attach them together, will they take the same length of time? Think about a sphere split into two halves: Two two halves of the sphere would fall at the same speed as each other. So if you dropped them next to each other, they'd fall together. And dropping them next to each other isn't going to be any different from screwing them together and dropping them together. That is, there won't be any force on the screws. So the combined (or screwed together) sphere has to fall at the same rate as the split sphere.

Because force 'pushing' object closer to earth is proportionally bigger for 'heavier' object. But heavier object is also have higher gravitation force.

So these two factors perfectly compensate each other: Yes you need more force for a set acceleration, but more force is here due to heavier mass.

Lets say two separate mass $M_1$ and $m_2$ where $M_1$ >> $m_2$, both fall, from the same instant in a gravitational field

Force on $M_1$ is $F1$ = G $M_{\text{earth}}$ $M_1$/ $R^2$

Force on $m_2$ is $F2$ = G $M_{\text{earth}}$ $m_2$/ $R^2$

Therefore the forces are $F1$ >> $F2$

So, most people think $M_1$ should accelerate much faster than $m_2$

But as you wrote above a = F/m and substituting $F1$, $F2$, $M_1$ and $m_2$ into that formula we find:

$F1$/$M_1$ = $F2$/$M_2$ = G $M_{\text{earth}}$/ $R^2$

Therefore the acceleration is independent of the masses we drop, and is a constant.

EDIT Bother, by the time I had written this down, it was answered, obviously the other authors had a different acceleration in their typing.

• What happens when M1 >> m2 is false? Does m2 pull on M1 mean acceleration grow with m2? – Lorenzo Boccaccia Mar 7 '14 at 1:04
• I wonder if there is an "intuitive" GR way of explaining this. One could say that in the model of a large mass warping space-time that only the distance from the center of mass that creates the disturbance effects the acceleration of another mass toward it's center, but that doesn't really say "Why?". Also, if this wasn't the case, we couldn't predict orbital periods without knowing the mass of the orbiting object. Is the best we can do is say that the answer to "Why?" is "Because that is what we observe."? – Jack R. Woods Apr 2 '18 at 16:35

protected by Qmechanic♦Oct 19 '14 at 20:08

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