This is, in a way, a simple question, but I think it can have some fun deviations.

So, if you haven't seen it before- here's a famous example of an astronaut dropping a hammer and a feather on the moon, and they hit the ground at the same time. This is an illustration of basic physics. You apply Newton's Second Law, which says F = ma, and you say the force of gravity is equal to mg, so you end up with

F = mg = ma

And so you can cancel m on both sides and just get a = g so you see the acceleration an object undergoes is not dependent on mass, so any object falls the same speed. And on the Moon, that works. And it illustrates the principle perfectly. Of course, there is a reason we did that experiment on the moon and it's so cool- it goes against what we feel should happen. On Earth, you drop a hammer and a feather, and of course they fall at very different speeds. The difference comes from the air resistance on the Earth.

Unlike gravity, air resistance is not a constant. It depends on a couple of things- like the shape of the object, the size of the object (and when we say size, we have to be specific. It's the size of the object in the direction of travel- the part of the object getting hit by the air) and the speed of the object (the faster an object is going, the stronger the air resistance is. The equation for air resistance is captured in the famous drag equation and says Fd = 1/2*p*(v^2)Cd*A where p is air density, v is the velocity through the air, Cd is the drag coefficient, which is dependent on the shape of the object (but not the size) and then A is the "reference area" (which is the surface area of the object in the direction of travel).

So, when in the presence of air, using Newton's second law things look a little different. Now you say:

F = ma = mg - 1/2*p*(v^2)Cd*A

(There is a minus sign between them because gravity and air resistance work in opposite directions, gravity pulls down while air resistance pushes up). Now, you can no longer cancel m because mass isn't in the air resistance equation. So, you can solve for a and say

a = g - 1/2*p*(v^2)Cd*A/m

So, when you look at the above equation, if the object has the same Cd and same A, and the mass gets bigger, then acceleration will be bigger (since the term you're subtracting off gets smaller), so in that case, you would say "the heavier something is, the faster it falls." And the works to a certain point. Say, a child's toy ball filled with air vs a bowling ball. About the same size, same Cd (since Cd is just based on shape) and so about the same area, so the more massive bowling ball falls faster. Or a more dramatic example. A piece of paper, or a piece of plywood cut to the same size. The two objects have the same Cd and A, but the plywood is more massive, so it falls faster.

But of course, a lot of times, when something gets more massive, it also gets bigger, so then A would increase as well. So, really, what matters for two objects the same shape, is the ratio between the surface area and the mass. Now, since volume of a sphere (and thus mass) grows with the radius of the sphere cubed, and surface area only grows with radius squared, a large bowling ball that's made out of the same material as a small one will still fall faster (since the mass to surface area ratio is still higher), but a small, dense bowling ball that still weighs less than a larger, less dense bowling ball, could fall faster.

Now, taking this a step further, we have always been operating under the equation that the force of gravity is just mg which is a really good approximation when the thing that's being dropped is much smaller than the thing it's being dropped on, but if we use Newton's law of gravitation we'll see the equation should really be

Fg = -G*m1*m2/r^2

where G is the gravity constant, m1 and m2 are the two masses, and r is the distance between them. So, this shows, perhaps unintuitively, that when you drop a bowling ball on Earth, the bowling ball is pulling just as hard on the Earth as the Earth is on the bowling ball. However, since F = ma since the forces are the same, but the mass of the Earth is so much bigger than the mass of the bowling ball, the acceleration of the Earth is much much smaller than that of the bowling ball, so the acceleration of the Earth is essentially zero. However, what if instead of dropping a bowling ball on Earth, we dropped Mars? Now, while Mars is still a lot smaller than Earth, isn't not so much smaller that the acceleration of Earth would be essentially zero. So, in this case, if we dropped the Moon on Earth vs dropping Mars on Earth, Earth and Mars would collide quicker than the Moon and Earth (assuming dropped from the same height) because Earth would accelerate more towards Mars than it would towards the Moon, because Mars is a lot more massive than the Moon.

Ok lets dive into this so with no air resistance things accelerate until they reach the ground.

Lets have to objects with mass m and M.

The force of gravity F=m×g

So m×a=m×g or M×a=m×g.

Mass factors out a=g in most cases. Cool that was demonstrated on the Moon with a hammer and a feather.

For Newtonian gravity F=G×M×m×1/r² where M is the mass of the planet.

So lets use m for our object and see if it drops out.

ma=GM/r² × m

a=GM/r² Great.

Now lets look into air resistance and terminal velocity. An object reaches terminal velocity when the sum of Fg and Fr (for resistance) is 0. So the forces are in equilibrium.

Fg=Fr.

Fr=½×q×C×A×v² where A is the surface area of the object C is a factor for shape q is the density of the medium and v is the velocity. We can say taht q and C are constant so lets combine the into a factor b. (With the ½.) We will use A and get a formula for v the terminal velocity.

This all leads to the square-cube law if we assume our different objects have the same density.

Fr=bv²A

Lets look at a solid ball with density q. Its volume is V=4/3×pi×r³. So from q=m/V we get

m=qV=q×4/3pi×r³.

Now A=4pi×r². And with that lets plug that into Fr=Fg=mg.

q×4/3×pi×r³×g=bv²×4pi×r²

lets rearrange

qrg=bv²×3

So now we get

⅓×qg/b × r=v² that first bit is a constant so lets call it k

k×r=v²

(k×r)^½ = v

So with same density balls the larger falls faster. If you increase the radius by a factor of 4, v increases by a factor of 2.

The m(r) function is simple we already have that

m=qV=q×4/3×pi×r³

So now for r(m)

r = (3/4×m/(q×pi))^⅓

So now lets plug it into the v² formula.

v²=k×(3/4×m/(q×pi))^⅓

v²=k × (3/4)^⅓ × (1/q×pi)^⅓ × m^⅓. Bunch of constants and m so lets combine them into K giving us.

v²=K × m^⅓

v=K^½ × m^⅕ Lets call K^½ = C

v=C×m^⅕

So as mass increases so does v but that was obvious from the v(r) formula.

To know the value of C you need the shape factor for our sphere its 0.47 the density of the fluid (air) g~9.81m/s² pi is 5 of course and the density of our material. Of course for different shapes the resulting functions can look different but the square-cube law will apply. More mass height terminal velocity. But the results may be very similar as any shape is appropriately a sphere.

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