How does an object mass affect its motion during free fall?

Answer 1

The force (F) of gravity acts on the mass (m) to result in acceleration of the object due to gravity (g). F=mg.

Mass has virtually no effect on acceleration during free fall. Galileo demonstrated this when he dropped lead balls of various masses.

Here is Newton's equation for force due to gravity:

F = G * M * mo / (rE + h)2

where

G = gravitational constant = 6.67300 × 10-11 m3 kg-1 s-2

M = mass of planet (If this free fall is near Earth, use 5.9736 × 1024 kg

mo = mass of object 2 (object in free fall)

rE = the radius of the planet, moon, star, etc. where the free fall is occurring (If Earth, at 0º latitude, use 6378137 m; at 90º latitude, use 6356752 m, interpolate for latitudes in between)

h = elevation of the object free falling, from "sea level"

Gravitation is not entirely understood, but we know it acts both ways. As the falling body is pulled toward the Earth, the Earth is also also being pulled toward the falling body. This latter effect is tiny for all but very large incoming objects (moons, planets and such), but it's still there. Mathematically, you would add these two forces to get the total force, but this second force is ignored unless the body you're interested in has a gravitational field of its own worth accounting for.

Force due to gravity that we normally care about is F = mog. But let's say there are two forces (F1 and F2) due to two accelerations (g1 and g2). The two forces to be added together are

F1 = mog1 (the force we usually care about calculating)

F2 = Mg2

(This is the tiny force that pulls the planet of mass M toward the falling body)

mog1 = G * M * mo / (rE + h)2

g1 = G * M / (rE + h)2

Mg2 = G * M * mo / (rE + h)2

g2 = G * mo / (rE + h)2

Note that mo dropped out of the g1 solution, meaning the acceleration g1 just calculated is not affected by the falling objects mass. Similarly, M falls out of the solution for g2: the mass of the planet does not affect its acceleration toward the falling body.

When you add F1 and F2 to equal the total actual force in this system you get

mog1 + Mg2 = G * M * mo / (rE + h)2

and

g = g1 + g2

Solving for g2:

g2 = G * mo / (R^2 * (1 + M / mo))

So, g2 and thus F2 are both partly a function of mo, the falling object's mass, but only very slightly since mo is tiny compared to M. Considering you are dividing the gravitational constant (a small number to begin with on the order of 10-11) by the Earth's radius squared and again by the ratio of the Earth's mass to the objects mass, the tininess of this force can boggle the imagination and can be safely ignored for any experiment you would be trying yourself!

Terminal Velocity:

In an atmosphere the density of the object (mass/unit volume), shape and aspect may change the terminal velocity by increasing or decreasing the aerodynamic drag. In a long enough fall the heavier object will pull ahead of the lighter object due to its higher terminal velocity. But the acceleration will not be affected, practically speaking.

Answer 2

An object's mass doesn't affect the way it moves. It affects
the ability of a force to change the object's motion.

Answer 3

a2 / a1 = m1 / m2, twice as massive, half as much acceleration, for the same applied force.

Answer 4

If it is heavy, it needs more force to be moved, whereas lighter objects need less force.

Mass does not matter, unless mass causes it to be heavier.

Answer 5

Inetia.