How does the mass of an object affect it's acceleration during free fall?

Answer 1

Not at all. All objects accelerate in free fall with the same acceleration. If they're dropped from the same height above the surface of the same planet at the same

time, they hit the ground at the same time. If any difference is noticed, it's strictly

the effect of air resistance. Try it in a tube from which the air has been pumped

out, and a feather and a Bowling ball really do fall together.

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!

Answer 2

Not at all. All objects accelerate in free fall with the same acceleration. If they're dropped from the same height above the surface of the same planet at the same

time, they hit the ground at the same time. If any difference is noticed, it's strictly

the effect of air resistance. Try it in a tube from which the air has been pumped

out, and a feather and a Bowling ball really do fall together.

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!

Answer 3

If you are talking about small things, such as skydivers or feathers, free falling towards big things, like the Earth or the Moon, then the effect of mass on acceleration is very tiny and it's probably not worth worrying about. It's not ZERO, but it would be close to zero. You decide. In most practical cases you can safely ignore the difference because it's really tiny (calculations below may explain how tiny) where air resistance and measurement error would easily hide any difference.

However, if you are talking about two similarly massive bodies freely falling towards each other (such as two helium atoms, two stars, or the Milky Way and Andromeda galaxies), then the acceleration due to gravity (g1 + g2, see below) could be significantly impacted by the mass of both bodies.

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 object 1 (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 if 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 shape of the object may change the terminal velocity by increasing or decreasing the aerodynamic drag, and this is independent of mass. A fat guy with a parachute will fall more slowly than a skinny guy without one. But acceleration due to gravity will not be affected, practically speaking.

Answer 4

Not at all. All objects accelerate in free fall with the same acceleration. If they're

dropped from the same height above the surface of the same planet at the same

time, they hit the ground at the same time. If any difference is noticed, it's strictly

the effect of air resistance. Try it in a tube from which the air has been pumped

out, and a feather and a bowling ball really do fall together.

Answer 5

It does not affect its acceleration in a vacuum, that is, no air resistance. The acceleration s the acceleration of gravity which is constant

Answer 6

It has no effect whatsoever.