I just wanted to say that if two objects that had the same mass while an artificial gravitational field was pulling on the smaller ball if it was rolling down a ramp at the same time as a bigger ball, then the smaller ball would roll down first if the artificial gravitational field was strong enough. For example, if a ball was dropped on the moon at the same time as a smaller one was dropped on earthm then the one on the Earth would fall faster relative to the smaller ball on the moon.
Well, if you drop a ball from 10 feet off the ground on the moon, it won't fall as fast as a ball dropped from 10 feet off the ground on the earth would. Is that what you mean?
Of course, air resistance complicates things slightly.
(I'm pretty sure I'm right about that, acceleration due to gravity being smaller on the moon than on the earth.)
If this isn't what you mean I really can't understand.
Schneibster
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The equation is,
F(g) = gmm'/d^2
Which in English is, "The force of gravity varies directly as the gravitational constant and the mass of the two bodies feeling the force, and inversely as the square of the distance." Thus, the amount of force a larger body feels is greater by the ratio of its mass to that of a smaller body. But also, remember Newton's First Law:
F = ma
Which in English says, "The force applied to an object varies directly as both the object's mass and the acceleration the object undergoes as a result of the force." Thus, the greater force accompanied by the greater mass results in the same acceleration; and thus, objects with the same size and shape and dropped from the same height but with different masses fall at the same rate toward a third object to which they are attracted. For instance, two cannon balls, one hollow, both fall to the ground at the same time when dropped from the Leaning Tower of Pisa. (If you buy that apocryphal tale about Galileo.)
F(g) = gmm'/d^2
Which in English is, "The force of gravity varies directly as the gravitational constant and the mass of the two bodies feeling the force, and inversely as the square of the distance." Thus, the amount of force a larger body feels is greater by the ratio of its mass to that of a smaller body. But also, remember Newton's First Law:
F = ma
Which in English says, "The force applied to an object varies directly as both the object's mass and the acceleration the object undergoes as a result of the force." Thus, the greater force accompanied by the greater mass results in the same acceleration; and thus, objects with the same size and shape and dropped from the same height but with different masses fall at the same rate toward a third object to which they are attracted. For instance, two cannon balls, one hollow, both fall to the ground at the same time when dropped from the Leaning Tower of Pisa. (If you buy that apocryphal tale about Galileo.)
StaticEngine
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Two objects dropped (in vacuum) from the same height in the same gravitational field will fall at the same rate, independant of their mass.
Two objects dropped from the same height in different gravitational fields will fall at different rates, dependant only on the strengths of the fields, and not on the masses of their objects.
So yes, if a marble were dropped from 10 ft. off the ground in a vacuum chamber on Earth, and a golf ball were fropped from 10 ft. off the ground of the Moon, the marble would fall faster.
Two objects dropped from the same height in different gravitational fields will fall at different rates, dependant only on the strengths of the fields, and not on the masses of their objects.
So yes, if a marble were dropped from 10 ft. off the ground in a vacuum chamber on Earth, and a golf ball were fropped from 10 ft. off the ground of the Moon, the marble would fall faster.
fishbob
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Gravity is only a theory. Intelligent Falling should be taught as an alternative.
Pragmatist
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As a general principle, no.
Using only classical mechanics:
Consider two equal point masses initially at rest with respect to each other in free space and in the absence of other extraneous force fields (i.e. gravitation, electromagnetic etc.). The masses are separated by a distance we can call d. The centre of mass of the system lies equidistant between the two masses at a point d/2 as measured from either mass. Both masses will experience a mutual gravitational attraction which will cause both of them to accelerate toward point d/2.
The force experienced by both masses will be: F = G * m1 m2 / d2
Now the acceleration experienced by each individual mass will be a = F / m
Substituting the first equation into the second the acceleration experienced by m1 will be equal to G m2 / d2 (because the m1's cancel out)
The acceleration experienced by m2 will be equal to G m1 / d2 (because the m2's cancel out)
But in each case this is acceleration toward point d/2. The acceleration of either body observed from the other body will be the sum of the accelerations toward the common centre of mass, so the equation for the acceleration of either body as observed from the other body will be:
a = (G / d2) * (m1 + m2)
Therefore the acceleration of one body as seen from the other body will depend on the masses of both.
Now, in practice, we normally observe things from the surface of the Earth and we usually observe the fall of bodies much lighter than the earth. So if we represent the mass of the earth as M and the mass of the falling body as m, the equation becomes:
a = G / d2 * (M + m)
And because usually m << M, it is negligible in practice and we can say empirically that objects tend to fall at the same rate toward the earth. But it is obviously not true if m is large and it is not strictly true in any event.
If however we take two different masses (of much smaller mass than a planet) and drop them side by side in vacuum on to the surface of a planet, we will observe that they will both impact at the same time. The reason for this is because the force vectors between each individual mass and the earth are lying parallel and nearly coincident, so the centre of mass is effectively a vector composite of that of the 3 bodies and all the objects are drawn toward it, but the planet is drawn by both the other masses while each of the other masses are to all intents and purposes drawn individually by the planet (because the gravitational force between the two falling objects is probably infinitesimal). Because each of the two test masses experiences the same acceleration due to the planet's gravity they will fall in equal times. Although strictly, the overall time of fall will be determined by the sum of the mass of the planet plus both of the test masses.
So if you drop two objects side by side in vacuum on to the surface of a planet they will both impact at the same time, but this does not strictly speaking imply that objects of different masses will fall at the same rate in a given planetary gravitational field independent of mass as a general principle. Because if you dropped those same objects of different mass simultaneously (ignoring relativity) from the same height on opposite sides of the same planet (or at least not in parallel), or if you dropped each mass independently of the other at different times and timed the fall, then the more massive object would impact the planet first.
You are thinking of two different objects either on the earth or without a gravitational field. I am thinking of two objects on two different planets falling on planets with different gravitational fields. One planet (The Earth) has a stronger gravitational field from (The Moon). If two objects were dropped on the Earth and the same experiment was done on the moon similtaneously, then the two objects with different masses on the Earth will fall faster then the two objects with different masses on the moon. I'm not talking about the inverse square law where the amount of gravity decreases with distance as you move further away from the Sun. I'm talking about two objects falling at the same time on one body (The Moon) each object hypothetically has different masses on the moon and on (The Earth) relative to the moon those two other different objects with different masses will fall faster on the Earth then if they fell on the moon.
If you include the inverse square law with what I was talking about. Then if a star elsewhere in the universe had a huge mass larger than our sun and if another planet the size of our moon was orbiting it in a different orbit and two objects fell on each of them and if those two objects had different masses, then the two objects would fall at different times on those different planets. If you add the inverse square law in there and got the planets to orbit that hypothetical star, then the motion of those planets around that star would be chaotic like in chaos theory.
Schneibster
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Hmmmm, something strikes me wrong here. Either you have made an incorrect assumption, or I have. I'm too lazy to figure my way through your presentation, so let me present my argument briefly, and if you have the time, could you please point out where you believe my incorrect assumption lies?
OK, this is where I had a problem.
Note that
F(g) = gmm'/d² (Eq. 1)
Where,
F(g) is the force of gravity,
g is the gravitational constant, a constant of proportionality that describes the operation of the gravitational force everywhere in our universe that we know of, which allows us to convert to our units of measurement,
m is the mass of the mass of the object,
m' is the mass of the planet, and
d is the distance between the object and the planet.
Note also that
F = ma (Eq. 2)
Where,
F is the force applied to an object,
m is the mass of the object, and
a is the acceleration of the object.
This second equation can be rearranged as
a = F/m (Eq. 2a)
Substituting equation 1 for F,
a = (gmm'/d²)/m (Eq. 3)
Simplifying,
a = gm'/d² (Eq. 3a)
And thus we see that the acceleration of gravity for an object depends solely upon the gravitational constant, the mass of the planet, and the distance between the object and the planet. IOW, the acceleration of gravity for any object relative to a particular planet is not dependent upon the object's mass and is the same for objects of all masses for that planet.
On the other hand, it is also clear that the acceleration of gravity is dependent upon the mass of the planet, and thus we see that for all objects, the attraction of the moon will be less than the attraction of the earth, and a marble will indeed fall faster on earth than a cannon ball on the moon.
While the attraction between two objects dropped simultaneously will theoretically affect them in addition to the attraction to the planet, this attraction is so small that even measuring it requires instrumentation of extreme delicacy. For objects of reasonable mass (near that of a person, within an order of magnitude or two) the strength of this force is many, many orders of magnitude below that of the planet. In fact, I would go so far as to say that under these circumstances, it is extremely unlikely that any difference could be detected between each object dropped alone and the two dropped together unless the fall time were many seconds, and the distance thus many hundreds or even thousands of feet; and in a vacuum, as well, since the difference would be swamped by differences in air and chaotic effects (turbulence) of the interaction with air during the fall.
Could you please tell me where you believe this is wrong?
Not so much "incorrect" as "oversimplified."
When I drop a hammer, the Earth moves up to meet it. (Action and reaction -- just as the hammer is moved by the Earth's gravity, so is the Earth moved by the hammer's) Not, granted, by very much. I'm probably justified in ignoring the movement of the Earth unless I want to be pedantically accurate.
If I drop an anvil, the Earth moves up to meet it, too. And since the anvil is more massive, the Earth experiences a greater force and moves up more. So the anvil will technically strike the Earth after slightly less time than the hammer would.
And, of course, if I drop a planet on the Earth, the planet would cause the Earth to move significantly.
Schneibster
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Ah-hah, I see. I've ignored the acceleration of the earth because I only put Newton's First Law into the equation once. Yes, this is correct. Thanks.
Pragmatist
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I see Dr Kitten beat me to it - thanks Dr K.
Pragmatist
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I haven't got a clue what you are talking about, and without intending to be rude, I suspect neither do you...