Why don't heavy objects fall more SLOWLY?

Saw

Oops. I correct again: where I said momentum, I should have said change of velocity and hence of momentum in a given time (collision time) = impulse, isn't it?

Saw

I have been thinking that the concept of “gravitational mass” may be somehow misleading. Really the underlying issue is: does the composition of the bodies involved in a gravitational interaction have any bearing in the outcome of a gravitational interaction? And, well, there is no need to mess that up with the concept of mass. One could perfectly let the sum of the masses of the “relative acceleration” term be divided out by the sum of the masses of the “reduced mass” term and still ask that question.

Of course, then the issue is, how could you make that factor (composition of the bodies) relevant, if (speculating about the idea) you wished to? In the context of collisions (what we could call contact force), such factor is no doubt relevant and it shines up mathematically in the coefficient of restitution, which is a dimensionless number. In the context of springs (elastic force), it appears as the spring constant, which is measured in N/m. In the context of gravity, we have the constant G, measured in Nm^2/kg^2. Thus… if we considered the possibility that G were multiplied by a certain (dimensionless) coefficient, would that be simply illogical, would it clash against physics principles?

It seems it would have to be a “reductive coefficient”, ranging between 0 and 1, like the coefficient of restitution (let’s call it “e”). Thus we could have less but not more relative acceleration than in the standard formula. On top of that, it would be empirically constructed, like I gather that it happens with “e”, on the basis of the properties (composition) of the two involved bodies. In fact, it would be “e”.

For example, if we have two bodies “dropped” from a relative distance R and acquiring a certain relative velocity at collision time and if the collision is partially elastic and partially plastic, the bodies would bounce off each other with outgoing relative velocity = (1+e) times their incoming relative velocity… Then they separate until running out of outgoing kinetic energy due to the gravitational force between them. Since that outgoing KE is now lower (part of it has been dedicated and is kept busy heating or deforming the bodies), gravity will stop them and revert their direction sooner, before they reach the initial distance R. Thus they’ll return with lower velocity, which will diminish even more after collision. And so on until the bodies are brought to rest with each other.

But does all that mean that the gravitational interaction is fully “elastic” (it only transfers to the bodies energy that is useful for motion, that is either kinetic or potentially convertible into kinetic energy)? In principle, I do not se why it should be necessarily so. Note this: when the bodies are separating from each other, they are being stretched, due to the non-uniformity of the gravitational field. That is the well-known phenomenon of tides. And when they descend or get closer to each other, I think that they progressively lose that condition, because (as the force increases due to the inverse square law) the difference of strength of the field between the two extremes becomes narrower. Given this, should the bodies perfectly recover their original form? Why should they? They will not when they are later compressed due to the collision. Should the gravitational stretching be different? Shouldn’t it also generate, due to the composition of the bodies, some conversion of the energy transferred by gravity into other non-mechanical forms? I wonder if there is any logical, not purely experimental reason banning that.

boit

If I lived in a planet where more massive objects were falling much slower, I will create a perpetual motion energy generator in no time. How? Since by implication more massive object will have less potential energy, I'll rig the contraption to haul stones in a single bag and then throw them down sepatately. On Earth surely 1/2(1+1+1)10 m^2/s^2 can not be less than 1/2(3)10 m^2/s^2 [from E=1/2mv^2]. This is because whatever you do the centre of mass will be a point source. The trick could be to make your more massive object so tall that the COG is in a zone where accelaration is say half 'g'. But unfortunately you can't make them rigid enough.

Saw

Boit, I understand that this is a little confusing, since the OP is actually that, but in the end we are not discussing that, we drifted to a related issue, namely equivalence between inertial and gravitational mass. In this context, I simply asked something like: if a collision between planets is imperfectly elastic (so that the outgoing relative v of separation < the incoming relative v of approach), we assume nevertheless that the subsequent gravitational interaction is perfectly elastic (the relative v of return after the path closes = the relative velocity of departure), but does it HAVE TO be so? Of course, the relative return v can't be higher than relative departure v, but can't it be lower, even in the absence of external forces?

If it were so and that deceleration happened in inverse proportion to the masses, the velocity of the CoM of the system would not change = conservation of momentum would be respected. Yes, Kinetic Energy, at the end of the cycle, would not be conserved but that is not unsual: that happened in the collision and it's not dramatic. The sacred principle is that Total Energy must be conserved, but that is also respected if the KE converts into any another form, even if it is not potential energy that is reusable for motion. The only principle that would be breached is that gravity ensures 100% conservation of mechanical energy. And... although this principle is quite useful for calculations and true in almost all practical situations where gravity is the only intervening force, is it necessarily always applicable or rather a practical assumption, which could however break at large scales?

Flustered

If I jump in the air and come back down due to gravity, you guys are saying that my speed in which I fall back to the ground is the same speed in which satellites orbit the Earth? Both bodies have gravity acting upon it so wouldn't it be the same speed?

Saw

Not really. The principle refers to acceleration, not to speed. All objects falling towards the Earth, regardless their different masses, suffer the same acceleration, as long as you leave air resistance aside. This is true as measured from or with regard to an inertial reference frame or (if you disregard as negligible the acceleration caused onto the Earth by the object in question) with regard to the Earth itself. And as you know acceleration means change of speed or direction per time unit.

You suffer the same acceleration as the satellite but your speed is lower. You started falling (at the point of maximum height) with speed zero and have fallen a short path and hence have accelerated (your speed has grown) during a short time. Instead, the satellite had to be launched with a high speed so as to enter into orbit. It is true that once it is in orbit, its speed does not change much; actually, if the orbit were circular, its speed would not change at all, because all the work of gravity woud be used changing its direction; yet the satellite's speed, for the reason explained above, will always be higher than yours, if you just jump and fall back.

Dali

Well - it can't be so. Just think of the situation if heavier objects really had fallen slower, what would have happened if you divided the heavy object in two parts? Should those parts start falling quicker then? Why? Or divide it in 1000 small light parts! Should speed depend on if those parts where held together or not? How tight do the parts need to sit together to fall slower than they would individually? It obviously could not work that way.

Best way to think about it is that gravity acts equally on each atom in every object. No matter how many atoms there are, or if they are bound together or not.

Buckleymanor

The marble and the ball only fall with the same acceleration if you "drop" them.If you placed the marble and ball on the floor of the spaceship when it was accelerating and then decelerated or stoped the craft, the bowling ball and marble would accelerate away from the floor at different rates.Gravity is similar to the force that pushes you back in car seat when you step on the gas but not the same.

TurtleMeister

If what you mean by "stop the craft" is to simply turn off the thrusters then that's not true. If you disregard the elasticity of the objects then both balls will remain on the floor. To bring them off the floor you would need to reverse the thrusters. And even then they would accelerate away from the floor at the same rate. That's because the balls will not really accelerate at all. The ship will accelerate from the balls.

Buckleymanor

No it would not, the craft would just have to be accelerating or travelling at a rate less than maximum speed the bowling ball had attained when it was first accelerated by the spacecraft.The objects are elastic I don't see how you can disregard this unless they are made of unobtanium.
You might as well conclude that the balls are stationary and the craft is doing all the movement and acceleration around them.
If the balls were placed on your car seat and accelerated and the car crashed the marble would be accelerated faster and travell further than the bowling ball if it's exit through the winderscrean was unristricted.

Jasso

How do you manage to get this? Ignoring elastic rebound, they would appear to accelerate in the opposite direction as the rocket, with the same magnitude.

And how would you tell the difference between constant accelerating and gravity in a soundless, windowless car?

Unless the direction of acceleration were to change, then the objects would not come off of the floor, even if the magnitude of acceleration were to decrease. That is because the spaceship will always be moving faster at a time t+dt and the only way the object will do that is if the floor of the spaceship is in contact with it.

Ah, I see how you managed to get that earlier. It's an idealization, used to ignore whatever is irrelevant to the situation. And it's irrelevant since once the object loses contact with the floor, its elasticity doesn't matter; it will continue with constant velocity because of Newton's First Law.

That's pretty much the entire point of the equivalence principle, that there is no difference between being on a spaceship with constant acceleration, or being on the surface of a planet with the gravitational acceleration.

No, they would continue on with the same velocity they had right before they lost contact with the seat.

TurtleMeister

Buckleymanor, I agree with Jasso's objections to your comments. An analogy to ignoring the elasticity of the objects is in the thought experiment for the universality of free fall. In that scenario we commonly ignore the effects of air resistance and buoyancy when we say that two objects will accelerate at the same rate regardless of there difference in mass. We do that because, as Jasso pointed out, those effects are irrelevant to the point of the thought experiment.

To demonstrate, here's another thought experiment. Imagine two balls of different mass sitting on a table in an accelerating spacecraft (or sitting on the surface of earth). Because of the elasticity of the objects, the two balls, and the table they are sitting on will deform from the force of acceleration. When the thrusters are turned off (or earths gravity is turned off) the balls and the table will resume their normal shape (rebound), causing the balls and the table/spacecraft (or earth) to accelerated away from each other. But once the balls and the table are no longer in contact, the acceleration stops, and they continue on away from each other at a constant speed. This is the effect that we want to ignore.

If we ignore the elasticity of the objects then the balls will remain on the floor when the thrusters are turned off. The only way to bring them off the floor would be to reverse, or change direction of, the thrusters. If we reverse the thrusters, then the spacecraft will accelerate away from the balls.

Buckleymanor

Which would be different for the bowling ball and marble.Substitute car and seat for catapult.
Fire bowling ball, then marble, with the same amount of force, which travells furthest and fastest.

russ_watters

You're confusing acceleration and force.

Buckleymanor

Don't forces come into play when onboard accelerating spacecraft or cars.
I ain't sure if it's just a question of blanking out the windows and making them soundproof.

Buckleymanor

Completly agree as to ignoring the air resistance and buoyancy effects on different objects in free fall.To enable us to come to the conclusion that they accelerate at the same rate regardless of there different mass.
The difference between that and the anology you are making is that it is possible to conduct an experiment with two different objects free falling within a vacuum to check your results.
Ignoring elasticity in a thought experiment is not possible experimentally so might as well substitute it for green cheese.

russ_watters

Of course - whenever you have one (net force, that is) you have the other. But you defined your scenarios such that in one case the accelerations were equal (thus, forces different) and in the other the forces were equal (thus accelerations different).