Why don't heavy objects fall more SLOWLY?

[Saw]

That’s a good point, bringing in the concept of reduced mass! It converges with the progress I was making in the domain of collisions (see the thread “Acceleration in an elastic collision”), where the conclusion seems to be that the force intervening in the collision can also be defined as:

F = \mu (1 + \varepsilon )v_{rel}^{initial}

where:

<br /> \begin{gathered}<br /> \mu = \frac{{m_1 m_2 }}<br /> {{m_1 + m_2 }} = {\text{reduced mass}} \hfill \\<br /> \varepsilon = \frac{{v_{rel}^{final} }}<br /> {{v_{rel}^{initial} }} = {\text{coefficient of restitution}} \hfill \\<br /> a = (1 + \varepsilon )v_{rel}^{initial} = {\text{relative acceleration}} \hfill \\ <br /> \end{gathered} <br />

On the other hand, if you want to consider the possibility that gravitational masses (included in the relative acceleration term) differ from inertial masses (those included in the reduced mass term), then you have to set an upper limit to the former.

In the collision formula, that limit, derived from the conservation of kinetic energy, is that relative acceleration cannot exceed 2 times relative velocity, since the coefficient of restitution can range from 0 to 1. Certainly, if you know that coefficient based on the ratio final/initial relative velocity, the formula is not very useful. But the idea is that you could also obtain it empirically as a coefficient inherent to the material that each body is composed of. In particular, one would have to use, together with “reduced or effective mass”, the “effective coefficient” of the two materials, as if they were one = k₁k₂/(k₁+k₂). I still haven’t thought, however, how to translate this to the gravity formula.

 

[Saw]

I've realized that what I called above, referring to collisions, the relative (or total) acceleration and force would actually be velocity increase and momentum of the system, respectively. To really get the acceleration and hence the force I should have divided (1+e)v_rel by the collision time.

 

[Saw]

I've realized that what I called above, referring to collisions, the relative (or total) acceleration and force would actually be velocity increase and momentum of the system, respectively. To really get the acceleration and hence the force I should have divided (1+e)v_rel by the collision time.

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]

If I lived in a planet where more massive objects were falling much slower

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]

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?

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]

This is a simple question, and I'm sure it has a trivial answer, but this thought occurred to me just now.

How is it that gravity is able to accelerate things at the same rate regardless of its mass, while one definition of mass is the measure of resistance to acceleration? If two objects are 100 miles from Earth, I would expect that the Earth's gravity would be able to accelerate the lighter object more quickly, while the heavier object would resist the acceleration and fall more slowly. Why is this not the case?

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]

If you are in a rocket ship out in space undergoing acceleration and you "drop" a marble and a bowling ball in your ship, they will "fall" with the same acceleration. Einstein's equivalence principle is saying that gravity is the same thing as the "force" that pushes you back in your car seat when you step on the gas.

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 stopped 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 you placed the marble and ball on the floor of the spaceship when it was accelerating and then decelerated or stopped the craft, the bowling ball and marble would accelerate away from the floor at different rates.

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]

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.

No it would not, the craft would just have to be accelerating or traveling 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]

.If you placed the marble and ball on the floor of the spaceship when it was accelerating and then decelerated or stopped the craft, the bowling ball and marble would accelerate away from the floor at different rates.

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.

Gravity is similar to the force that pushes you back in car seat when you step on the gas but not the same.

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

No it would not, the craft would just have to be accelerating or traveling at a rate less than maximum speed the bowling ball had attained when it was first accelerated by the spacecraft .

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.

The objects are elastic I don't see how you can disregard this unless they are made of unobtanium.

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.

You might as well conclude that the balls are stationary and the craft is doing all the movement and acceleration around them.

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.

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.

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 Earth's 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]

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.

That's the point you can't ignore elastic rebound.

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

By using elastic rebound to note the difference speeds of acceleration upon different objects.

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.

Would that not depend on the speed the magnitude of acceleration decreased.

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.

There is.

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

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 travels furthest and fastest.

 

[russ_watters]

You're confusing acceleration and force.

 

[Buckleymanor]

You're confusing acceleration and force.

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]

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 Earth's 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.

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]

Don't forces come into play when onboard accelerating spacecraft or cars.

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).

 

[Buckleymanor]

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).

If I were to hit a bowling ball with a tennis bat and then a marble, both with the same amount of force won't one travell further than the other.
The marble and the bowling ball are separated on the floor of the craft.

 

[russ_watters]

Yes...

 

[Drakkith]

If I were to hit a bowling ball with a tennis bat and then a marble, both with the same amount of force won't one travell further than the other.
The marble and the bowling ball are separated on the floor of the craft.

In your spaceship you are accelerating them both at the same time to the same speed. Both of them are accelerating at the same rate and once you reach your maximum velocity both will be traveling at that velocity. Each one takes a different amount of force to accelerate, but since they are both on the floor of the ship at the same time they both have to accelerate at the same rate.

This is kind of similar to hitting both the bowling ball and the marble with the bat at the same time, except that hitting them with the bat has much more complicated physics come into play such as elasticity and such. Imagine you had them both sitting on a catapult at the same time and simply launched them into the air with it. Both would accelerate at the same rate to the same speed.

 

[Buckleymanor]

This is kind of similar to hitting both the bowling ball and the marble with the bat at the same time, except that hitting them with the bat has much more complicated physics come into play such as elasticity and such. Imagine you had them both sitting on a catapult at the same time and simply launched them into the air with it. Both would accelerate at the same rate to the same speed.

Yes if they were both sitting in or on the catapult together the bowling balls mass and the marble mass would each have an influence on each other and the catapult, so they would accelerate at same rate.
If they were separated and on the floor of the spacecraft , or placed on separate seats in the front of car when the craft decelerated or the car crashed and elasticity was 'allowed' to do what it does.
They would still accelerate at the same rate to the same speed.The less massive marble would not have as much elastic effect when first accelerated as the more massive bowling ball.
Thanks.

 

[Drakkith]

If they were separated and on the floor of the spacecraft , or placed on separate seats in the front of car when the craft decelerated or the car crashed and elasticity was 'allowed' to do what it does.
You would be able to tell that the objects are being accelerated and not in a gravitational field.

No, the floor of the spacecraft is like the catapult, accelerating both equally. Same for a car.

 

[Buckleymanor]

Yes I must have been editeing at the same time you were posting thanks.

 

[Buckleymanor]

In your spaceship you are accelerating them both at the same time to the same speed. Both of them are accelerating at the same rate and once you reach your maximum velocity both will be traveling at that velocity. Each one takes a different amount of force to accelerate, but since they are both on the floor of the ship at the same time they both have to accelerate at the same rate.

This is kind of similar to hitting both the bowling ball and the marble with the bat at the same time, except that hitting them with the bat has much more complicated physics come into play such as elasticity and such. Imagine you had them both sitting on a catapult at the same time and simply launched them into the air with it. Both would accelerate at the same rate to the same speed.

To be honest I am haveing trouble trying to understand this.
If you just had the the bowling ball on the floor of the spaceship and you accelerated it to your maximum velocity and then applied the brakes or reverse thrust and measured the distance and the speed it traveled at.
Then you repeated the same experiment with just the marble won't the distances and speeds of the two objects be different.

With respect to this question I posted.

If I were to hit a bowling ball with a tennis bat and then a marble, both with the same amount of force won't one travell further than the other.

And the answer I received from russ_watters

Yes...

If that is the case then won't it be possible to distinguish between acceleration due to gravity and acceleration due to acceleration.

 

[Drakkith]

To be honest I am haveing trouble trying to understand this.
If you just had the the bowling ball on the floor of the spaceship and you accelerated it to your maximum velocity and then applied the brakes or reverse thrust and measured the distance and the speed it traveled at.
Then you repeated the same experiment with just the marble won't the distances and speeds of the two objects be different.

There is no maximum velocity. Your velocity will continue to increase as long as you apply thrust. What you can do is measure the velocity relative to something else, like the Earth, for the amount of fuel you expend. Accelerating the ship with the bowling ball will get you to a lower velocity than with just the marble for the same amount of fuel.

Also, you say we should stop accelerating and apply the brakes. But that IS acceleration, it's just in the negative direction to your current vector. If you applied the same acceleration in both instances both the bowling ball and the marble would be traveling the same speed upon measurement. This is because once you reverse the thrust you are only accelerating the ship, and the motion of the ball and marble relative to the ship would be the same because the ship will accelerate the same in both cases.

And you cannot measure the distance the bowling ball or marble travel. Both will travel to the other side of the ship at the same velocity and hit the wall. Remember that in order to say that acceleration is the same as gravity we cannot "cheat" and look outside to see what is actually going on. We are forced to measure things inside the ship.

 

[TurtleMeister]

To be honest I am haveing trouble trying to understand this.
If you just had the the bowling ball on the floor of the spaceship and you accelerated it to your maximum velocity and then applied the brakes or reverse thrust and measured the distance and the speed it traveled at.
Then you repeated the same experiment with just the marble won't the distances and speeds of the two objects be different.

No. When the spacecraft reverses thrust, whatever objects are sitting on the floor will simply continue traveling at the velocity the spacecraft obtained before it reversed thrust (acceleration has been removed from the objects so Newton's first law applies). While the spacecraft (and the floor of course) will accelerate away from the objects (in the opposite direction).

If I were to hit a bowling ball with a tennis bat and then a marble, both with the same amount of force won't one travell further than the other.

Yes, one will travel faster than the other. But in that case you are hitting the objects separately at different times. That scenario is not analogous to the spacecraft scenario. Ask yourself this question: What would happen if you attached the bowling ball and the marble together and hit them both at the same time? That would be more analogous to the spacecraft scenario. You don't have to even attach them together, just hit them both at the same time.

 

[Buckleymanor]

And you cannot measure the distance the bowling ball or marble travel. Both will travel to the other side of the ship at the same velocity and hit the wall. Remember that in order to say that acceleration is the same as gravity we cannot "cheat" and look outside to see what is actually going on. We are forced to measure things inside the ship.

Correct me if I am wrong but that would only apply if the bowling ball and marble were on board the craft at the same time.If you took the bowling ball up first and measured it's velocity it would be slower than the marbles.

 

[Drakkith]

Correct me if I am wrong but that would only apply if the bowling ball and marble were on board the craft at the same time.

As long as you decelerate the ship at the same rate it doesn't matter if you have one or the other or both.

If you took the bowling ball up first and measured it's velocity it would be slower than the marbles.

Measured its velocity relative to what?

 

[Buckleymanor]

Measured its velocity relative to what?

You accelerate the craft for a given amount of time with maximum thrust with just the bowling ball on board then decelerate for another measured amount of time with maximum reverse thrust.
Then repeat with just the marble on board and then compare the time it took for the marble and bowling ball to leave the floor and hit the far side of the ship.

 

[TurtleMeister]

You accelerate the craft for a given amount of time with maximum thrust with just the bowling ball on board then decelerate for another measured amount of time with maximum reverse thrust.
Then repeat with just the marble on board and then compare the time it took for the marble and bowling ball to leave the floor and hit the far side of the ship.

The time would be the same. The mass of the bowling ball and marble have nothing to do with it (the time period you are talking about). What affects that time interval is the thrust of the rocket and the mass of the spacecraft (minus the bowling ball and marble). As I explained in my previous post, once the spacecraft reverses it's thrust the only thing that accelerates is the rocket. The bowling ball and marble (regardless of whether this exercise is done with them together or separate) will simply obey Newton's first law and "coast" until the other side of the spacecraft accelerates into them.

 

[Drakkith]

You accelerate the craft for a given amount of time with maximum thrust with just the bowling ball on board then decelerate for another measured amount of time with maximum reverse thrust.
Then repeat with just the marble on board and then compare the time it took for the marble and bowling ball to leave the floor and hit the far side of the ship.

It would be equal. In both cases the only thing you are decelerating is the ship, not the marble or bowling ball. So applying maximum thrust would cause both to reach the other side of the ship after an equal amount of time.

 

[Buckleymanor]

It would be equal. In both cases the only thing you are decelerating is the ship, not the marble or bowling ball. So applying maximum thrust would cause both to reach the other side of the ship after an equal amount of time.

Are you sure won't they have different closing velocities.
The marble and the ball would have attained different velocities with the initial acceleration.
Because you are only decelerating the ship the marble would have a greater intial velocity than the bowling ball because the ship would be lighter.

For instance if it were possible to stop the ship the craft that weighed more could not have attained the speed of a lighter craft with the same amount of thrust.

 

[Drakkith]

Are you sure won't they have different closing velocities.
The marble and the ball would have attained different velocities with the initial acceleration.
Because you are only decelerating the ship the marble would have a greater intial velocity than the bowling ball because the ship would be lighter.

For instance if it were possible to stop the ship the craft that weighed more could not attained the speed of a lighter craft with the same amount of thrust.

No, we are measuring the velocity relative to the ship, not an outside source. In such a case the velocity of the ship relative to an outside object is irrelevant, only the acceleration upon braking matters.

 

[Buckleymanor]

No, we are measuring the velocity relative to the ship, not an outside source. In such a case the velocity of the ship relative to an outside object is irrelevant, only the acceleration upon braking matters.

I don't see how that will prevent the lighter object reaching the other side of the ship more quickly.
The ship would have to accelerate at the same velocity for both which it won't becuase it's more massive with one object on board than the other.

 

[Whovian]

I don't see how that will prevent the lighter object reaching the other side of the ship more quickly.
The ship would have to accelerate at the same velocity for both which it won't becuase it's more massive with one object on board than the other.

No, the engine is accelerating the ship, but is not doing any work on the ball until it hits the edge. The ship and its contents are more massive, but it's what the engine is doing work on or accelerating that matters, and it isn't doing either to the ball initially. And aren't we starting to get into the basics of General Relativity here?

 

[Jasso]

I don't see how that will prevent the lighter object reaching the other side of the ship more quickly.
The ship would have to accelerate at the same velocity for both which it won't becuase it's more massive with one object on board than the other.

Ok, let's use some actual formula's for this. Suppose a rocket with an object inside had been constantly accelerated and then instantly reversed the direction of acceleration. At the moment that the acceleration changed, it had a velocity v_0, and we label the location as the origin. The distance the base of the rocket travels at a time t later with new acceleration -a_ris:

d_r = v_0 t - \frac{1}{2} a_r t^2.

Now, up till now, the object was in contact with the base of the rocket, so it also had a velocity of v_0. However, once it loses contact with the ship, it will keep traveling at that velocity until it makes contact again. So the distance that the object travels in the same time t is simply:

d_b = v_0 t.

If the chamber where the object is had a length l, and t was the time it took for the object to travel to the other end, then the object would have to travel a distance of d_r + l to make contact with the other side of the rocket (since both the ends of the chamber will have moved the same distance). So setting those equal:

d_b = d_r + l

Replacing d_b and d_r with the formulas and solving for l:

(v_0 t) = (v_0 t - \frac{1}{2} a_r t^2) + l
(v_0 t) - (v_0 t - \frac{1}{2} a_r t^2)= l

Collecting terms and simplifying:

(v_0 - v_0) t + \frac{1}{2} a_r t^2= l
(0) t + \frac{1}{2} a_r t^2= l
\frac{1}{2} a_r t^2= l

Solving for t:

t= \sqrt{\frac{2 l}{a_r}}.

Note that this doesn't depend on either the mass of the object or the velocity that it and the rocket are traveling at when the acceleration changes.

 

[Drakkith]

I don't see how that will prevent the lighter object reaching the other side of the ship more quickly.
The ship would have to accelerate at the same velocity for both which it won't becuase it's more massive with one object on board than the other.

No, the engines are only exerting a force on the ship, not the ball. The ship is not more massive because the ball is there. Only once the ball reaches the other side of the ship will it's mass have to be taken into account in regards to the acceleration.

Imagine you are in a ship floating in space with a ball in the center of a room just floating there. You fire your thrusters at a constant rate until the ball hits the side of the room. The rate of acceleration is independent of the mass of the ball because the ball is NOT having any work done upon it until it hits the wall. If I were an observer floating next to you looking in the window I would NOT see the ball move at all while the ship moved around it until it touched the wall.

This is exactly the same as when we stop firing the thrusters and begin to decelerate. As soon as the thrusters stop firing, neither the ship nor the ball are accelerating any longer. They are in the same situation as the above example, except with the ball floating at one edge of the room instead of the center. When I begin to fire my thrusters and decelerate the ship, I am NOT doing anything to the ball until it reaches the other side of the room, at which point it will touch the wall and we will have to expend energy to accelerate it as well, at which point it's mass comes into play.

 

[Buckleymanor]

No, the engine is accelerating the ship, but is not doing any work on the ball until it hits the edge. The ship and its contents are more massive, but it's what the engine is doing work on or accelerating that matters, and it isn't doing either to the ball initially. And aren't we starting to get into the basics of General Relativity here?

Of course it is, from the moment the craft leaves from wherever it takes off from it is doing work on every part of the ship including every thing on board.

 

[Buckleymanor]

No, the engines are only exerting a force on the ship, not the ball. The ship is not more massive because the ball is there. Only once the ball reaches the other side of the ship will it's mass have to be taken into account in regards to the acceleration.

From the moment the ship takes off it will expend more fuel and accelerate less if a heavy object is on board if this was not the case you could take a mountain on board and use no more fuel than if a marble was there.
What is mass if it is not the resistence to a force applied.

 

[Drakkith]

From the moment the ship takes off it will expend more fuel and accelerate less if a heavy object is on board if this was not the case you could take a mountain on board and use no more fuel than if a marble was there.
What is mass if it is not the resistence to a force applied.

Ignore the ship taking off. Focus on the deceleration part. That is where your confusion lies. If the ball is NOT in contact with the ship then the ship cannot exert a force upon it, meaning that it takes no extra energy to decelerate the ship while the ball is not in contact with the ship.

 

[Buckleymanor]

Ignore the ship taking off. Focus on the deceleration part. That is where your confusion lies. If the ball is NOT in contact with the ship then the ship cannot exert a force upon it, meaning that it takes no extra energy to decelerate the ship while the ball is not in contact with the ship.

I agree it won't exert a force on the ball if it is not in contact with it but it will still have to use more energy to decelerate a faster moveing ship, and that depends on which ball was on board when it was first accelerating.

 

[Drakkith]

I agree it won't exert a force on the ball if it is not in contact with it but it will still have to use more energy to decelerate a faster moveing ship, and that depends on which ball was on board when it was first accelerating.

We are not decelerating to a stop, we are only decelerating until the ball hits the opposite wall. The ship will not take more time to decelerate for either one.

 

[TurtleMeister]

I agree it won't exert a force on the ball if it is not in contact with it but it will still have to use more energy to decelerate a faster moveing ship, and that depends on which ball was on board when it was first accelerating.

It appears that your misconception is that the speed relative to the point of take off is relavent to this time interval. Well, it is not. The time interval in question remains the same regardless of what speed the spacecraft has attained relative to the start point and regardless of how much energy it took to achieve that speed.

 

[Jasso]

I agree it won't exert a force on the ball if it is not in contact with it but it will still have to use more energy to decelerate a faster moveing ship, and that depends on which ball was on board when it was first accelerating.

F = ma
The accelration only depends on the total force applied and on the mass that is being accelerated. Since there is no force on the ball when not in contact, it will not be accelerated, so the only thing being accelerated is the ship. Since the only thing being accelerated is the ship and the thrusters have the same force in both cases, it will have the same acceleration in both cases until the ball hits the other side.

 

[Buckleymanor]

It appears that your misconception is that the speed relative to the point of take off is relavent to this time interval. Well, it is not. The time interval in question remains the same regardless of what speed the spacecraft has attained relative to the start point and regardless of how much energy it took to achieve that speed.

So for a spaceship traveling at allmost lightspeed the time intervall will be the same as craft traveling at 1000Kph.

 

[Whovian]

So for a spaceship traveling at allmost lightspeed the time intervall will be the same as craft traveling at 1000Kph.

I think we're talking in the reference frame of the ship(?), in which case it won't matter what velocity it is traveling at relative to the takeoff point. The ship observes the takeoff point as time dilated and length contracted, the takeoff point observes the ship as time dilated and length contracted.

 

[TurtleMeister]

So for a spaceship traveling at allmost lightspeed the time intervall will be the same as craft traveling at 1000Kph.

Yes, that is correct (in the reference frame of the spacecraft ).