Why don't heavy objects fall more SLOWLY?

Whovian

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

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 [itex]v_0[/itex], and we label the location as the origin. The distance the base of the rocket travels at a time [itex]t[/itex] later with new acceleration [itex]-a_r[/itex]is:

[tex]d_r = v_0 t - \frac{1}{2} a_r t^2[/tex].

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

[tex]d_b = v_0 t[/tex].

If the chamber where the object is had a length [itex]l[/itex], and [itex]t[/itex] was the time it took for the object to travel to the other end, then the object would have to travel a distance of [itex]d_r + l[/itex] 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:

[tex]d_b = d_r + l[/tex]

Replacing [itex]d_b[/itex] and [itex]d_r[/itex] with the formulas and solving for [itex]l[/itex]:

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

Collecting terms and simplifying:

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

Solving for [itex]t[/itex]:

[tex]t= \sqrt{\frac{2 l}{a_r}}[/tex].

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

Drakkith

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

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

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

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

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

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

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 acheive that speed.

Jasso

[tex]F = ma[/tex]
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

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

Whovian

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

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