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From: Timo Nieminen on 27 May 2010 15:14
On May 27, 9:54 pm, boltar2...(a)boltar.world wrote:
> Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> >On May 27, 9:26=A0pm, boltar2...(a)boltar.world wrote:
> >> On Thu, 27 May 2010 03:48:06 -0700 (PDT)
>
> >> Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> >> >Even if the rocket is moving faster than the exhaust speed, it still
> >> >works. In this case, the rocket gains more KE than the motor is
> >> >providing, but the exhaust is losing energy, not gaining energy, since
> >> >it ends up moving slower than it started.
>
> >> That doesn't make sense. The KE can only come from the energy the motor
> >> is providing. It can't come out of nowhere.
>
> >All it is is (energy provided by motor) = (change in KE) = (change in
> >KE of rocket) + (change in KE of exhaust). If (change in KE of
> >exhaust) is negative - that is, the exhaust is losing energy - then
> >(change in KE of rocket) can be more than (energy provided by motor).
> >Energy provided = energy gained, no energy from nothing.
>
> In that scenario the rocket will at some point start taking all the KE from
> the exhaust which will then have a speed of zero (ignoring relativity) so
> then what? The rocket according to F = MA will still have to accelerate but
> theres not enough power from the motor to provide the extra KE it needs in the
> time required.

Maybe it's time to sit down and do the calculation? Try it, and you'll
see that this isn't a problem.

The faster the rocket goes, the more KE the unexhausted exhaust starts
with. When the rocket goes faster than the exhaust, the change in the
KE of the exhaust is larger than when the rocket goes at the speed of
the exhaust, even if the final KE of the exhaust in not-zero.

If this troubles you, do some examples yourself. Many examples. Try
the completely inelastic collision problem (i.e., where the two
colliding bodies stick together). Start with the same relative speed
between the two bodies, but with different speeds for the bodies. For
the different speeds, the KE of each body, and the total KE, will be
different. The final KE will be different. But in all cases,
KE_initial - KE_final = the same value, the amount of energy lost in
the collision. (Where "lost" means converted to other forms of
energy.)

What the instantaneous speed of the rocket (or the colliding objects)
is results from human choice of what to measure velocities relative
to. You can always choose a coordinate system so that the spaceship is
at rest, or so that the spaceship is moving at the exhaust speed, or
whatever speed you choose. This human choice doesn't affect the
physics (i.e., the conservation of energy), but it does affect the
human description of it (e.g., how much KE each body has). This is
exactly the same as the effect that choosing where the origin is
(keeping it stationary for simplicity) has on the motion of a particle
- it affects the coordinates, but not the velocity or acceleration.
The trajectory is the same shape, just shifted.

This isn't at all new. The equivalence of coordinate systems was
discussed by Newton and Galileo. But it has been confusing students
for hundreds of years. You don't have to take it on faith; you can try
it for yourself, and caculate examples, solve problems, until you see
how it works. That's one purpose of those long lists of end-of-chapter
problems.

From: boltar2003 on 28 May 2010 05:02
On Thu, 27 May 2010 12:14:39 -0700 (PDT)
Timo Nieminen <timo(a)physics.uq.edu.au> wrote:
>Maybe it's time to sit down and do the calculation? Try it, and you'll
>see that this isn't a problem.
>
>The faster the rocket goes, the more KE the unexhausted exhaust starts
>with. When the rocket goes faster than the exhaust, the change in the
>KE of the exhaust is larger than when the rocket goes at the speed of
>the exhaust, even if the final KE of the exhaust in not-zero.
>
>If this troubles you, do some examples yourself. Many examples. Try
>the completely inelastic collision problem (i.e., where the two
>colliding bodies stick together). Start with the same relative speed
>between the two bodies, but with different speeds for the bodies. For
>the different speeds, the KE of each body, and the total KE, will be
>different. The final KE will be different. But in all cases,
>KE_initial - KE_final =3D the same value, the amount of energy lost in
>the collision. (Where "lost" means converted to other forms of
>energy.)

Ok , I understand now. A good explanation , thank you.

I still can't figure out how it works when what provides the force doesn't
have any mass itself such as some sort of field however since then the field
won't have any KE to start with to donate to the rocket.

>This isn't at all new. The equivalence of coordinate systems was
>discussed by Newton and Galileo. But it has been confusing students
>for hundreds of years. You don't have to take it on faith; you can try
>it for yourself, and caculate examples, solve problems, until you see
>how it works. That's one purpose of those long lists of end-of-chapter
>problems.

I know its true. I'm not one of these nutters trying to prove it to be false, I
just didn't understand why it was true :)

B2003

From: jbriggs444 on 28 May 2010 08:17
On May 26, 12:16 pm, JimboCat <103134.3...(a)compuserve.com> wrote:
> On May 26, 10:16 am, boltar2...(a)boltar.world wrote:
>
>
>
>
>
> > Before anyone asks , no this isn't some school project I'm working on. Its
> > a simple problem I can't figure out for myself.
>
> > If you have a rocket motor giving a constant thrust then the rocket - not
> > accounting for weight change due to fuel use - will accelerate at a constant
> > rate. However if say its accelerating at 1m/s/s then it will take twice as
> > long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 m/s.
> > Yet going by 0.5 mv^2 the kinetic energy for a rocket weighing 1 kilo at the
> > given speeds is:
>
> > 1 m/s = 0.5J
> > 2 m/s = 2J
> > 4 m/s = 8J
>
> > So in the first second it gained 1.5J but in the next 2 seconds it gained
> > 6J. I don't get it? The motor wasn't putting out any extra energy so where
> > did it come from?
>
> Kinetic energy is frame-dependent (because velocity is frame-
> dependent), and the rocket is continuously changing frames (more
> accurately, it isn't in an inertial frame at all). So the energy
> comparison is simply invalid: it doesn't mean anything.

A frame of reference is not something that the rocket can change.
A frame of reference is something that the person analyzing a
situation
can select. It is a free choice.

[Thinking back to physics 101, quite a number of students had the
mistaken impression that a frame of reference must be tied to
a physical object. I do not accuse you of making that mistake,
but your posting can be read that way].

If you wish to analyze the situation from a series of "tangent
inertial frames"
in which the rocket is momentarily at rest, you are free to do so. If
you wish
to analyze the situation from an "accelerating frame" in which the
spacecraft
is at rest, you are free to do so.

The OP wishes to examine the situation from the frame of reference in
which the rocket was initially at rest. That is a valid inertial
frame.

The energy comparison is valid. The laws of physics hold
in all inertial frames. The work-energy being delivered to the rocket
by
the rocket motor is increasing over time. You are correct that this
is
a frame-relative statement. But it is a frame-relative statement that
is
true regardless of what single inertial frame you choose to work in.

The resolution to this apparent paradox is as PD stated, to complete
the energy accounting by examining the work done on the exhaust
stream. The work done on the rocket is increasing over time. The
work done on the exhaust stream is decreasing over time and may
even go negative.
From: PD on 28 May 2010 09:46
On May 28, 4:02 am, boltar2...(a)boltar.world wrote:
> On Thu, 27 May 2010 12:14:39 -0700 (PDT)
>
>
>
> Timo Nieminen <t...(a)physics.uq.edu.au> wrote:
> >Maybe it's time to sit down and do the calculation? Try it, and you'll
> >see that this isn't a problem.
>
> >The faster the rocket goes, the more KE the unexhausted exhaust starts
> >with. When the rocket goes faster than the exhaust, the change in the
> >KE of the exhaust is larger than when the rocket goes at the speed of
> >the exhaust, even if the final KE of the exhaust in not-zero.
>
> >If this troubles you, do some examples yourself. Many examples. Try
> >the completely inelastic collision problem (i.e., where the two
> >colliding bodies stick together). Start with the same relative speed
> >between the two bodies, but with different speeds for the bodies. For
> >the different speeds, the KE of each body, and the total KE, will be
> >different. The final KE will be different. But in all cases,
> >KE_initial - KE_final =3D the same value, the amount of energy lost in
> >the collision. (Where "lost" means converted to other forms of
> >energy.)
>
> Ok , I understand now. A good explanation , thank you.
>
> I still can't figure out how it works when what provides the force doesn't
> have any mass itself such as some sort of field however since then the field
> won't have any KE to start with to donate to the rocket.

This is a misconception. Fields are deemed to be physical entities
partly because they do exhibit important physical properties like
stored energy and momentum. You might ask how they do that if they
don't have mass. You may have been misled that the momentum of
ANYTHING is defined as mass times velocity, or that the kinetic energy
of ANYTHING is half the mass times the velocity squared. If you got
that impression, then it was an unfortunate mistake. That rule for
momentum works only for massive objects and even then only for those
moving slowly (compared to the speed of light). There are different
rules for finding momentum for other kinds of physical entities.

>
> >This isn't at all new. The equivalence of coordinate systems was
> >discussed by Newton and Galileo. But it has been confusing students
> >for hundreds of years. You don't have to take it on faith; you can try
> >it for yourself, and caculate examples, solve problems, until you see
> >how it works. That's one purpose of those long lists of end-of-chapter
> >problems.
>
> I know its true. I'm not one of these nutters trying to prove it to be false, I
> just didn't understand why it was true :)
>
> B2003

From: BURT on 28 May 2010 15:32
On May 26, 7:16 am, boltar2...(a)boltar.world wrote:
> Before anyone asks , no this isn't some school project I'm working on. Its
> a simple problem I can't figure out for myself.
>
> If you have a rocket motor giving a constant thrust then the rocket - not
> accounting for weight change due to fuel use - will accelerate at a constant
> rate. However if say its accelerating at 1m/s/s then it will take twice as
> long to accelerate from 2 m/s to 4 m/s as it did to get from 1 m/s to 2 m/s.
> Yet going by 0.5 mv^2 the kinetic energy for a rocket weighing 1 kilo at the
> given speeds is:
>
> 1 m/s = 0.5J
> 2 m/s = 2J
> 4 m/s = 8J
>
> So in the first second it gained 1.5J but in the next 2 seconds it gained
> 6J. I don't get it? The motor wasn't putting out any extra energy so where
> did it come from?
>
> B2003

The energy of the fuel becomes the mass of the accelerating ship.

Mitch Raemsch
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