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From: Uncle Ben on 26 May 2010 12:44
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.
>
> Jim Deutch (JimboCat)
> --
> "the content of physics doesn't need the language of causality. I
> think it is analogous to the way evolutionary biologists talk about
> teleological arguments. Teleology is just shorthand for the way
> natural selection works. Causality is just shorthand for the way
> coupled systems behave." [Bernard B. Beard]- Hide quoted text -
>
> - Show quoted text -

You say the rocket is continuously changing frames. This shows that
you do not grasp the meaning of a frame of reference.

The rocket is continuously changeing speeds. That is physics at
work. But the choice of a set of coordinates by which to describe the
process is up the physicist, not the physics.

Uncle Ben
From: Sam Wormley on 26 May 2010 13:13
On 5/26/10 9:16 AM, boltar2003(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.

That's very much like an object falling in in constant gravitation.

Potential Energy + Kinetic Energy = Constant

Since acceleration, dv/dt is not zero, velocity is increasing
and kinetic energy, mv^2/2 is increasing as the square of the
velocity.


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
>
>

From: Igor on 26 May 2010 13:30
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?
>
> B2003

I'll give you a hint. It has something to do with a quantity called
power.

From: Timo Nieminen on 27 May 2010 01:25
On Wed, 26 May 2010 boltar2003(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?

OK, the rocket motor is putting out constant power (as you'd expect if
it;s burning the propellant at a constant rate). Where does this energy
go? Some of the energy goes into the rocket. But not all of it. The
exhaust goes backwards, and this has energy as well.

The rate at which the kinetic energy of the rocket is increasing is Fv,
where F is the thrust and v is the current speed of the rocket, as per the
usual power P = Fv. Meanwhile, if the exhaust speed relative to the rocket
is V, then the speed of the exhaust, relative to whatever we measure the
rocker to be moving at v, is V-v. As v increases, the rocket gains a
larger part of the total energy output, and the exhaust a smaller part.

If the mass lost per second in the exhaust is m, the force is F = mV. If v
is much smaller than V, the KE put into the exhaust per second is then

KE_e = (1/2) m(V-v)^2 = approx (1/2) mV^2 - mVv

and the KE put into the rocket per second is

KE_r = P = Fv = mVv.

The total power output, KE_e + KE_r, is constant.


From: boltar2003 on 27 May 2010 04:55
On Thu, 27 May 2010 15:25:27 +1000
Timo Nieminen <timo(a)physics.uq.edu.au> wrote:
>The rate at which the kinetic energy of the rocket is increasing is Fv,
>where F is the thrust and v is the current speed of the rocket, as per the
>usual power P = Fv. Meanwhile, if the exhaust speed relative to the rocket
>is V, then the speed of the exhaust, relative to whatever we measure the
>rocker to be moving at v, is V-v. As v increases, the rocket gains a
>larger part of the total energy output, and the exhaust a smaller part.
>
>If the mass lost per second in the exhaust is m, the force is F = mV. If v
>is much smaller than V, the KE put into the exhaust per second is then
>
>KE_e = (1/2) m(V-v)^2 = approx (1/2) mV^2 - mVv
>
>and the KE put into the rocket per second is
>
>KE_r = P = Fv = mVv.
>
>The total power output, KE_e + KE_r, is constant.

Ok , that makes sense. But what happens when you get to the point where v
is as close to zero as makes no difference and so the rocket is getting most
of the energy being created? Would the rocket continue accelerating at the
same rate?

Or I suppose a better example would be if there was no exhaust and the
spaceship was simply reacting against a magnetic field using a constant
power electromagnet. What happens once the power being supplied to the
electromagnet is less than the gain in kinetic energy required to continue
accelerating at the current rate?

B2003

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