3
$\begingroup$

The spacecraft with the same physical properties is moving in space. Let's assume that there are no resistance in space.

  1. A stationary observer observes a spacecraft whose speed varies from 100 m/s to 120 m/s.

  2. An observer moving at 100 m/s in the opposite direction to the spacecraft observes the situation 1. (The relative speed of the spacecraft to the observer will change from 200 m/s to 220 m/s.)

  3. A stationary observer observes a spacecraft whose speed varies from 200 m/s to 220 m/s.

It is intuitive to see that the fuel consumption of the two situations is the same, as number 1 and number 2 observe the same situation only by varying the observer's condition.

The problem is number 3

The velocity changes are the same, but the increase in kinetic energy is greater by 3 than by 1 and 2.

If the initial velocity is higher, if the velocity change (always increasing) is the same, the kinetic energy change is always greater. Fuel energy is converted to kinetic energy, so it seems fuel consumption will increase as the initial speed increases.

But I've searched for similar questions(Comparison of the spacecraft's fuel consumption under the following circumstances) and found that fuel consumption has little to do with the initial speed, but with the change in speed.

If so, does fuel consumption only affect speed changes in the case of automobiles in a situation where there is no energy loss due to air resistance or frictional force? I need to explain this in more detail. (Why we can get more kinetic energy with the same amount of fuel)

$\endgroup$
1

3 Answers 3

7
$\begingroup$

If so, does fuel consumption only affect speed changes in the case of automobiles in a situation where there is no energy loss due to air resistance or frictional force?

Exactly the same problem can be observed in case of automobile moving on the surface of the Earth. Suppose, a car of mass $m$ is moving at speed $v$. Then it burns some fuel and increases its speed to $2v$. Then the change of kinetic energy is $$ \frac{m(2v)^2}2 - \frac{mv^2}2 = \frac32mv^2 $$ However, if you consider this situation from a point of view of another car, moving alongside the first one with the speed $v$, then the initial speed of the car is zero, and the final speed is just $v$, so the change of the kinetic energy is $$ \frac{mv^2}2 - 0 = \frac12mv^2 $$ Thus, from this point of view the burned fuel produces 3 times less energy increase. What is going on?

You have forgotten the kinetic energy of the Earth. The burned fuel gives its energy to the whole system, which is not an isolated car, but a system of the car and the Earth. If we treat this system as a closed one, then its momentum should be conserved. When the car accelerates from speed $v$ to speed $2v$ (or, in the second case, from speed 0 to speed $v$), its momentum increases by $mv$. This change of the car's momentum is compensated by the change of the momentum of the Earth. To compensate the forward momentum of the car $mv$ the Earth gets an additional backward speed $u$: $$ Mu = mv,\qquad u=\frac mMv, $$ where $M$ is the mass of the Earth.

So, in the first case the initial speed of the car was $v$, the initial speed of the Earth was 0. The final speed of the car is $2v$, the final speed of the Earth is $-u$. The change of the kinetic energy is $$ \frac{m(2v)^2}2 + \frac{Mu^2}2 - \left(\frac{mv^2}2 + 0\right) = \frac{3mv^2}2 + \frac{M\left(\frac{mM}v\right)^2}2=\frac{3mv^2}2+\frac{m^2v^2}{2M} $$ Since the mass of the Earth is much larger than the mass of the car, $M\gg m$, the second term is almost zero and we get our old answer, $\displaystyle\frac{3mv^2}2$.

However, in the second case the situation is quite different. The initial speed of the car is zero, but the initial speed of the Earth is $-v$. The final speed of the car is $v$, the final speed of the Earth is $-(v+u)$. Thus, the change of the kinetic energy is $$ \frac{mv^2}2 + \frac{M(v+u)^2}2 - \left(0 + \frac{Mv^2}2\right) = \frac{mv^2}2 + \frac{M(v^2+2vu+u^2)}2-\frac{Mv^2}2= \frac{mv^2}2 + \frac{M(2vu+u^2)}2 $$ Now we substitute $\displaystyle u=\frac mMv$ in the second term: $$ \frac{mv^2}2 + \frac{M(2v\frac mM v+\left(\frac mMv\right)^2)}2 = \frac{mv^2}2 + \frac{2mv^2}2+ \frac{m^2v^2}{2M} $$ This is exactly the same answer which we got in the first case. The last term is again vanishing in limit $M\gg m$, so the simple answer is again $\displaystyle\frac{3mv^2}2$.

Surprisingly, in the second case a substantial part of the energy of the burned fuel is used to increase the kinetic energy of the Earth, not the car. We consume two times more energy to accelerate the Earth than the car itself!

The same reasoning is valid for your original question. You have forgotten about the kinetic energy of the jet. In order to accelerate the rocket you burn some fuel, its energy is used to accelerate both the rocket and the jet. If you carefully calculate the change of the kinetic energy of the rocket and the jet in both reference frames, you'll find no discrepancies.

$\endgroup$
4
$\begingroup$

Short answer

A short answer is, that this energy excess comes from the decrease in kinetic energy of the fuel itself. Kinetic energy is not absolute, but depends on the frame of reference. And since it is not a linear function of velocity, so does its difference. Chemical energy stored in the fuel does not depend on the frame of reference, as long as there are no relativistic effects included.

Longer answer

Let us assume that a rocket with mass $M$ burns a small mass of fuel $m \ll M$, which converts $Q$ energy from chemical energy to kinetic energy of the gases, which are then directed backwards with relative speed $w$. In response, the rocket accelerates from $v_1$ to very slightly larger $v_2$. The moving observer has velocity $u$ with respect to the stationary frame. Also to avoid problems with signs, let us assume that the rocket already has some speed, and that $v_1 > w$.

From the law of conservation of momentum, we immediately see that $M(v_2 - v_1) = mw$.

From the law of conservation of energy (and neglecting all thermodynamical losses, which are not important here) we know that $Q = \frac{mw^2}{2}$.

  1. case: The kinetic energy of the rocket increases, and the change is $$\frac{Mv_2^2}{2} - \frac{Mv_1^2}{2} = \frac{M(v_2^2 - v_1^2)}{2}.$$ However, not all of this energy comes from the chemical energy of the fuel. There is also some kinetic energy in the expended fuel, that was originally carried within the rocket. It decreased by $$\frac{m(v_1 - w)^2}{2} - \frac{mv_1^2}{2} = -mv_1w + \frac{mw^2}{2}.$$

  2. case: In this frame the kinetic energy increased much more, $$ \frac{M(v_2+u)^2}{2} - \frac{M(v_1+u)^2}{2} = \frac{M(v_2^2 + 2uv_2 + u^2 - v_1^2 - 2uv_1 - u^2)}{2} \\ = \frac{M(v_2^2 - v_1^2)}{2} + Muv_2 - Muv_1,$$ which is more by $Muv_2 - Muv_1$ compared to the previous case.

    At the same time, the kinetic energy of the burnt fuel decreased, $$ \frac{m(v_1 + u)^2}{2} - \frac{m(v_1 + u - w)^2}{2} = -mv_1w - muw + \frac{mw^2}{2}. $$ which is by $muw$ more.

    The key observation here is that in order to conserve energy, these differences should be equal: $$ Mu(v_2 - v_1) = muw, $$ which if of course true, because after cancelling $u$ we obtain $M(v_2 - v_1) = mw$, or the law of conservation of momentum. Since both momentum and energy are conserved, the paradox disappears.

  3. case: finally, this is mathematically exactly the same, except that $u$ is now interpreted as an addition to the speed of the rocket, and the observer is stationary.

$\endgroup$
3
$\begingroup$

You're missing the kinetic energy stored in the fuel itself. When a rocket accelerates from 100m/s to 120m/s, it uses some quantity of fuel. Different observers in different frames may say the rocket is moving at different speeds, perhaps 50m/s or 200m/s or anything else depending on reference frame, but they'll all agree that the rocket accelerated by 20m/s and that some fixed amount of fuel was used. The chemical energy in the fuel is constant, but the kinetic energy change from the 20m/s change depends on the observer's velocity. How can this be?

The key is to realize that the fuel carries both chemical energy and kinetic energy. An observer that sees the rocket moving at a high speed assigns a greater KE change to the rocket after acceleration - this "extra" energy doesn't come from the chemical energy of the fuel, which is constant, but from the kinetic energy of the fuel, which is frame-dependent.

Observers in different reference frames may assign different energy changes to the same velocity change. That difference is perfectly accounted for by the differences in the initial energy content in different reference frames. Two observers who see a rocket moving at different speeds see that the fuel tank carries the same amount of chemical energy, but different amounts of kinetic energy. Burning the tank yields different amounts of energy in different frames.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.