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Consider a rocket undergoing constant acceleration in a rectilinear path. Assuming the rocket is ideal (ignoring fuel mass and energy loss such as sound, radiation, vibration, etc.) a portion of the expended energy increases the rocket's kinetic energy, while the remainder becomes the kinetic energy of the exhaust.

Now, let us assume that the rocket has the ability to always redirect its thrust to be perpendicular to the direction of motion. The path of the rocket is now circular, with a constant speed.

Given that the speed is now constant, the rocket's kinetic energy does not increase. Accordingly, must not the exhaust now obtain the entirety of the spent energy?

From the rocket's frame of reference, would the exhaust gases be perceived as exiting at a greater velocity than during rectilinear acceleration, even though the throttle setting of the engine has not been increased? If so, from a mechanical perspective, what would cause this?

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This is an interesting question, because when it is moving in a circle, the magnitude of its tangential velocity is constant and its angular velocity is also constant. Therefore the total kinetic energy of the rocket is constant. So where is all that energy required to produce the thrust going? The answer is that it is going into the ever increasing angular kinetic energy of the exhaust plume. The angular kinetic energy of a particle with mass m is given by: $$1/2 \ I \omega^2,$$
where $\omega$ is the angular velocity and I is the the moment of inertia of the particle. For a particle at radius r the the moment of inertia is $I = m r^2$ so the above equation can be written as: $$1/2 \ m r^2 \omega^2.$$

For the exhaust particles spiralling outwards from the rocket, the radius is always increasing and as the rocket puts out a continuous stream of particles the the total mass of the exhaust plume is always increasing, so it follows that the kinetic energy of the exhaust gases is always increasing.

When a rocket is moving in a straight line with instantaneous velocity v relative to observer S, you could accelerate the exhaust particles to -v relative to the rocket so that the trail of exhaust particles appears to be stationary from the point of view of observer S, so that the final kinetic energy of the exhaust particles is zero. This is not possible when the rocket is moving in a circle. This is the main difference between the two cases.

Does a rocket moving in a circle expel exhaust at a greater velocity?

In conclusion, while the velocity of the exhaust leaves the rocket at the same velocity in the rest frame of the rocket in both the linear case and the circular case, the velocity of the exhaust relative to an external inertial observer is different in the two cases.

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    $\begingroup$ The exhaust particles don’t spiral out in this scenario. There is a spiral shaped plume of particles that travel radially out. $\endgroup$
    – Dale
    Commented Jul 13 at 16:18
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    $\begingroup$ @Dale I agree. Perhaps I should of made that clear. its more of an expanding spiral. $\endgroup$
    – KDP
    Commented Jul 13 at 16:28
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    $\begingroup$ P.S. It is not precisely radial in the reference frame of a non-rotating inertial observer, because she sees the exhaust particles leave the rocket with a radial component and a tangential component equal to the instantaneous tangential velocity of the rocket. To a co-rotating observer the particles initially depart radially then progressively fall behind the "line of sight". $\endgroup$
    – KDP
    Commented Jul 13 at 16:45
  • $\begingroup$ Why does it have to be angular kinetic energy? why not just their kinetic energy? $\endgroup$
    – user20574
    Commented Jul 14 at 17:46
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The energy of the fuel does become the kinetic energy of the exhaust.

Let us ignore the fact that the rocket becomes lighter as it burns fuel, and that it will eventually run out of fuel. The rocket motor burns fuel at a constant rate. The exhaust comes out at a constant speed. From an inertial frame, in the absence of other forces, the exhaust will keep its speed forever, flying radially outward.

Each second the rocket runs, it burns some amount of fuel and produces some amount of exhaust. The increase in kinetic energy of the exhaust does not come from increasing the velocity of exhaust that has already left. It comes from giving kinetic energy to new exhaust.

From the rocket's frame, the rocket and fuel are at rest. As the rocket burns, the exhaust is ejected at a constant speed.

In an inertial frame, the rocket goes in circles at a constant speed. The fuel is ejected perpendicular to this circle at a constant speed.

For a rocket accelerating in a straight line, the rocket and fuel are still at rest. The exhaust is still ejected at the same constant speed.

From an inertial frame, the rocket and fuel are traveling together at the same speed. The exhaust is ejected at the same speed relative to the rocket. This speed changes with time.

The total change in kinetic energy is the same for the linear path. But some goes into accelerating the rocket and some goes into accelerating the fuel. This is most easily seen from an inertial frame.

Note that the fuel is moving forward and has kinetic energy. This becomes exhaust. When the rocket is moving slowly, the exhaust moves backward. If the rocket is moving fast enough, the exhaust moves forward slower than the rocket.

From the rocket's frame, the force and energy calculations are more complex. Accelerated frames always have a fictitious force. Fictitious is an unfortunate name that makes people think "imaginary." See Trying to gain better intuition about fictitious forces for more about fictitious forces.

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  • $\begingroup$ In the case of the linear path, an external inertial frame observer would see energy split between the rocket and the exhaust. In the case of the circular path, an external inertial frame observer would see the entirety of the energy go to the exhaust, since the speed of the rocket is constant. Does this not imply that the observer sees the exhaust exit at a greater speed relative to the rocket, than in the straight line example? $\endgroup$
    – Trever Thompson
    Commented Jul 13 at 14:14
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    $\begingroup$ No. In all cases, the motor pushes out exhaust the same way, at the same speed. The speed of the exhaust relative to the inertial observer changes as the rocket speeds up. $\endgroup$
    – mmesser314
    Commented Jul 13 at 14:52
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Energy is frame dependent.

In a inertial frame that is momentarily comoving with the rocket, its kinetic energy is zero, and the kinetic energy of the gas is the same for both cases, if we suppose the same type and adjusting of the engine.

But as time passes for the same inertial frame, the kinetic energy of the rocket will increase in the first case. The kinetic energy of the gas will decrease until zero, and increase again.

In the second case, for the inertial frame of the centre of the circle, the kinetic energy of the rocket is constant. The kinetic energy of the gas is also constant and bigger than in the comoving frame.

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Accordingly, must not the exhaust now obtain the entirety of the spent energy?

Yes.

From the rocket's frame of reference, would the exhaust gases be perceived as exiting at a greater velocity than during rectilinear acceleration, even though the throttle setting of the engine has not been increased? If so, from a mechanical perspective, what would cause this?

Rather than the rocket's frame, let's use the frame where the center of revolution is at rest.

In the inertial frame watching rectilinear acceleration, the rocket (and the fuel) are gaining KE and the instantaneous exhaust (at least until the rocket is moving at $v_e$) is losing KE.

In the inertial frame watching the revolving rocket, the rocket and fuel have a constant KE and the exhaust has a constant KE.

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