Where does the energy go in a rocket when no work is done? While playing Kerbal Space Program, I wondered where my chemical energy would go when fired at 90° to the motion. It would do no work on the rocket, but all that energy has to go somewhere, right? Anyway, my question is, where does the energy go?  
 A: Let's say that the rocket is traveling in the $y$-direction at some velocity $v$, which may or may not be non-zero. A force - in this case, thrust provided by the engine - is applied perpendicular to the direction of motion, in the $x$-direction. This force produces an acceleration, which causes the rocket to move. Therefore, the force is not applied at a 90 degree angle to the motion of the rocket, and a non-zero amount of work is done.
Case 1: Force applied continuously in the $x$-direction
Here's how the angle between the motion and the applied force changes over time:


As alexgotsis noted (answer since deleted), the force will give the rocket angular momentum if the engine does not fire precisely against the center of mass of the rocket. I really should have drawn the picture better.
Let's do some math.
The initial velocity can be broken up into components $v_{x_0}$ and $v_{y_0}$. Given that there is zero force applied in the $y$-direction, we can conclude that $v_x=v_{x_0}$ for all times $t$.
Assuming a constant force $F=F_x$, we have some acceleration, $a=a_x$. We can calculate the velocity, $v_x$, at any time $t$ by using
$$v_{x_f}=v_{x_0}+a_xt$$
The angle between the direction of motion and the force is $90+\theta$. Here,
$$\theta=\tan^{-1}\left(\frac{v_y}{v_x}\right)=\tan^{-1}\left(\frac{v_y}{v_{x_0}+a_xt}\right)$$
The work done at any given time is
$$W=\int F\cdot ds\cos(90+\theta)=\int F\cdot ds\cos\left(90+\tan^{-1}\left(\frac{v_y}{v_{x_0}+a_xt}\right)\right)$$
To find $ds$, you need to figure out how far the rocket has traveled.
Case 2: Force applied continuously at some angle to the direction of motion of the rocket.
This all holds if the rocket is continuously fired in the $x$-direction. If it is fired at some angle to the rocket's motion at any given time, then you have different motion:


Here, the force happens to be applied at an angle near 90 degrees, but not quite.
The motion now is more complex, because the acceleration vector is dependent on $\theta$.
So where does the energy go? Into the rocket and its exhaust, as kinetic energy.
Case 3: Force applied continuously perpendicular to the direction of motion of the rocket
What if we fix the engine so it always thrusts perpendicular the direction of the rocket's motion? We get uniform circular motion:

In this case, all the energy goes into the exhaust of the rocket. The rocket itself maintains uniform kinetic energy, so long as it has uniform speed. This is the case I think you were talking about.
A: Very little of the energy from a rocket engine ever goes to the kinetic energy of the rocket. The only way you get perfect conversion to KE of the rocket is when the propellant is directed in the opposite direction of motion and when the ejection velocity is exactly equal to the speed of the rocket. In that case, the propellant winds up containing 0 kinetic energy, thus all the kinetic energy liberated in the process goes into the rocket.
If the propellant is fired perpendicular to the direction of motion, then the rocket sees 0 change in its own kinetic energy. For the record, this only applies to one frame of reference (probably that of the nearby planet).
In this scenario that you have described, all of the change in kinetic energy liberated by the rocket engine goes into the propellant.
Of course, the total energy of the reaction is much more, and a great deal of that goes to heat.
A: If a train is moving along the tracks at 60 mph and there's a crosswind at 30 mph the crosswind, in fact, does not do any work on the train. That's because the train's motion is constrained by the rails so that it cannot move in the direction that the wind is pushing it. If you remove the constraint its motion will change because the wind is doing work on it. That's why, for example, if you row a boat straight across a river you end up downstream from where you started: the current moved you downstream. In order to end up directly across from where you started you have to aim the boat upstream, to compensate for the current. Either way, the current is doing work on the boat. Same thing for your rocket.
