Is any energy required to deflect an asteroid, with force always perpendicular to its trajectory? Let's assume there is an asteroid traveling on a straight line (far from any gravitational source), and we need to deflect it from its actual trajectory, so we build a rocket motor on the surface and we make all arrangements in order to generate a continuous push force always perpendicular to its trajectory (obviously this is a heavily simplified case). Given this, we start the rocket motor and its force makes the asteroid to deflect in a big arc.
I am pretty sure the rocket will need energy proportional to the asteroid mass to make this happened. If this is true, the question is this: Why do we say the centripetal force does not make any work on a circular uniform motion? If we check this case, the effect of the rocket motor is to put a centripetal force over a mass, so results should be exactly be the same.
UPDATE:
This is an attempt to calculate this "energy" (maybe it should have another name):
Does the centripetal force do some work?
 A: (Summary: In this post I argue that you need at least an energy of $m v_1^2(1-\cos(\theta))$ in the idealized kinematics situation to deflect an asteroid of mass $m$ and velocity $v_1$ by an angle $\theta$ using rockets, without changing the magnitude of $v_1$.)
Energy isn't the be-all and end-all of motion. The problem is that momentum also has to be conserved. 
Let's say that our rocket is attached to our asteroid. So, the initial energy of the asteroid+Rocket fuel is $\frac{1}{2} (m+M) v_1^2$. But we're accelerating the asteroid with a reaction mass, so if the momentum of the fuel/rocket/asteroid system is $(m+M)\mathbf{v}_1$, then after some amount of time of the rocket firing (let's say all of M is expelled), we have to have: $m\mathbf{v}_1'+M\mathbf{v}_2 =(m+M)\mathbf{v}_1$ where $\mathbf{v}_2$ is the mass-averaged velocity of the gas expelled by the rocket. This system then has energy $\frac{1}{2}m v_1^2+\frac{1}{2}Mv_2^2$ (where mechanical energy isn't conserved, since we have fuel exploding, and we've assumed that $v_1'^2=v_1^2$ - we just change the direction of the velocity). So the energy we've had to give the system is $\frac{1}{2}m v_1^2+\frac{1}{2}Mv_2^2-\frac{1}{2} (m+M) v_1^2=\frac{1}{2}Mv_2^2-\frac{1}{2}M v_1^2$. Doing some algebra: $\mathbf{v}_2=\frac{(m+M)}{M}\mathbf{v}_1-\frac{m}{M}\mathbf{v}_1'$. Plugging in, and ignoring the factor of $\frac{1}{2}$, the energy is proportional to:$$M\left(\frac{m+M}{M}\right)^2 v_1^2+M\left(\frac{m}{M}\right)^2 v_1'^2-2 M \frac{m+M}{M}\frac{m}{M}\mathbb{v}_1\cdot \mathbb{v}_1'-M v_1^2$$
$$=v_1^2\left(\frac{m^2+M^2+2mM+m^2+(-2m^2-2Mm )\cos(\theta)-M^2}{M} \right)$$
$$=v_1^2\left(\frac{2m^2+2mM-(2m^2+2Mm )\cos(\theta)}{M} \right)$$
$$=v_1^22\frac{m^2+mM}{M}(1-\cos(\theta))$$
As $M\to\infty$ our system gets more efficient, but we never get close to zero energy! We're always expending a bit more than this amount of energy: 
$$m v_1^2(1-\cos(\theta))$$
Maybe that relationship can be derived through simpler means. It makes sense.
So yes, through using a reaction mass, it takes energy to deflect an asteroid. 
This does not imply that the sun expends energy to deflect the planets, because we explicitly assumed that we were using rockets.
To summarize:


*

*No work is done on the asteroid. 

*A minimum of $mv^2_1(1−cos(θ))$ is done on the rocket fuel.

*This result is due to conservation of momentum, meaning...

*If momentum isn't conserved (say we're modeling the sun as a fixed point w/ a 1r potential), then this result won't hold and it might not take energy to deflect the asteroid. THIS is the sense in which it doesn't take energy to deflect an asteroid, but it breaks momentum conservation.

A: 
Why do we say the centripetal force does not make any work on a
  circular uniform motion?

In your case, the rocket does not do any work...on the asteroid.  As others have noted, no such guarantee is provided for the accelerated propellant spewing out of the back end of the rocket.  
It is possible to apply a centripetal force to an object without expending any energy, but it is not guaranteed that just because you are in the process of applying such a force that no energy is transformed or transferred elsewhere in the system.
A: As long as the asteroid stays in a circle and has a constant speed, the centripetal forces does not do any work. But if you increase the centripetal force, the asteroid will no longer stays in the circle, in fact it will fall. This is equivalent to say that the asteroid develops a velocity in the direction of the centripetal force. So in this case, you do do work on the asteroid. 
A: After asking lots of people, finally the right answer arrived to my inbox from Henry Reich, the creator of MinutePhysics. We can summarize the answer on:


*

*You can use lots of energy to deflect an asteroid, but

*The work done is zero because of the definition of work. Work="change of energy"


According to Henry:

All physicists mean by “work” is the change of total energy of an
  object over time.

That is, no matter how much energy has been expended by making a circle, if the speed of the asteroid does not change, the energy does not change, and since work = energy, no work has been done.
The problem, he said, is in the use of the word "work":

This is why I hate the use of the word “work” by physicists. If I could abolish that word forever and just have everyone say “change of energy” then I think all sorts of confusion would be avoided, because the word “work” has many colloquial meanings, many of which are closer to “power” or “force” or other things than “change of energy”.

Details can be seen in The centripetal force work problem – solved!
Thanks to every one who answered this question, I wouldn't arrive to the right answer without you all.
