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?

  • $\begingroup$ To deflect it, applying the force perpendicularly is the most efficient. Anything not perpendicular will put some energy into speeding up or slowing down the asteroid. You're right that if you continually applied a constant perpendicular force you would deflect the asteroid into a giant circular path. $\endgroup$ Oct 24, 2014 at 15:37
  • $\begingroup$ @BrandonEnright - That is backwards. The best approach, if you have enough time is to first directly along or against the velocity vector. $\endgroup$ Oct 24, 2014 at 17:02
  • $\begingroup$ @DavidHammen, yes, you could make some work against the velocity vector. However if mass is big, that would costs lots of energy. As BrandonEnright says, perpendicular is most efficient. $\endgroup$ Oct 24, 2014 at 17:21
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    $\begingroup$ Absolutely not. Thrusting along or against the velocity vector changes the orbital period, and the asteroids velocity helps you, particularly at perihelion. Thrusting normal to the velocity vector is a non-strategy. You only use it when its too late to change the period, and by then, it's too late period. $\endgroup$ Oct 24, 2014 at 17:28
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    $\begingroup$ @DavidHammen the Oberth effect is irrelevant. In the isolated rocket + asteroid inertial frame of reference the rocket will have the same efficiency any direction it is fired. The Oberth effect only matters when the rocket / object is affected by gravity. $\endgroup$ Oct 24, 2014 at 21:40

4 Answers 4


(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.
  • $\begingroup$ I have not checked the detailed math, however when you say "it requires energy" you are right (my equations -not exposed here- shows that too). However, the big question here is: Why we need energy if what we have here is a uniform circular motion, supposed to require not energy to keep the mass moving on a circle?. As you can see, if we stop the rocket, te big arc ends and we have a straigh line again! By the way, congratulations, you are on the right track. $\endgroup$ Oct 25, 2014 at 17:53
  • $\begingroup$ @stalinbeltran I was satisfied to stop my investigation here: No work is done on the asteroid. A minimum of $mv_1^2(1-\cos(\theta))$ is done on the rocket fuel. This result is due to conservation of momentum. If momentum isn't conserved (say we're modeling the sun as a fixed point w/ a $\frac{1}{r}$ potential), then this result won't hold and it might not take energy to deflect the asteroid. $\endgroup$
    – user12029
    Oct 26, 2014 at 8:31
  • $\begingroup$ Remember we have no gravity here. Just mass and speed. $\endgroup$ Oct 27, 2014 at 15:48
  • $\begingroup$ @stalinbeltran Hence the "if". $\endgroup$
    – user12029
    Oct 27, 2014 at 20:34

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.

  • $\begingroup$ If I understand you right, that means that energy is not required to make the asteroid to follow the circular path. However, if you stop the rocket, the asteroid follow a straight line. That makes me think the energy is required to deflect the asteroid, which sounds unusual for circular motion. $\endgroup$ Oct 24, 2014 at 19:35
  • $\begingroup$ Energy is required to accelerate the propellant of the rocket. The confusion arises because there are two effects going on - the asteroid is turning, which requires no energy, and the propellant is being accelerated, which does require energy. If you choose a different method of applying the force, like using a tether to another object, then you will find there is no need to expend energy. $\endgroup$
    – Brionius
    Oct 24, 2014 at 19:53
  • $\begingroup$ that sounds good. However, consider the asteroid frame of reference. What asteroid 'see' is that there is a rocket pushing it in some direction, and that force is continuous. So force applied over a distance give us energy. So what makes the circular orbit is energy, because without that energy there is not force possible. The 'tether' is an option, but that will means there are two kinds of circles: ones that require energy to exist, and ones that do not require any energy. $\endgroup$ Oct 24, 2014 at 20:59
  • $\begingroup$ First of all, you can't expect your assumptions to all work out if you're switching to a non-inertial reference frame like the asteroid's. If you're in the asteroid's reference frame, then the asteroid never moves - not sure how you think work is being done (of course a specious argument to counter a specious argument; we need to account for the acceleration of the reference frame). Second, no energy is expended TO ROTATE THE ASTEROID. Energy is expended to accelerate the fuel. If you hold an object still in your hands, no work is done, but if you sing while doing so, you're using energy. $\endgroup$
    – Brionius
    Nov 9, 2014 at 23:02
  • $\begingroup$ Basically the "no work is required to exert a centripetal force" principal isn't a magic phrase that exempts you from the laws of physics. If you pick a foolish method of exerting that force like a chemical rocket, you're going to use up extra energy. That doesn't change the fact that no work was done on the asteroid. $\endgroup$
    – Brionius
    Nov 9, 2014 at 23:04

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.

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    $\begingroup$ That is what we learn from physics. However, the rocket will stay pushing the asteroid, and so using energy. Where does this energy go? Is it work going nowhere? If so, why when the rocket stops the asteroid stops traveling on a circle? It sounds like the rocket energy is what make it happends, which has non sense for circular motion. $\endgroup$ Oct 24, 2014 at 16:57
  • $\begingroup$ This answer does not address the question. $\endgroup$
    – garyp
    Oct 24, 2014 at 17:52
  • $\begingroup$ Let me explain my post in detail. As you know, if the net force provides just enough acceleration, the asteroid is moving in a circle uniformly, that is, the net force of the asteroid is given by $mv^2/r$. But what if you increase the net force (assuming it still points to the center of the orbit)? You provide too much force, so the asteroid will fall to the center and therefore, the net force does work. $\endgroup$ Oct 24, 2014 at 20:32
  • $\begingroup$ Another way to think of it is to imagine after increasing the net force, the asteroid will still move in a circle, but since $F=mv^2/r$ and the speed will not change assuming the net force does no work, then $r$ decreases to compensate the increase in net force. This implies that the asteroid falls and the net force does work. $\endgroup$ Oct 24, 2014 at 20:35

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:

  1. You can use lots of energy to deflect an asteroid, but
  2. 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.


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