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Say a spaceship is traveling though space in a uniform circular motion.

It's not orbiting any planet, it just flies in circles in an empty space.

The only force working on the spaceship would be the centripetal force caused by the ship's engine.

Thus, the work would be $0$, as the force would always be perpendicular to the ship's path.

But that sounds counterintuitive to me, it would seem that the spaceship must do some work, otherwise it would just float in a straight line.

Can anyone point out the error in my reasoning?

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  • $\begingroup$ When doing thought experiments with rockets, it helps me to remember that the center of mass of the complete system (the rocket and everything that used to be part of the rocket) never accelerates. Whatever the rocket does is an expenditure of stored internal energy to increase the kinetic energy of the system by shooting bits of itself away in different directions such that momentum (0 in the initial rest frame of the rocket) is conserved. $\endgroup$
    – g s
    Mar 18, 2022 at 16:47
  • $\begingroup$ "Whatever the rocket does is an expenditure of stored internal energy to increase the kinetic energy of the system by shooting bits of itself away in different directions." But doesn't this statement apply to the "straight-line" movement as well? Would that mean that if the spaceship is flying in a straight line, the total work of the whole system is also zero? $\endgroup$
    – nietup
    Mar 18, 2022 at 18:30
  • $\begingroup$ Mechanical work is a scalar: the total amount of system internal energy converted into system kinetic energy or vice versa. When we talk about positive and negative work, we aren't talking about the direction in which work is done, but whether we're increasing system internal energy while decreasing system kinetic energy, or decreasing system internal energy while increasing system kinetic energy. $\endgroup$
    – g s
    Mar 18, 2022 at 19:15

4 Answers 4

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You are right in saying that the centripetal force doesn't do any work, in fact the kinetic energy of the system doesn't increase as the absolute value of the velocity $|\vec{v}|$ stays constant.

I guess what you find counter intuitive is that the spaceship has to burn some fuel to keep rotating, so where does this energy go? Simply it is in the fuel.

To keep the rotation, the spaceship will need to keep ejecting mass, in particular if the centripetal force you need is $\vec{F}$, then from Newton's law, every $dt$ you need the change in momentum $\vec{F}dt= md\vec{v}$, where I neglected the loss of mass of the spaceship for simplicity. Since the system is isolated the momentum has to be conserved, therefore for every $dt$ you need to eject some fuel carrying that much momentum (with vector pointing outwards). This ejected fuel will also carry the energy we were looking for.

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  • $\begingroup$ "with vector pointing outwards" the vector is perpendicular to the radius and calculated that way by definition of force work done is zero. The diminution of mass cannot be ignored, the force F has to change magnitude to keep the radius constant as I discus in my answer, $\endgroup$
    – anna v
    Mar 18, 2022 at 7:38
  • $\begingroup$ Why should the vector pointing outwards be perpendicular to the radius? Regarding the diminution of mass (if I got what you meant) the radius would change if we don't account for the change of mass of the spaceship and there would be work done on the spaceship, and I agree, I just neglected this effect for the sake of simplicity as the point of the answer was "finding the lost energy". $\endgroup$
    – Andrea
    Mar 18, 2022 at 16:01
  • $\begingroup$ I read the question as "why is the work zero", from definition of work. In a circular path , the dp/dt keeping it there should be tangent to the circle. If it were in the direction of the radius there would not be a circular path, imo. $\endgroup$
    – anna v
    Mar 18, 2022 at 17:25
  • $\begingroup$ The momentum of the spaceship $\vec{p}$ is tangent to the trajectory but the differential $d\vec{p}$ has to point towards the center of rotation or the object would keep going straight and it wouldn't be circular motion. Maybe I didn't get what you mean. $\endgroup$
    – Andrea
    Mar 18, 2022 at 18:42
  • $\begingroup$ I am trying to understand. I visualize the satellite as a straight line rocket tangent to the trajectory. A burst of ions goes off from its tail, tangent to the trajectory, it is carrying a dp/dt. How can this be perpendicular to the trajectory? I think it is the incremental change. in the mass that changes F and it is important not to be ignored for simplicity. $\endgroup$
    – anna v
    Mar 19, 2022 at 4:38
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Without a force the spaceship would be floating in straight line at constant velocity. The reason why work is zero and the object is accelerated comes from its very definition.

From the second Newton's law:

$$\mathbf F = m\frac{d\mathbf v}{dt}$$

Making a dot product with an infinitesimal displacement:

$$\mathbf {F.dr} = m\frac{d\mathbf v}{dt}\mathbf {.dr} = m\mathbf {v.}d\mathbf v = d\left(\frac{1}{2}m\mathbf {v.v}\right) = d\left(\frac{1}{2}m|v|^2\right)$$

Because $\mathbf {F.dr} = dw$ by definition, it is zero if there is no change in the modulus of the velocity, even if the velocity vector change.

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This is a continuation of Andrea's and Claudio's answers:

From this link

Work refers to an activity involving a force and movement in the direction of the force. A force of 20 newtons pushing an object 5 meters in the direction of the force does 100 joules of work.

Energy is the capacity for doing work. You must have energy to accomplish work - it is like the "currency" for performing work. To do 100 joules of work, you must expend 100 joules of energy.

italics mine

Energy is a conserved quantity, work is not a conserved quantity, as its definition relies on the vector direction of the force, as Claudio states.

So energy is conserved as discussed in the answer by Andrea, but work in this particular problem is zero if the loss of mass is ignored.

If the loss of mass is not ignored, the magnitude of F should change because of the diminution of mass, so as to stay in the same r radius circle,and then there is work done.

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Thanks everyone for answering!

I find you explanations compelling, I would like to just add one more example in which it's much more intuitive for me to understand that no work is done in circular motion:

Say that our spaceship is attached to a string, and that the other end of that string is attached to the center of the hypothetical circle.

Once the shuttle is in motion, the string will keep it going in circles forever, no engines required, no work done!

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