As question states, I start and then stop a spaceship in vacuum by using orientable thrusters. In the acceleration phase, I am doing positive work, but in the deceleration phase I am doing negative work. These two should cancel out and I would do zero net work.

Seems kind of strange that I do zero net work although it make sense because the spaceship's kinetic energy didn't change overall. The space ship used up a bunch of chemical potential energy which is now transferred to the "thruster dust" floating about in space.

Does this make sense?

  • $\begingroup$ W=Fd. You exerted a thrust force for some distance. That looks like work to me. $\endgroup$ – electronpusher Jun 28 at 20:20

Yes it makes sense based on the work energy theorem which states:

“The net work done on an object equals the change in its kinetic energy.”

So if the spaceship starts at rest and ends at rest the change in its kinetic energy is zero and therefore the net work done is zero.

Positive work was done in the beginning to accelerate it and give it kinetic energy. An equal amount of negative work was done at the end to decelerate it and take that kinetic energy away.

As far as the use of chemical energy is concerned consider that the spaceship may have traveled from rest at one location in outer space to another location at rest in outer space millions of miles apart. That requires energy.

Hope this helps

  • 1
    $\begingroup$ Of course, all the exhaust still has kinetic energy, so work was done. $\endgroup$ – Jon Custer Jun 28 at 20:43
  • $\begingroup$ @JonCuster I would look at the kinetic energy of the dust as being responsible for transporting the spaceship from one point to in outer space to another beginning and ending at rest. But the net work done on the spaceship is still zero. Would you agree? $\endgroup$ – Bob D Jun 28 at 21:01

Net work on the spaceship is zero, yes. But don't forget all the "thruster dust", as you mention, the exhaust gas. Non-zero net work has definitely been done on that - those particles are at full speed moving away from the spaceship - so the overall net work of this entire system is non-zero.

It therefore just depends on what you are looking at; what you define as your system.

Furthermore, even though the net work done on the spaceship is zero, that does not mean that no energy was lost, of course. The thruster was pointing backwards in order to accelerate the ship, and pointing forwards in order to slow down the ship. Although this causes different signs of work to be done on the ship, a non-zero amount of chemical energy in the form of fuel is released in both cases. That energy is now converted into the kinetic energy of the exhaust gas particles as well as into heat, undoubtedly.

  • $\begingroup$ Steeven see my comment to Jon Custer as another perspective. $\endgroup$ – Bob D Jun 28 at 21:02

Place a bubble or balloon around your spaceship and start the engines. The jets hit the balloon, the ship hits the balloon on the opposite side and at the end the ship stays where it is.

Instead of imagining a balloon, you must always calculate the work for a closed system or, in an open system, the outgoing and incoming forces.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.