How is the work done to push a planet over 1m with 1N the same as pushing a feather over 1m with 1N? [duplicate]

Assume there are no other forces acting and the rocket+fuel described do not weigh anything. Also, by rocket I mean engine/thruster, not space shuttle.

Suppose you have a planet, say of mass 1,000,000,000 kg and you push it with a rocket which exerts a force of 1 N. This will accelerate the planet by $10^{-9}\, m/s^2$.

Then, suppose you have a feather of mass 0.005 kg and push it with the same rocket which exerts a force of 1 N, accelerating it by $200\, m/s^2$.

It follows that the feather will traverse a distance of 1 m much faster than the planet and thus spend much less rocket fuel to do so. So, the energy used to propel the planet must be much more than the energy used to propel the feather.

But, work done = force × distance, $1 \times 1 = 1\, J$ for both the planet and the feather. I think I have misunderstood something. How is this possible?

• TL;DR By definition. It's force * displacement. Fun-fact: change in their momentum will be the same. May 24, 2017 at 11:40
• "Work" isn't "how fast you did it", it's "what you did". It doesn't depend on time. If you carry a box up the stairs it doesn't matter if you took 30 seconds or all day to do it, you still carried a box up the stairs. May 24, 2017 at 12:44
• @JasonC My question concerned time as over time more energy is needed by the rocket to burn and apply the force of 1N. Read the answers below. May 24, 2017 at 15:33

So, the energy used to propel the planet must be much more than the energy used to propel the feather.

Yes. But that is because the rocket is a special thing. See below.

It follows that the feather will traverse a distance of 1m much faster than the planet [...]

Work is not about duration. It does not depend on time or on how fast. Pushing with 1 N and moving the planet 1 m requires the same work by the force regardless of it taking 1 hour or 1 year or 1000 years.

This might not seem very intuitive at first. If it takes a long time, isn't more energy then spent since the force must be upheld longer?

The answer is no. Think of a table holding up an apple with it's normal force. No energy whatsoever is spent by the table to uphold this force. It can do this forever. Force does not require energy - apart from in specific special "machines".

And that's the thing. The root of your confusion, if I'm right. A rocket propulsion engine (most types of engines to be fair) is such a kind of machine. It takes energy to create the force it exerts. It takes fuel. Fuel is burned at a never-zero rate and so the energy consumption of a rocket does depend on time.

The human body with its contracting and expanding muscle fibres is such a kind of machine as well. Eventually you will get tired of pushing, but the wall doesn't.

When thinking of the work formula $$W=\int\vec F\cdot d\vec x$$, I like to compare a wall with a balloon (essentially what you are doing, but in a slighty simpler setup):

• Pushing on a wall doesn't really make any difference even when pushing hard. Nothing moves. No work is done, you are just wasting your time.
• Pushing on a balloon is easy - it moves far. But it was so easy that no one would say that you did any significant effort. We wouldn't say that you did any significant work to move it.

The first example is your situation.

• No work is done, you are just wasting your time. No work is done on the wall, true, but the biomechanical and biochemical inefficiencies of the body do result in an energy demand on the body pushing against the wall. It's just not formal work. May 24, 2017 at 4:04
• Very good analogies! May 24, 2017 at 6:20
• @hBy2Py As mentioned already, yes, the human body does spend energy within in order to create the force that does work. Work is done within the body to create this force. This is not the work I am talking about. I am talking about the work done by the body as a complete machine, and whatever happens inside is inside. If this is unclear, then don't think of a person pushing, rather think of a thrown stone that hits the wall and exerts a force in this impact. May 24, 2017 at 9:22
• Mmm, Steeven, I'm not disagreeing with you - I just thought that it bore emphasizing that the non-conservative methods of force generation in a biomechanical system do behave qualitatively differently that conservative methods such as via gravity. May 24, 2017 at 12:03
• the muscles in the body are always trying to return to their original position, but unlike a spring, they cannot be chemically hooked onto anything. Energy is released constantly as the fibres contract in bursts and therefore are constantly being displaced. Some of this energy is also inevitably wasted as heat and in other indirect biological processes required to contract the muscle
– user86425
May 27, 2017 at 6:10

Let's take a Closer Look at the Whole System: It consists of the Planet / Feather (the accelerated object), the engine / thruster, and (and this is important) particles acellerated by the thruster in direction reverse to the pushing force acting on the planet. Your massless engine, however it works, uses the energy of the massless fuel to accelerate a) The Planet / Feather, and b) some other particles (usualy the burned fuel) in order to satisfy conservation of momentum. The whole Energy of the System is given by: $$E = E_{planet} + E_{burned fuel} + E_{fuel}$$ The whole momentum is given by $$P = P_{planet} + P_{burned fuel}$$

To make it short: On your one meter acceleration way, the energy that the planet and the feather get are the same. This isn't true however for momentum: The planet will carry more momentum --> There will be more "burned-fuel-particles" carrying more momentum, and thus also carrying more energy. And that's the energy that is missing in your equation. The whole energy of the system will be conserved.

To look at it more detailed: Force is the time derivative of momentum. The thruster pushing the planet / feather with one 1 N means, that every second there is generated a backward flying "burned-fuel-particle" carrying 1 kg $\frac{m}{s}$ momentum. Since the planet takes (as you noticed) more time to move 1 m, the there will be more particles accellerated backwards. They carry most of the energy that is provided by the fuel.

• Thanks for your answer, I wish I could mark two responses as correct, I'll attempt to improve Steeven's answer with yours asap. Thanks for clearing things up! May 23, 2017 at 22:53
• Joined physics.se just to +1 this: you made me realise that the work is being done right in front of our eyes but sometimes we Just dont see it because we're too focused. May 24, 2017 at 5:51

The work done on the planet (or feather) is equal to the change in KE of the planet (or feather). This is the work-energy theorem. If they start from rest, they will both have the same KE after the 1m of pushing. The planet will have a huge mass and very small speed. The feather small mass and large speed. The KE will be the same. The way you "produced" the force doing the work does not enter into the work-energy theorem. You can use more or less energy depending of your "machine", the system doing the work. But this does not affect the change in KE of the system your machine does work on.

• The question was about how energy does not seem to be conserved, but yes, the KE must be the same for both of them. However, the input energy is different since the planet's rocket fires for a long time while the feather's for a very short while. Thanks for the answer though. May 23, 2017 at 14:06
• Energy conservation in what system? You are comparing two systems. Energy conservation refers to comparing a system in two different states (initial and final). If you look at any of the two systems, for example rocket and planet, the energy at the beginning of the pushing and the energy after pushing 1m, the total energy is conserved. You should of course include all parts of the system (ejected gases too) and all forms of energy, not just KE. Conservation does not mean energy in different systems must be the same or has to be distributed in the same way.
– nasu
May 23, 2017 at 14:36
• Since the KE in both systems is identical in the end after they have traveled the 1m with a force of 1N pushing them, the input energy must be the same for both of them and that equal to the KE since we are not taking into account inefficiencies, right? But, if the rockets use different amounts of fuel as they fire for different amounts of time, the input energy must be different right? So, how can this be? BTW, by rocket I mean rocket motor/engine, not space shuttle. May 23, 2017 at 14:44
• You confuse "system" with just one part of it. The feather and the planet have the same KE in the end. So the work done on them is the same. This is all that work-energy theorem has to say about it. Think about a weight suspended from a hovering helicopter and the same weight sitting on the table. In both cases the work done on the weight is zero. But in the first case the helicopter uses a lot of fuel and accelerates lots of air (so increases its KE). In the second case there is no energy spent. The KE of the weight are the same in both cases but the final KE of the systems are not.
– nasu
May 23, 2017 at 14:55
• So no, you don't know that the KE of the systems are the same in the end. Just that the planet and the feather have the same KE.
– nasu
May 23, 2017 at 14:56

You might find this a bit simplistic, but I just wondered if maybe it helps you if I highlight some basics you may already be aware of...
that work is done to change velocity (accelerate), and an object continues at constant velocity unless work is done to it.
Work achieves a speed, not a distance. If an object has a non zero speed and no external forces interfere then it will continue to travel until it eventually achieves a great distance.
Compared to the motion subject to friction and air resistance that we typically experience on earth motion in vacuum can intuitively seem "a bit of a freebie". It is indeed cheaper.

• "Work achieves a speed, not a distance" This is not entirely accurate. Firstly, even if final speed becomes zero because of other forces, then work may still be done by the force we are looking at. And secondly, distance appears in the very definition of work: $W=\int \vec F \cdot d\vec x$ which actually is the question. May 23, 2017 at 20:58
• yeah so work has achived a speed of zero. F=ma, it's completley accurate in relation to this problem. May 23, 2017 at 21:17
• I'm not sure I understand that argument @JMLCarter. You could reach a very high speed very quickly and still have done almost no work at all. The feather is an example. But push something larger, and it reaches a much smaller speed even if you only push so much that the same work is done. Just like the planet. Work and speed are not really related. Work and distance are, though - work can't be done unless a distance has been traveled. May 23, 2017 at 21:43
• Thanks for you reply, nice profile picture! I have a African Grey and a Cockatiel :) May 23, 2017 at 22:41
• So you are suggesting that you can push the planet for a much smaller amount of time and allow it to continue through the 1m? In that case the planet will have less kinetic energy than if you were to push it for longer (assuming the force is the same). In order for work to be done, a force has to be applied over a distance so yes moving after the 1m push doesn't require work. May 24, 2017 at 6:27