# Do orbiting planets have infinite energy?

I know that planets can't have infinite energy, due to the law of conservation of energy.

However, I'm confused because I see a contradiction and it would be great if someone could explain it.

Energy is defined as the capacity to do work. Work is defined as Force x Distance. Force is defined as Mass x Acceleration. Thus, if we accelerate a mass for some distance by using some force, we are doing work, and we must have had energy in order to do that work.

In orbit, planets change direction, which is a change in velocity, which is an acceleration. Planets have mass, and they are moving over a particular distance. Thus, work is being done to move the planets.

In an ideal world, planets continue to orbit forever. Thus, infinite work will be done on the planets as they orbit.

How can infinite work be done (or finite work over an infinite time period, if you'd like to think of it that way) with a finite amount of energy?

Where is the flaw in this argument?

• I would appreciate it if people who downvote this question let me know what I can do to improve it or why it's not a good question for the site. I'm new to this site, so it's especially hard to see downvotes without even learning how to improve. – Pro Q Feb 26 '19 at 18:57
• There is little wrong with the question. I think those that did simply saw the title and the text without equations, then down voted without reading the question. What you have is a perfectly valid argument, it is just based on unsound premise which leads to the contradiction. – user400188 Feb 28 '19 at 2:10

Your definitions are incorrect. Force is rate of change of momentum and is a vector. More importantly, the work done by a force is not force x distance, it is the force resolved in the direction of the displacement x the magnitude of the displacement. This is more formally known as the scalar product of force and displacement.

In the case of a (circular orbit), the centripetal force supplied by gravity is at right angles to the displacement, so no work is done.

• Perhaps you should add what happens in an elliptical or hyperbolic orbit – magma Feb 24 '19 at 23:14

Power expended when moving in orbit $$\vec {F}.\vec {v}=-\nabla \phi .\frac {d\vec {r}}{dt}=-\frac {d\phi}{dt}$$ , $$\phi$$ is gravitational potential. Hence the work of gravitational forces is $$W=\int {\vec {F}.\vec {v} dt}=-\int {\frac {d\phi}{dt}dt}$$. For a periodic motion, the integral $$W$$ over the period is zero. For hyperbolic motion, the integral $$W$$ over the entire time of motion is zero.

In an ideal world, even then it is impossible to supply infinite energy.

I think the flaw here is that work done is only a measure of the change in energy. Obviously the satellite responds to the force of gravity acting on it (which in turn implies that the satellite does not have infinite energy).

What you have to understand is that work done on an object is the dot product of the force and displacement vector. This is where the $$cos\theta$$ term becomes vital. Remember that $$\theta$$ is the angle between the force and the displacement vector. The object makes displacement, true. But that is a horizontal displacement. The force cannot change the magnitude as such, because you have to apply a force in the direction of motion to cause the change in speed. Since the force acts perpendicularly, the object would have to move towards the planet. Which it does, but it does not fall to the planet either. Because the direction of force changes here. And hence, there is a circular motion.

Because the satellite has that constant speed $$v$$ (I am assuming simplest idea of a circular orbit), it has a kinetic energy. Because the satellite is attracted by gravity, there is a corresponding potential energy.

• The speed of an orbiting body is only constant if the orbit is circular. See en.wikipedia.org/wiki/Elliptic_orbit – PM 2Ring Feb 25 '19 at 0:02
• @PM 2Ring Yes. I have edited it. – Karthik Feb 25 '19 at 0:16

Remember that work is force times displacement, not distance. Displacement is a vector, which means when a planet moves a full circle, its overall displacement is zero, resulting in a work of zero.

• No, it isn't. It is the scalar product of force and displacement. The last paragraph is misleading and unhelpful. – Rob Jeffries Feb 24 '19 at 23:00
• Yes, I just realized that now. For some reason I had a brain fart and thought that work was a vector. – Jason Chen Feb 24 '19 at 23:03
• Answer is incorrect unfortunately. You can have an overall displacement of zero but a work that isn't zero, if the field isn't conservative. The gravitational field is conservative, but it need not be in general. – Allure Feb 25 '19 at 0:48
• I'm afraid this is just wrong. If something moved in a circle, but kept speeding up because of work being done on it, the net displacement could be zero, but with non-zero work. – Dawood ibn Kareem Feb 25 '19 at 2:21
• But it is not correct for the right reasons though. – Karthik Feb 25 '19 at 11:20

I am a new to this PLEASE DON'T DELETE THIS! (David Z I am talking to you!) It depends on what you mean. There is a lot of research that still needs to be done on certain parts of this subject as for some parts of it are not well known due to the inability to test such theories that some people may have and your questions is sort of ambiguous. If you simply mean that the act of orbiting an object is free energy then that would be incorrect and it does not gain energy that is known, it does not go faster in other words. Other things with orbiting and things close to this need to be researched more in order to be found correct.

Edit. THANKS for not deleting this and leaving me helpful suggestions! I have realised that your problem here is that you are misunderstanding the meaning of work.

• -1 This is confusing and disjointed. An answer should specifically address the physics concerns of the question with some level of detail rather than hand-waving about "other things" and "inability to test such theories." This is close to no answer at all. – Bill N Mar 2 '19 at 23:22
• An example of a specific theory and what test would need to be done to show that that theory is more likely to be correct compared to a different theory, and how that result would would apply to my question, would be appreciated. – Pro Q Mar 3 '19 at 17:13
• I don't think this answers the question. OP asked what the flaw in their argument was, and the answer is that they have misunderstood what work is. No "extra research" needed here. – Dawood ibn Kareem Mar 3 '19 at 21:47
• hey Pro Q, PLEASE read through a physics chapter on gravitation. The Newtonian version of gravitation has been understood since the mid 1750's, and no more work needs to be done on this subject in order to understand it. – David White Mar 12 '19 at 18:20