# What is the essence of the fact that the work in the potential field of forces to move to some point does not depend on the trajectory? [closed]

Is it something like a guarantee that somehow arriving from point A to point And you will have the same energy? For example, having a spaceship in the solar system with a fuel tank, you can fly to B and spend half, or you can fly further for the entire volume of the tank and then, when the gravity of the sun attracts the ship to point B from some distance where the ship managed to fly, the ship will have kinetic energy equal to just this half of the tank fuel?

Let me start with the simplest form of potential: the potential of a uniform, unidirectional force.

Here on Earth, close to the Earth surface, gravity at the floor of a room, and at the ceiling of a room, is effectively the same, and it is also effectively the same from one side of the room to another.

So: for any desktop experiment Earth gravity acts as a uniform, unidirectional force.

We can go from the ceiling to the floor anywhere in the room: the difference in potential is the same.

If we move sideways no work is done.

We can start at a particular point at the height of the ceiling, descend a little, move sideways a little, descend a little, etc.

The net work done is not affected by the distance that you move sideways. To evaluate the work done from start point to end point it is sufficient to add up the work done each time you descended a little.

Next we consider moving diagonally. We can think of moving diagonally as a superposition of descending and moving sideways. The sideways component of the motion does not affect the work done. To evaluate the work done it is necessary and sufficient to consider only the motion component parallel to the direction that the force is acting in.

So:
In the case of a uniform, unidirectional force there is simple no opportunity for the work done to be somehow dependent on the path. Try as you might, you can't make it dependent on the path.

Next level up is a spherically symmetrical force. You then need to define what counts as moving sideways. Moving sideways must then be defined as: 'to move perpendicular to the local direction of the force.'

With that in place the same reasoning as in the case of a uniform, unidirectional force applies.

In my reasoning I mentioned the concept of superposition. We can think of that as a principle of mechanics: superposition of forces and motion.

Stated in terms of vectors:
All the vector properties of mechanics have the following in common: they can be added and subtracted as vectors of Euclidean space.

Force vectors, position vectors, velocity vectors, acceleration vectors; all can be added and subtracted: no self-contradiction arises.

Interestingly, in a text that predates the writing of the book 'Principia', Isaac Newton used a set of 6 laws of motion, the concept of superposition of motion one of them. When it came to composing the Principia Newton opted to present the well known set of three. The other three of the earlier set of 6 ones are also stated in the Principia, but they are presented as corollaries.

But really, Newton could just as well have opted for a set of 4 laws of motion, with the superposition principle as one of them. If Newton would have done that then today we would have had 'Newton's 4 laws of motion'.

Whatever the case, the superposition principle is an absolute necessity in theory of motion.

Work is an indicator of the change in the potential by definition. $$W = -\Delta U$$ And potential does not depend on the path thus work does not depend on the path that

About your example, it can be explained by the conservation of energy principle Let the tank have an energy of E in them and suppose that while burning the fuel there is no loss in the energy. If the rocket uses half of the fuel it will have a speed of (m is the mass of the rocket) (let r be the initial distance from the sun and R be the final distance from the sun) $$\frac E2 - \frac{GMm}{r}= \frac{mv^2}{2} - \frac{GmM}{R}\\ v^2 = \frac{E}{m} + \frac{2GM}{R} - 2GM/r$$

If the rocket uses all of the fuel it will have a speed of $$E - \frac{GMm}{r} = \frac{mv^2}{2} - \frac{GMm}{R}\\ v^2 = \frac{2E}{m} + \frac{2GM}{R} - \frac{2GM}{r}$$

So yes rocket will have kinetic energy that is E/2 more than the first situation.