B Does gravity defy the law of conservation of energy?

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If we have an object in space (deep space where it is under no other gravitational influence) and we push it a little so that it gains some velocity and after some time comes into the influence of a planet's gravitational field and crashes on it. Where is that energy from the crash coming from? It seems like the object just gains energy out of nowhere.
 

davenn

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Where is that energy from the crash coming from? It seems like the object just gains energy out of nowhere.
Of course not

what happens when an object gets caught in gravity ?
Consider meteors coming into the Earth's atmosphere
or
if you drop an object from a height say a stone from your hand
what is the form of energy the stone has before you drop it ?
what is the form of energy it has after your drop it ?


lets go from there for a start :smile:


Dave
 

sophiecentaur

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after some time comes into the influence of a planet's gravitational field
It's always under the influence. You can never get too far away, in your simplified Universe and the Potential Well of the planet has a value, everywhere.
 
I think there's something to be learnt here.
Let's think of a very simple and slightly strange universe that consists of one very big planet and one tiny object. In this universe the gravitational force suddenly becomes zero at a certain distance (unlike our own universe) and the object happens to be just outside this distance from the planet.
If now, due to random fluctuations, the object suddenly moves a tiny amount towards the planet and it starts feeling its gravitational pull, the planet will start doing work on the object.
This work gives the object the kinetic energy with which it eventually crashes into the planet.
The question the OP asked is where this energy was to start with. Remember that originally the object did not feel any force from the planet.
The conclusion has to be that the potential energy of the object has the same value at every position outside the reach of the gravitational force.
Inside the reach of the force the potential energy gets smaller and smaller the closer the object gets to the planet and the kinetic energy gets bigger, but the sum of the two is always the same.
This also means that outside the reach of gravitation the potential energy of the object has its maximum value, and it has this value although the object doesn't even feel the planet when it's so far away from it.
You see that zero force doesn't imply that potential energy is at its minimum.
 
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It certainly does imply a local minimum.
And your simplified universe with arbitrary cutoffs allows all forms of mischief involving non-conservative fields for which a potential is not even defined.
Not a useful construct I think.
 

jbriggs444

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It certainly does imply a local minimum.
Zero force implies an extremum. Could be minimum. Could be maximum.

I think you can find a local maximum at the center of a rigid cube with eight gravitating point masses at its corners. Obviously, the gravitational force at the center is zero.
And your simplified universe with arbitrary cutoffs allows all forms of mischief involving non-conservative fields for which a potential is not even defined.
@Philip Koeck was careful to specify a cutoff which still allowed the field to be conservative. The potential in his toy universe is well defined.

In general, you are correct that waving a magic wand and removing gravity in an arbitrary region might allow the resulting field to be non-conservative. However, hand-waving away gravity in a region that is bounded by an equipotential surface is not a problem.

The "cavorite" gravity shield proposed by H. G. Wells was indeed a non-starter.
 
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In my opinion, the question raised is not about physics, but a logical question. Because it is evident that the questioner knows that the sum of potential and kinetic energy is constant. But he asks the question that the object is so far away that gravitational force has no effect, as if nothing else existed outside the object. And then the object begins to move in a straight line with uniform motion - complying with Newton's first law - and it's time to fall into the gravitational space of a celestial body. At first, it had minimal kinetic energy, its potential energy was nothing (because it was not in the gravitational space), and then it gained enormous energy.

However, we have to decide whether to build our conclusions on mathematically formulated physical laws or on measurable physical reality. In the first case, the underlying assumption is wrong, because in a mathematical sense gravity is not diminished at any distance. In the second case, however, there is no basis for the assumption that an object that is so distant that the effect of gravity is physically ##0## will, due to the impact of a tiny force, actually carry out a straight line of uniform motion for such a long time to reach the gravitational space.

Unfortunately, in my opinion, similar thinking mistakes occur in real science, in different disciplines.
 

jbriggs444

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In the first case, the underlying assumption is wrong, because in a mathematical sense gravity is not diminished at any distance.
Gravity does not become zero at any distance, yes. Even so, when integrated over any distance, the resulting potential remains finite. The mathematical model works.

[If one contemplates gravitating point masses, there are problems lurking at very close ranges, however]
 

sophiecentaur

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If the speed of an object is high enough it can follow a hyperbolic orbit. This is a once off orbit which takes the object further and further away from the star and its path gets nearer and nearer to a straight line.
 

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