Perpetual Motion In The Gravitational Field

[izemabdo]

ohhhhh sorry i confound this with electrostatic...soorry again i decline

 

[GarageDweller]

Assuming you are talking about the same attractive potential (like gravity) in both you cases, then your first case is wrong. An attractive potential increases with distance regardless of whether you are close to the source (planet) or very far away.



This is also wrong. You can always add any constant to the potential energy and still obtain the same dynamics (only differences in energy are observable), so you can add a positive constant that will make the energy positive (in fact, this is why U=mgh is positive while U=-GMm/r is negative - when doing the linear approximation on the latter formula, there is a negative constant that is ignored).

The sign of the overall potential energy doesn't matter, only whether it increases or decreases with distance is important here.

Yes i know this, but this is usually not explained in high school level physics since they are not famaliar with th gradient operator

 

[izemabdo]

Please tell me that was a typo

Assuming you are talking about the same attractive potential (like gravity) in both you cases, then your first case is wrong. An attractive potential increases with distance regardless of whether you are close to the source (planet) or very far away.



This is also wrong. You can always add any constant to the potential energy and still obtain the same dynamics (only differences in energy are observable), so you can add a positive constant that will make the energy positive (in fact, this is why U=mgh is positive while U=-GMm/r is negative - when doing the linear approximation on the latter formula, there is a negative constant that is ignored).

The sign of the overall potential energy doesn't matter, only whether it increases or decreases with distance is important here.

we conventionaly take this constant equal 0 constant=0
due to this even r--->infinity; U=0
or U=GMm/r +constant ...r-->infiity
we conclude constant=0

 

[GarageDweller]

The constant makes no physical difference due to F=-Grad(U)

 

[gabbagabbahey]

we conventionaly take this constant equal 0 constant=0
due to this even r--->infinity; U=0
or U=GMm/r +constant ...r-->infiity
we conclude constant=0

You are missing a negative sign. U=-GMm/r + constant. Taking the constant to be zero, U is a negative number that approaches zero, as r goes to infinity. The potential increases with distance.

 

[izemabdo]

The constant makes no physical difference due to F=-Grad(U)

yes, true but according to this question maybe he or she want to understand how can make to an object motion around the earth...is in it?,

 

[Vorde]

yes, true but according to this question maybe he or she want to understand how can make to an object motion around the earth...is in it?,

You're not making any sense, and it doesn't seem like the OP is currently online.

 

[izemabdo]

You are missing a negative sign. U=-GMm/r + constant. Taking the constant to be zero, U is a negative number that approaches zero, as r goes to infinity. The potential increases with distance.

no sign -

the primitive of -1/r² are 1/r !

 

[GarageDweller]

Comedy gold
Perhaps we should change the flat space interval to dxdtdydz=ds^2 just for kicks too

 

[davenn]

Comedy gold

isnt it ! LOL

will some one please just link to a factual www site with the correct formula and put every one at ease

Dave

 

[gabbagabbahey]

no sign -

the primitive of -1/r² are 1/r !

\mathbf{F}=-\mathbf{ \nabla } U = - \mathbf{ \nabla } \left( -\frac{GMm}{r} \right) = -\frac{GMm}{r^2}\hat{r}

The force of gravity is pointed radially inward toward the Earth. The negative sign is necessary.

 

[Vorde]

isnt it ! LOL

will some one please just link to a factual www site with the correct formula and put every one at ease

Dave

No need, gabby's said it a couple times, he's just not listening.

 

[GarageDweller]

\mathbf{F}=-\mathbf{ \nabla } U = - \mathbf{ \nabla } \left( -\frac{GMm}{r} \right) = -\frac{GMm}{r^2}\hat{r}

The force of gravity is pointed radially inward toward the Earth. The negative sign is necessary.

Thank you, i wanted to put it this way but did not know how to type that

 

[izemabdo]

isnt it ! LOL

will some one please just link to a factual www site with the correct formula and put every one at ease

Dave

sorry for this language accent, but i used to speak french well

 

[physio]

Okay, let's define potential energy.

If you have a region with a massive body, every object in the region will be attracted to that body with an acceleration depending on how far away from that body the object is (gravity).

If you put an object at a certain distance away from the body without any starting kinetic energy (i.e. motionless), it will gain motion (and gain kinetic energy) and start moving towards the body (because of gravity). Obviously kinetic energy is not conserved or this would be impossible.

So what we do is add in another type of energy, called potential energy. If you put the object right next to the body, it won't move (akin to being on the surface of the earth), if you move the object away from the body you would have to exert energy to move it away from the source of gravity (because it's being pulled in the opposite direction).

The potential energy of an object a certain distance away from a source of gravity is defined as the amount of energy it takes to bring the object to that distance when you start with the object right at the source of gravity (actually it's defined as the work required in lifting the object not the energy required in the lifting the object, but if you don't know what work is then thinking of it as energy is fine).

If you understand that, realize that no matter how little the acceleration from the massive body is (i.e. how far away you are from the source of gravity), it will always take some energy to move further away from the body: meaning that potential energy will always increase with distance.

Okay...so if I have an object the size of Earth and another object taken from the planet and I leave the object in space (considering no other objects in space), then will it start directly falling towards the larger object and vice versa with an initial low velocity and then as the distance decreases the velocity increases?

Am I right in saying so? Have I understood??

I have further 2 questions..

1) Why do you say it's the work required and not the energy required? Are they both not mathematically the same?

2) Can I safely generalize that force (any of the fundamental forces) is the prime reason for potential energy then?

 

[voko]

Okay...so if I have an object the size of Earth and another object taken from the planet and I leave the object in space (considering no other objects in space), then will it start directly falling towards the larger object and vice versa with an initial low velocity and then as the distance decreases the velocity increases?

If the initial velocities of the two objects are zero relative to each other, then yes, they will move toward one another with ever increasing velocity. If we treat both objects as point masses, then we even get infinite velocities at the impact time; this, of course, will not happen with real objects.

If, however, the mutual velocity is not zero, and it is not directed along the line between the two objects, then they will orbit about their mutual center of mass. Again, this is completely true only for point masses. Real objects may collide if the orbits are not well separated. The separation depends on the initial position and velocities. If the separation is adequate, they will move "perpetually" if nothing else acts on them and we assume that the objects do not undergo any changes.

 

[physio]

I don't know why I am digressing but (and this is completely off topic) if in the universe everything attracts everything then shouldn't the universe sort of collapse i.e everything will come together and smash us into extinction...How is it expanding? Also if the universe is expanding, then it is expanding into what? A big thanks to everyone whoever has helped me understand the concept of P.E.

 

[Vorde]

I'd start a new topic. There's lots of stuff to explain.

 

[sophiecentaur]

I'd grab a book

I totally agree. Half of this thread has consisted of telling the OP what he could have found in a basic Physics textbook. Trying to learn something from scratch in a series of mis-directed questions, followed by answers that are then mis-construed is not the best way to get to know a formal subject like Physics.
It doesn't even need to involved spending money on a textbook. There are dozens of .edu web pages that have all the information, well described and accurate - for free! It just requires a bit of time and effort to take it on board.

 

[physio]

I am reading lectures on physics by r.feynman as for the .edu web pages I think you are right...:)

 

[Vorde]

I am reading lectures on physics by r.feynman as for the .edu web pages I think you are right...:)

People might totally disagree with me here, but I wouldn't say that's the best place to start. The Feynman Lectures are much more mathematical and rigorous than a everyday physics book, but less so than a real textbook. I find that his lectures are fantastic (the best, really) for review, but not the best for learning stuff (remember the lectures are based off of a freshman physics course designed by him that was a self-admitted failure).

I'd start with some of the physics-orientated books for the everyman that are all around if you want to just know what goes on in the world. If you are more serious and care about the mathematics I'd get a real textbook (there are free ones online) or try to find a class you can take in introductory physics (that's vague, but I don't know your age).

 

[physio]

I have resnick and halliday but I don't quite like that book. I wanted to know the basics of force, momentum, energy etc. By reading R.Feynman I actually "UNDERSTOOD" what force really is. I understood why ƩF=ma, how momentum is closely related to the concept of inertia and how Newton has not done much but has taken other peoples results and integrated them into one whole. But above all I have understood there are no really's in science, only models (really!). :) Resnick and halliday doesn't quite give me that kind of an insight, it doesn't engender a joy of learning physics. But if you can suggest some other books which explains the basics really well than that would be wonderful. Thanks!

 

[Vorde]

If Feynman is working for you that's great, I only shared my experience in case you were in a similar boat as me. Best of luck to you!

 

[physio]

Vorde, your explanation really helped clear a lot of stuff...Thanks again!

P.S. Which books have you used to advance your physics knowledge.??

 

[Lsos]

Don't know if this is too little and/ or too late, but here's my .02:

OP, even IF the gravitational force from the Earth was 0, that doesn't mean there is no more potetnial energy. The energy is still there, and you can easily get it back by simply pushing the object back towards the Earth's gravitational well.

E=mgh is an approximation assuming an even gravitational field. As you know, for larger distances g will change, and therefore we have to modify the formula to take this into account (I'm not good at calculus, but I imagine we need to integrate). Basically we solve many E=mgh equations for all the different values of g, and we add them all together. This means that as you get further away, the gain in potential energy might diminish, but the potential energy will ALWAYS go up. Even IF at some large distance, mgh=0, this value is simply added on top of all the other values. Those other values never go away.

As for your off-topic question of why the universe is expanding despite gravity attracting everything...that's a VERY good question. It's a very good question becaue indeed it doesn't make sense, and many people are scrambling trying to figure out the answer.

 

[physio]

Thanks Lsos I understood the answer from yours and Vorde's explanation, that even if the object is very far away there will be some potential energy because F=Gmm/r^2 and F=0 only when the distance is infinite...Thanks!