Perpetual Motion In The Gravitational Field

[physio]

Can perpetual motion be possible in a gravitational field? If I take an object and into space from earth, the potential energy increases but as we go further away from the Earth the forces which contribute in the work done decreases, thus it seems to me as if energy is destroyed. Because the kinetic energy of the object is directly supplied by me, the potential energy mgh starts to decrease as we move away from the ground, thus it seems that potential energy is decreasing. Please explain why this reasoning is wrong. Thanks!

 

[Vorde]

Because potential energy increases as distance from the potential increases. Think about it, what happens if you double h in mgh.

 

[physio]

But acceleration due to gravity is 0 for an object not under the influence of Earth's gravity. Hence mgh=0.

 

[Vorde]

You'll have to rephrase, I don't really understand. The gravitational field is infinite in reach.

 

[physio]

Assume the object is at a height of 5 metres, the potential energy is mgh=5mg. Now I give it sufficient kinetic energy such that it starts ascending. As it is moving up the potential energy decreases and also the height is increasing from a reference position, but at a certain stage it escapes Earth's gravitational field and now there is no acceleration due to gravity thus potential energy has been made 0. Also the object is moving because I have supplied the kinetic energy hence it seems to me that the potential energy has been destroyed. There is some flaw in this argument which I cannot discern...

 

[fedaykin]

I don't understand what relation your other questions have to perpetual motion. If you take an object and place it some distance away from Earth in space, and no other forces change the objects motion, the object will have kinetic energy equal to the work you did to move it when it returns to the point you started moving it, e.g. the surface of the Earth.

http://physics.info/gravitation-energy/ might give you some insight.

It is true that the potential energy per unit distance decreases as you get farther away from the Earth's center of mass, but you'll never be able to completely leave the influence of Earth's gravity in a finite amount of time. Note that any mass still has an escape velocity where an object will never return. An object that leaves the surface at escape velocity would only hit zero velocity after an infinite amount of time. The flaw in your argument seems to be the idea that you can escape Earth's gravity well.

If you're thinking of astronauts in microgravity as having escape Earth's gravity well, they have not. They are merely in freefall, but moving tangentially at an equal rate. Since all the objects around them are also falling, they are not pushed upward as we are by the surface of the Earth. You can experience a little bit of freewall in an elevator by jumping shortly before the car begins descending. Be careful though.

 

[Vorde]

Acceleration due to gravity is never zero. It gets exponentially smaller and approaches zero but never reaches it. Just think mathematically, graph mgh where mg is a constant and h is a variable and see what happens as h (distance from ground) goes to infinity. (Hint: potential energy also goes to infinity)

 

[physio]

What I have understood is that when you take an object to a certain height it's potential energy increases because of the acceleration i.e. the velocity of the object is greater at the lowest point when dropped from a height h2 than the velocity of the object when dropped from height h1 where h2>h1. So what I am driving at is this: can we move the object in a closed path in such a way that the object's velocity increases in short can we obtain perpetual motion in a gravitational field?

Also if we have given an object sufficient energy to break away from the Earth's gravitational field then what happens to it's potential energy because now according to my understanding it cannot "fall" back in an elliptical fashion or directly since there is no "sufficient" force acting on the object to bring about the change.

 

[Vorde]

But its potential force is still rising. It doesn't matter how fast the object is going (i.e. if it will ever return). Potential Energy is strictly a function of distance, therefore and distance increases potential energy will increse without regard to acceleration and velocity.

 

[physio]

which formula are you using?

 

[russ_watters]

Sounds like you are describing escape velocity. If an object is traveling above escape velocity, it will continue to travel away from Earth forever. All this means is that it has enough kinetic energy that it will never be able to trade all of that for gravitational potential energy.

 

[Vorde]

which formula are you using?

The same formula you've been using the whole time. But it would work with any formula.

 

[physio]

So you mean to say potential energy keeps on increasing as we move the mass from the Earth's surface into space? But I thought the value of g kept on decreasing...

 

[izemabdo]

HI! to request this (PHYSIO) look...there s no perepetual motion as you expected..but all depend on launching speed...so we got three cases
first in this speed are weak it can undergo a free fall
second if this, are so important , the object in space will search an equilibre position due to attractive forces applied by other planet..etc
here if we want a perpetuel motion we have ( as done in launching fusil) to communicate to this object a very important speed over space (after launching from earth)in other terms ,at space the object receive another speed (from a capsule for example that re fixe in space)..consequently , our object going to do an elliptical trajectory which the Earth occupates one of this pole...i guess it s clear for you ...else you can contact me on my account facebook or in my gmail just to make things right
there is https://www.facebook.com/abdelkoddoussiz

 

[gabbagabbahey]

The formula U=mgh is only valid relatively close to the Earth's surface. It is a first-order (linear) approximation of U=-\frac{GMm}{r} (where r is the distance from the centre of mass of the Earth - assuming that the Earth is a sphere, this is just its centre).

Clearly, as the distance from the center of the Earth increases, so does the potential energy (its absolute value decreases, but since it is negative, its signed value increases)

 

[izemabdo]

gabbagabbahey ,right but don't forget other forces applied by other planets...in space there is a solar system ..

 

[GarageDweller]

Perpetual motion is impossible only in a more realistic context, if you approximate everything in central force problems, you can have non-physical perpetual motion.
For example, the orbits calculated using the effective potential continue until infinity. This is because we have approximated the system and ignored the rest of the universe
To be honest i find the idea very ill defined, motion is only meaningful when it is measured with respect to some observer, it is always possible to choose a frame in which an object is in motion.

 

[Vorde]

gabbagabbahey ,right but don't forget other forces applied by other planets...in space there is a solar system ..

You're complicating a simple situation. You're right, of course, but considering outside forces makes the mathematics much more annoying and yields the same end result.

 

[izemabdo]

great, in reality all physics are approximations...& pereptuel motion in reallity doesn't exist
but according to our approximations we expect the possible motion.

 

[GarageDweller]

The formula U=mgh is only valid relatively close to the Earth's surface. It is a first-order (linear) approximation of U=-\frac{GMm}{r} (where r is the distance from the centre of mass of the Earth - assuming that the Earth is a sphere, this is just its centre).

Clearly, as the distance from the center of the Earth increases, so does the potential energy (its absolute value decreases, but since it is negative, its signed value increases)

The original poster is obviously not famaliar with taylor series, this will just confuse him more

 

[Vorde]

I don't think a lack of familiarity with taylor series is a deal-breaker for that explanation.

Moreover I think another poster highlighted the flaw in the OP's reasoning, he is under the impression that you can 'escape' the gravitational influence of a body. This is impossible, the influence will grow weaker and weaker, but never disappear.

To address izemabdo, I'm not sure what the OP was talking about with perpetual motion, but for clarifications sake: perpetual motion machines don't exist.

 

[physio]

I understood F=GMm/r^2 where "r" the distance can be anything large i.e. there will be yet some force even when two objects are separated by a large distance but what I do not understand is that how is the potential energy increasing i.e. how can it have a positive value??

 

[Vorde]

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.

 

[izemabdo]

come on ! in space the formula of potential energy is (+or -)GMm/r where +or - depends on force's nature ie(is it attractive or repulsive)
there is 2 cases
first if that radius r---> infini ie distance between object and planet are so important
U--->0 potential energy will decreased
else r is small U so important than U increase all dépend on r

 

[GarageDweller]

I understood F=GMm/r^2 where "r" the distance can be anything large i.e. there will be yet some force even when two objects are separated by a large distance but what I do not understand is that how is the potential energy increasing i.e. how can it have a positive value??

Newtonian gravitatioanl potential is never positive, it starts at negative infinity and heads for 0 at a distance infinitely away.
As you move a mass farther from another, you are doing work, when you release that mass the field pulls it back and does the same work you did putting it there. This follows from the conservation of energy. And of course, the farther it is, the more work you have done to put it there, hence a higher potential energy

 

[GarageDweller]

come on ! in space the formula of potential energy is (+or -)GMm/r where +or - depends on force's nature ie(is it attractive or repulsive)
there is 2 cases
first if that radius r---> infini ie distance between object and planet are so important
U--->0 potential energy will decreased
else r is small U so important than U increase all dépend on r

Please tell me that was a typo

 

[izemabdo]

sorry i don't got you??

 

[izemabdo]

Please tell me that was a typo

typo?

 

[GarageDweller]

How could there be repulsive gravitational fields?

 

[gabbagabbahey]

come on ! in space the formula of potential energy is (+or -)GMm/r where +or - depends on force's nature ie(is it attractive or repulsive)
there is 2 cases
first if that radius r---> infini ie distance between object and planet are so important
U--->0 potential energy will decreased
else r is small U so important than U increase all dépend on r

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.

Newtonian gravitatioanl potential is never positive, it starts at negative infinity and heads for 0 at a distance infinitely 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.

 

[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...:)