Is the First Law of Thermodynamics True?

Kind of like a Lava Lamp, but you can play with the blobs. I would love to be an astronaut, but I wouldn't get any work done. I would be playing all the time.

True, it's one of those Calculus things, weight approaches zero as distance becomes infinite. The amount of "potential energy" is the limit of integral of force (weight) integrated from the surface of the earth to distance D as D becomes infinite. The force is inversely proportional to the square of distance, so there is a finite amount of energy (area under the curve).

http://www.sliderbase.com/spitem-739-2.html

Several posters have confirmed, but just to elaborate - integrating force*displacement from Earth's surface to ∝ gives finite energy, which can be used to calculate the escape velociy.

Comes to sqrt(2*M*G/R) (I'm sure you know what the letters stand for!) giving about 11200m/s.

Thanks for confirming my memory from many years ago!

No problem!

"to the point at which gravity of the earth vanishes"

no such point

Sorry John, didn't see your post.

Is it 'great minds ..............'

or

'fools never ..........'?

;-)

Exactly....

Idealizing that, if we lived on the only planet in the universe. no matter how far away an object was moved away from that solitary planet, then, there would still be one really, really, really... minuscule, but still calculable, pull-of-gravity, on that object...

Since gravity does vary with distance from it's source, maybe the equation should look more like:

P.E. = m ( ^_{_}/_h1h2g(h)dh ) h₂ (?)

_{Please forgive my clumsy attempt at an integration symbol...}

Yes, but in the normal limits, the g(h) term is very nearly constant, because the lift is "infinitesimally" small relative to the radius of the earth, and the integral reduces to the familiar term we are accustomed to. Certainly when considering the general case, yes, the entire functionality is to be considered.

Actually, I should correct myself by saying something more like:

assuming the system is solely of a single planet, and a single object, farrrrrrr from said planet, all within in infinite Euclidean space, then:

P.E. = m_object*( ( (integral of g(h) from h1 to h2) ) dh ) * (h2 - h1)

where h1 is the radius from the gravitational center of the planet to the surface of the planet

and h2 is the distance from the gravitational center of the planet to the (exorbitant?) location of the object...

Idealizing a little more, if said object was (raised) to a (ridiculously) large, and geosynchronous type of orbit around said solitary planet, AND all forms of energy were conserved (i.e.: no thermal, chemical, physical, mechanical, electrical, etc., loses occur),

then P.E. = M_object * ((( integral of g(r) from r₁ to r₂ ) dr ) * ( h = r₂ - r₁ )

and F_object = m * a = M_object * g(r) = G * ( M_planet* M_object ) / ( ( r₂ - r₁ )^2 )

thus, g(r) = G * ( M_planet ) / ( ( r₂ - r₁ )² )

and P.E. = M_object * ((( integral of (G * ( M_planet ) / ( ( r₂ - r₁ )² ) ) dr ) * ( r₂ - r₁ )

=> P.E. = G * M_object * ( r₂ - r₁ ) * ((( integral of ( M_planet ) / ( ( r₂ - r₁ ) ) dr )

...which is all well and good, and calculatable, (within) infinite euclidean space...

However, what kind of (strangeness) happens in the time-space continuum remains to be determined...

To elaborate a bit on John and Stuart's (correct) comments:

As you "lift" the object farther and farther from the Earth, the gravitational attraction between the Earth and the object gets weaker and weaker, but it never reaches zero. If you move the object to a point where the gravitational attraction is one millionth of the the attraction at the Earth's surface, stop the object from moving with respect to the Earth, and then release it, it will then experience an acceleration of one millionth of a "g" toward Earth (ignoring all other objects in space). That is very small, but it is not zero!

Due to that very small acceleration, the object will, very slowly at first, start falling toward Earth. If unobstructed, the object will gradually get closer to Earth, and thus experience a greater attraction, and thus a greater acceleration. Just before reaching the earth, it will be traveling roughly 25,000 mph (escape velocity from Earth).

"and if we push the block a little so that it comes under the gravitational field of earth how it suddenly gets large amou[n]t of kinetic energy ?" The object does nothing suddenly, until it crashes into the Earth after very gradually accelerating from the little push speed up to escape velocity.

As others have noted, gravity does not vanish at some point.

However, enough kinetic energy can certainly be added to an object such that the diminishing gravity will never reverse it's course. That is still not problematic when considering potential energy.

Potential energy is not a guarantee or promise that the potential will be realized in that specific amount or in that form. Furthermore, potential energy is not exclusive nor a characteristic of a specific object in and of itself as it is.

Potential energy describes the energy made available or the energy required in changing from one specific state to another.

Conversion down one potential path can certainly alter the potential of other paths. If someone is moving 20 lbs of TATP up a hillside in a wheelbarrow, after ascending 100ft hits a jarring bump resulting in complete rapid conversion of the TATP to gasses, the potential energy of the TATP gained in the ascent is going to be difficult to account for....not really that problematic in the grand scheme though.

• "E=mgh" is an approximation for use near mean sea level for small values of h.
• As suggested by the original post, if h is large enough then there is the possibility of stuff remaining in orbit at various orbital velocities for all values of h.
• It is possible for stuff to remain above the same spot on the surface for a particular value of h, a "geostationary orbit".

There is a load of information on orbital topics in Wikipedia.

Back on earth, if I lifted this mass above my head I would have done work on it, once stationary above my head, no more work is done on, but to hold it steady without doing work on it knackers me - from where I am standing it seems like a lot of work to me - doing nothing but becoming exhausted doing it.

You should talk to this guy.

That site you linked to seems to be overtly bigoted in its sectarian evaluation of ancient Greek traditional beliefs as 'mythology', instead of providing the same leeway given to other supernatural belief systems that comes with designation as 'religion'.

Linking to such a radically intolerant site might, in light of the UN's assembly's adoption of rules to eliminate all religious intolleranve, might run afoul of cr4's rule prohibiting linking to sites that run counter to law......though I am no lawyer, so this is, at best, lay conjecture.

Ts ts. Are we talking Wikipedia here?

Is that a reverse ad homenem fallacy, or is that the one called appeal to authority? I get them confused.

'This couldn't be sectarian bigotry....this is Wikipedia' isn't a constructive rebuttal.

You might want to talk plain English with me because I am sooooo sloooow.

It might benefit other readers too.

But one thing for sure asking a question is not a rebuttal even so I can see that it sounds like a appeal to authority.

Simply I was too lazy to find a web site that explains that concept of Atlas holding up the sky in a truthful fashion.

Quarrel not, as we are on the same side. Lets not just cut the rope that holds up the sky.

Merely using the format of a question shouldn't really allow assertions to be made with impunity. Neat trick if you can get away with it, though.

To be more clear, I am not quarreling in earnest. Just poking a bit of fun at how it is perfectly acceptable to call currrently out of fashion religious stories 'myths', when doing the same with stories currently popular religions would likely earn one the title of bigot.

Excellent point!

There is some mythical stuff about CO2 and what it does?

What only one OT for this one?

Applying a force to hold an object stationary is not doing work on it. Work happens only when a force moves thru some distance.

Exactly, so why do I get exhausted without doing any work, when holding a weight steady?

'...so why do I get exhausted....'

You get exhausted because you have completed your time in the work area, done your work, and have now moved past that part. Remaining in the work area isn't a good option as you would just get in the way of newcomers doing their part. Your fuel has been mostly burned up (the amounts that remain, the catalysts will catch almost completely).

Your pressure and temperature continue to fade, making the intensity you had when you were really fired up at your peak, seem almost like fantasy compared to now. As you get colder and more viscous you begin to look forward to finally being completely exhausted without the confinements that restrain all living things, yet still feel a tinge of remorse at the anticipation of no longer being part of this whole process.

.

You get exhausted because you have been efficient in converting your potential energy into more usable types of energy.

Thanks TINAC (apologies for the acronym - I have to conserve typing energy). I think I understand. I must lay down for a moment - phew!

This is actually a good question. What changes in the block after you raise it up and it has potential energy? What is potential energy and where is it stored until it is returned as kinetic energy? I submit that the block is microscopically identical whether it has potential energy or not, whatever potential energy is.

We talk about potential energy but it's really an accounting trick. "Potential energy" is really energy that is stored in the gravitational field. If you raise the block h meters, you add energy "mgh" to the earth's gravitational field. When the block falls h meters, "mgh" energy is removed from the gravitational field and returned to the block as kinetic energy, just like releasing a spring.

Just my viewpoint.

In actuality, since gravity is the effect of the curvature of space time, there will be points where a body will cross a summit in the curvature of space time and the net gravity attraction to earth will be zero. At the point that the body crosses to another gravity well, then all the net potential energy vanishes. I could see that in the infinite ledger book of energy, that potential remains in some sort of balance equation between the three bodies in some sort of gravitic menaj au tois.

I have always wondered how you could inject potential energy into a system without applying force to the masses in the system, as in space inflation/deflation or directly tweaking the curvature of space/time.

It sounds like you are describing Lagrange points.

https://en.wikipedia.org/wiki/Lagrangian_point

If you have a small body (e.g., a spacecraft) in the neighborhood of much larger objects (e.g., earth, moon, etc), the larger bodies will define the spacetime curvature landscape for all practical purposes. Of course, this landscape changes as planets and moons move, and this is used advantageously to send spacecraft to the outer regions of the solar system. So you can assume the small body will traverse a straight trajectory through the spacetime defined by the larger bodies.

To a much, much smaller extent, the spacetime landscape is defined by all the masses, big and small. The earth's gravity is strengthened very, very slightly when a meteor falls to earth, the gravity well becoming very slightly deeper. It's complicated and nonlinear, and the general 3 body problem cannot be solved.

As Wheeler put it:

"Spacetime tells matter how to move; matter tells spacetime how to curve."

(Wheeler's succinct summary of Einstein's theory of general relativity, in Geons, Black Holes, and Quantum Foam, p235)

What if we lift the block to the point at which the gravity of the earth vanishes and then release it it does not show any motion, where had all our mechanical energy gone.

This is how I see it.

1.Energy is required to lift the block.

2. Energy is required to release the block. The block could not be released unless energy was expelled.

3. Even if you got so far out in space where there was no gravity from the earth the absence of the earths gravity would have no effect on the kinetic energy that the mass had been subjected to due to the lifting motion that got the mass there in the first place.

If you lifted an object into zero gravity, that object is attached to the lifting force, this force stays constant, the only way to separate the object from that force is to expend a greater force upon that same object, in this way, the energy is then transferred to the mass.

This is how I understand it.

Ignoring the atmosphere and all other objects in space except the Earth and the block, the block could be shot upwards with exactly the Earth's escape velocity. All of the lifting force would be applied during the time the block was in the "gun's barrel". The object would then move away from Earth very rapidly at first, and gradually slow down as it got farther away. Since the original kinetic energy was due to exactly the Earth's escape velocity, the object will take forever to slow to zero speed. Any error in original velocity will change the result: a slight excess energy will leave the object moving away from Earth forever; a slight deficiency of energy will make it like any ordinary projectile - it will eventually fall back to Earth.

Thus there is no "release", other than perhaps the exit from the "gun barrel".

"If you lifted an object into zero gravity, that object is attached to the lifting force, this force stays constant..."

There is no reason or need for the lifting force to remain constant. If the object is carried outward by a rocket, the lifting force MUST change. The lifting force must exceed the gravitational attraction at liftoff, then either equal or exceed the gravitational attraction all the time the object is being lifted. By the time the rocket reaches the edge of the solar system, that lifting force is minuscule, but not zero.

If at that point the rocket were to release the object, it would indeed require energy from the rocket to actuate the release mechanism, but if done properly that release process would not need to change the energy of the object in any way. Of course after release, gravitational attraction between the rocket and the object would affect the object slightly.

I sort of, but don't completely understand your response. I don't see in the op questions anything about a gun barrel or a rocket or escape velocity. My understanding of the question was that there was " lift " , not any amount of it. If I was to stand on the earths surface and throw a ball towards space at a given speed and there was no atmosphere to create drag or gravity to reduce the speed, then the ball should continue at the same given speed, unless a force acts upon the lift that I provided when I threw it. The force that stops the ball would need to be equal to the force that I applied to give it motion or greater to change its direction.

Of course the lifting would change if there was a rocket involved and not remain constant, but again I didn't see any mention of a rocket in the original question.

The OP said "what if we lift the block to the point at which gravity of the earth vanishes ..."

Virtually everyone here knows there is no such point, but at a very large distance from Earth, it almost vanishes. The only method we currently have available for lifting anything to such distance involves using a rocket. If we ignore the atmosphere (pretend it does't exist), then some form of gun, including a rail gun, becomes an option, in thought experiments, to propel (lift) the object.

The work/energy required to lift the object can be applied in many ways, but it can NOT be a constant force throughout the entire distance. That force MUST be larger near the Earth, and smaller farther away. It would certainly be possible to calculate the average force required, but that average force would not even begin to lift the object from the Earth's surface.

The larger force near the Earth can be applied in several ways. Currently, most objects that are lifted to such distances are lifted by multiple-stage rockets, possibly including some form of propulsion that exerts a very small force over a very long distance (ion, EM, solar sail, etc). These multiple stages make the calculations for force required at each point along the path quite complex. Exerting all the force over a relatively short distance, like in a rail gun, simplifies the calculations significantly.

"If I was to stand on the earths surface and throw a ball towards space at a given speed and there was no atmosphere to create drag or gravity to reduce the speed...". We do commonly simplify approximate calculations by ignoring atmospheric drag, but in this case, we can't ignore gravity, because that's what the whole concept of gravitational potential energy depends on...

"The force that stops the ball would need to be equal to the force that I applied to give it motion...". Definitely FALSE! A larger force over a smaller distance, or a smaller force over a larger distance can produce the same end result. In fact, by the time the object reaches whatever point is considered to have negligible gravity, the Earth will have been pulling back on the object, slowing it down to the point that it's velocity is nearly zero, and only a very slight, essentially instantaneous, nudge will stop it.

I don't think releasing a payload requires energy input from the rocket.

Consider a joint that is held tight and restrains a compressed spring that would push the payload away from the rocket. The joint is an interference fit of two materials with diffrrence rates of thermal expansion such that at or above 20 C the joint is steady, but below 0 C the joint releases the spring.

Alternately, a payload might be held against the rocket using the force of the rocket as it accelerates the payload. When the rocket runs out of fuel, the atmosphere, though rarefied, has more effect on the low density spent rocket than on the package. Also any mechanical rebound from material elastic de formation will come into play.... Viola! Separation.

If you want to control the instant (or distance) at which release occurs, then there must be some form of trigger mechanism. I agree that the release energy could be stored in one or more springs, or many other forms of energy storage, but that trigger to tell it when to release will require some form of energy.

Perhaps using the term "rocket", was a mistake. At this point in the trajectory, I was thinking of the "rocket" as whatever remains after all the fuel has been expended. Clearly it is possible to release the load without burning any fuel.

It still needn't be a utilization of energy stored by spring, fuel, etc. Separation could be motivated by a loss of heat to the surroundings.

One possibility would be actuators containing a fluid which expands upon freezing.

Another possibility might have a pyroelectric materials that motivate separation ( the release could be handled by differential thermal expansion rates) when cooled sufficiently to establish large like charge potentials on the rocket and payload.

There is no such thing as zero gravity anywhere in the universe.

I don't believe the OP has an engineering background based on the question.

Although Rixter has come close, I don't think this question has been answered in enough of an elementary way so as to satisfy his need for an answer.

Let's assume the entire earth is made up of blocks the size as the one you are trying to remove. Each block has an equal mass, and equal gravitational force. All blocks together make up the gravitational force you experience when you walk on the earth, or pick up your block. If you remove one block from the group, and send it to the realms of deep space, you are removing one block worth of gravitational force, which remains with that block until it attaches to another pile of blocks. Assuming you remain on earth, you will now experience one block less of gravitational force when you pick up your next block. If you repeat this until 75% of the blocks are in deep space, you will find it really easy to pick up a block.

" Each block has an equal mass and equal gravitational force."

According to your answer, each block has an equal mass, but what you left out is that each block has a percentage of the total gravitational force.

If I had ten blocks on a platform and each block weighed 1 pound, not counting the weight of the platform, I would have a total weight of 10 pounds.

If I removed 9 of the blocks, the one block remaining would weigh 1 pound.

The last block would still remain as heavy as the first and have the same mass, the gravitational force on the last block would be the same as the first one.

"...If I removed 9 of the blocks, the one block remaining would weigh 1 pound....'

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Weight describes the gravitational force on an object. If the entire mass were comprised of 10 blocks and 9 were removed (from the closed system), the remaining would be weightless.

.

'...The last block would still remain as heavy as the first and have the same mass, the gravitational force on the last block would be the same as the first one....'

Not so much. While the mass remains the same, the weight is dependent on other mass in the vicinity.

'....each block has a percentage of the total gravitational force....'

That doesn't describe the effect accurately. Gravitational force depends heavily on other masses in n the vicinity. Modeling it as discrete amounts per mass doesn't allow useful predictions.

Newtonian gravitational force is calculated as G x m1 x m2 ÷ r2

Assuming comparison at the same radius it is easy to see that the contribution to gravitational force isn't well described as discrete contributions per unit mass.....

.... consider a total mass of 10 equal blocks. Compare; if m1 is 5 blocks and m2 is 5 blocks, the gravitational force is more than 2.5 times what it would be if m1 were 9 blocks and m2 were 1 block.

You might be wasting your time.

I can't argue with good math. I stand aside.

You are correct. From Newton's law, F=(G*M₁*M₂)/r² , where M₁ is the mass of the earth that's left and M₂ is the mass being lifted to very deep space. The average amount of mass left, M_1, is 1/2 the mass of the earth at the start. It follows that the average force and average energy to remove the mass would be reduced to 1/2 the values for the first mass to be removed.

... since we're talking about what-ifs?

What would happen if we place this block at the center of the Galaxy?

Center by volume? Mass? Popular interest/attention?

I suspect in each case, one result would be it would become unbearable to be around the block.

'I'm the center of the galaxy!'

"Yeah, and it is also unbearably uncomfortable around you, block."

We'll the center of the universe is off limit's Apparently it's nonexistent!?!?!

..There must be something there that ... matters

Hmm, possibly. Of course if the topology of the universe is toroidal (or coffee-cupal), the center by aether volume or mass won't matter.

The center of interest does keep catching my attention,

Then the super-massive black hole at the center of our galaxy would have consumed it.

Game over. Actually, game, set, and match.

The First Law of Thermodynamics holds, but this does not preclude the possibility of recovering energy from another 'brane, as illustrated by NASA's new engine (even though presently it only produces micro-newtons of force).

The mind boggling amount of energy within every cubic centimeter of our universe, is truly a vast amount, 10¹³¹ J/cm³, thus removing even a sliver of that into our reality would seem to (appear to) be pure magic, and in direct violation of the law of conservation of energy. We have yet to master this, but if our race ever does, I fear that we will fail to use it responsibly, and burn ourselves up.

"The mind boggling amount of energy within every cubic centimeter of our universe, is truly a vast amount, 10¹³¹ J/cm³"

That's one I hadn't heard! Where does that come from?

Does that value hold whether or not there is any matter in the cm³?

Boggled my mind too! James, can you explain?

I have to retract, it turns out I was off my some 18 orders of magnitude! See this:

vacuum catastrophe

The number is 10¹¹³ J/cm³, so it is still a vast quantity of energy to tap into, essentially with no penalty in our frame of reference, as long as we don't over do it.

I think the whole topic is interesting, as I am very interested in Heisenberg traps of particles such as protons on a lattice, then introducing a strong perturbation of the lattice. The lattice could be metal crystal such as nickel, other metals, or could it perhaps be something like graphene, where the known proton permeability and transport (through planes of graphene) are high.

I suspect that protons like to be trapped between graphene layers of graphite. Calculations upon the terminal velocity of charged carbon particles in water, revealed a considerable perturbation energy during collisions of oppositely charged macro-particles.

Could protons in such a state suddenly have access to sufficient energy to undergo nuclear electron capture, leading to cold neutrons? Cold neutrons are a fascinating topic of their own. Absorption cross sections have already been proven to be many orders of magnitude higher (across a spectrum of acceptor nuclei) than for thermal neutrons.

One company out in California is already exploiting this by showing a demonstrated energy gain by (apparent nuclear transmutations) discharging high pulsed current in hydrogen exposed nickel wires. There seems to be a progression of captures, leading to ⁴H, (quadrium) which Beta decays to ⁴He.

I think that as time progresses, many such Heisenberg traps will be found that permit access to the vacuum energy.

...one more confirmation that I'm an old classical physicist, and apparently destined to remain such!

I'm still learning stuff that's new to me, and am open to new ideas that can be tested, but this one's beyond me.

Don't feel alone! I have been trying to get my rather limited mind around this ever since I first heard of it. In fact, at first, I dismissed this as bogus mumbo jumbo talking about zero point energy. Perhaps, I will be convinced if I ever see any calorimetric evidence of something happening in a chemical experiment that appears to be nuclear related. Of course, it would be an added plus to detect some light, maybe some radiation or something not in the normal background.

Thanks for that, not something I'd heard of. I haven't studied it fully (not that I'm likely to follow all of it) but one thing in your link caught my eye.

Such arguments are usually based on dimensional analysis and effective field theory. If the universe is described by an effective local quantum field theory down to the Planck scale, then we would expect a cosmological constant of the order of M p l 4 {\displaystyle M_{\rm {pl}}^{4}} . As noted above, the measured cosmological constant is smaller than this by a factor of 10⁻¹²⁰. This discrepancy has been called "the worst theoretical prediction in the history of physics!".

This strikes me as very similar to the ultraviolet catastrophe which paved the way for quantum theory. Classical physics predicted that the energy of electromagnetic waves should be infinite, as each wavelength should contribute energy, and wavelengths down to zero should contribute (I haven't described it too well, but it's on Wiki). It turns out (as I'm sure you know) that below a certain point the energy at each wavelength decreases, so the total energy is finite. Although the Planck length is > zero, it's so far below e.g. nuclear dimensions (10^-20 x proton dia) that perhaps at that scale the energy drops off. But I'm sure physicists have already thought of that.

Not sure about "the worst theoretical prediction in the history of physics!". The Rayleigh-Jeans formula in the ultraviolet catastrophe was out by a factor of infinity!

It is a huge discrepancy between what the vacuum energy is calculated to be and what it is measured to be --- a factor of 10¹²⁰.

Would you mind elaborating a little bit more on that? Who measured the vacuum energy, and how did they do it?

I think he is referring to the upper band of estimates of zero point energy.

But how is it estimated? I'd like to know

Obviously, more matter in that volume means somewhat more energy.

It is there where you are, also here where I am, and everywhere else. We just do not have access to it (normally, and certainly not efficiently). A release of even one cm³ worth would have devastating consequences, would it not?

NASA''s new microwave thrust engine is based on this fact, thus far incredibly inefficient, but real.

What has this to do with 1st (or indeed the zeroth, 2nd or 3rd) law of thermodynamics?

Thermodynamics is about heat (and its conversion to and from mechanical energy) among rather a lot else.

Quite.

The first law of thermodynamics is a LAW. It must be untrue or we wouldn't need a law to make people give it lip service.

ROFLMFAO!

PE=mgh, --> see here, that PE is dependent with 3 independent variables not only height and mass but gravity as well.

First Law simply tells you about the accounting of energy or the law of the conservation of energy.

You need to study the concept of systems and surroundings. You need to isolate things and probably identify the energy sources.

You need to ask your teacher on this.