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I know this is tangential, but as lay person regarding physics I am stuck on the following thought experiment regarding gravity: We know about gravitational po

by phn 6y ago

I know this is tangential, but as lay person regarding physics I am stuck on the following thought experiment regarding gravity: We know about gravitational potential energy. We harness it all the time with dams and whatnot. Generally speaking, when we lift matter high above, we get some of the used energy back when it falls back down. But when we talk about sending say, a spaceship into space, it may fall on a different planet, with a different gravity (a function of the planet's mass), changing the amount of energy yielded by the fall. The question is: Where is the energy stored in the meantime? Where does the extra energy, if the ship falls into a heavier planet, come from? It seems that the energy is being stored somewhere, and that matter itself allows us to tap into that, but is not the place where it is stored. Is there anything I can read to better understand what we know about this?

simonh 6y ago
That turns out to be a really interesting question. In physics the gravitational energy of a massive body is considered to be negative. That takes a bit of getting used to. If you consider the universe after the Big Bang, it consisted of extremely thinly dispersed clouds of hydrogen and helium atoms. Over billions of years these collapsed into galaxies and stars. During the process of that collapse these atoms accelerated towards each other, impacted and heated up considerably due to the release of kinetic energy. This kinetic energy has to come from somewhere, so we consider that the energy of the gravitational field of a star or planet is negative energy. It's an energy deficit. As an side, it is estimated that if you add up the energy in star, including the matter it's composed of, and deduct the negative energy of it's gravitational field they cancel out. It seems likely that the net energy of the universe may well be very close to zero. https://en.wikipedia.org/wiki/Zero-energy_universe https://en.wikipedia.org/wiki/Zero-energy_universe
- goldenkey 6y ago
  I am happy that you provided the only realistic answer here. Gravity is the negative to light...radiation pressure pushes and gravity pulls. We live in a yin yang universe. Like Alan Watts would say, the universe is in the middle of cosmic dance, and eventually all positives and negatives will neutralize and we will be back where we started. This bides well for cyclic theories of cosmology. https://en.m.wikipedia.org/wiki/Radiation_pressure https://en.m.wikipedia.org/wiki/Radiation_pressure Radiation pressure - Wikipedia
  - pestaa 6y ago
    I listen to Alan Watts quite regularly, and always fascinated by the broad range of sciences he can tap into. Even more interesting given the criticism he has towards academia.
    - dmix 6y ago
      aka Hippies feigning science for some pseudo-religious ends.
      - pestaa 6y ago
        Are you sure he misrepresented science? On the contrary, I get the impression that Alan took scientific discoveries and matched those onto his philosophical stances. He objected to the personal beliefs of the stereotypical scientist, and put forward an argument that the science works even if you stop viewing the universe as a purely materialistic mechanism. Which is not to say it then automatically possesses meaning, an intelligent designer or purpose.
  - saagarjha 6y ago
    > This bides well for cyclic theories of cosmology. But the second law of thermodynamics really doesn’t. I should note the radiation pressure does not have much to do with gravity, either, unless the only thing you’re focusing on is whether a star is going to collapse.
  - kleer001 6y ago
    I'd love to hear a refutation to this (however simple) rather than an anonymous and empty downvote.
    - simonh 6y ago
      Gravity and light are unrelated phenomena, one being a curvature of space and the other a fluctuation in electromagnetic fields. They are in no way the opposite of each other. It’s like saying apples are the opposite of oranges. It’s not even intelligibly refutable.
- pdonis 6y ago
  > it is estimated that if you add up the energy in star, including the matter it's composed of, and deduct the negative energy of it's gravitational field they cancel out No, this is not true. If it were, the star's mass would be zero. What "gravitational potential energy is negative" means is that the mass of the star is less than the sum of the mass of all of its constituents. Or, to put it another way, if we make a star out of a highly diffuse cloud of atoms as you describe, the process will have to release energy (by emitting radiation into space that escapes the star system). But the energy left over is far from zero: the mass of the star is energy. > It seems likely that the net energy of the universe may well be very close to zero. This is a speculative hypothesis that is not the same as the usual concept of "gravitational potential energy" being negative. The fact that it is often framed in similar terminology is misleading.
  - simonh 6y ago
    >No, this is not true. If it were, the star's mass would be zero. All it means is that if we were to take all the constituent particles in a star and spread them out at 'infinity' from each other (effectively all across the universe), pushing them apart working against the gravity of the star holding it together, the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc). The energy cost of disassembling a star equals the energy in the star, so the net energy of the star is zero. It's just a matter of doing the relevant calculations to see this, which was first done by Pascual Jordan in the 1940s. I realise it's a ridiculously counter-intuitive result. Apparently when Einstein was told this, he stopped dead in his tracks while crossing a busy road.
    - saagarjha 6y ago
      It doesn’t actually make much sense to do that for each individual particle, because the strong force will prevent some of them from being separated due to color confinement.
    - pdonis 6y ago
      > f we were to take all the constituent particles in a star and spread them out at 'infinity' from each other (effectively all across the universe), pushing them apart working against the gravity of the star holding it together, the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc). This would require that all of the particles in the end state of the disassembly were massless (so all of the rest mass goes away in the disassembly process); but if they are massless, they can't also have zero kinetic energy (the only kind of energy a massless particle can have in the absence of gravity) or they don't exist at all. So I don't think the scenario you are implicitly relying on here is possible. > It's just a matter of doing the relevant calculations to see this, which was first done by Pascual Jordan in the 1940s. Do you have a reference? (And no, I don't mean the pop science references in the "Zero energy universe" Wikipedia article, I mean an actual reference to the published paper by Jordan where he makes the calculations you refer to. Or a more recent paper where someone else makes similar calculations.)
- gpderetta 6y ago
  Apart from the zero-energy universe, which is quite an interesting conjecture, as far as I understand, considering gravitational energy to be negative it is only a convenience. There is really no absolute energy value, the only thing that matter is energy deltas and you can pick a zero point arbitrarily. IANAP of course.
  - lambdatronics 6y ago
    It's a little different in general relativity, because the mass-energy equivalence implies there really is a zero point -- the gravitational energy of the planet would itself add to the apparent mass of the planet, in the same way that the binding energy of nuclei make the nuclei have measurably different masses compared to the sum of the constituents.
    - gpderetta 6y ago
      Can you elaborate on that? I understand that there is a self interaction component in general relativity, but how does that lead to an absolute zero point? Note that 'because the math says so' can be an acceptable answer as I wouldn't be able to understand the math.
      - lambdatronics 6y ago
        I'm not an expert on GR, but basically: energy density and mass density are summed up in the Einstein field equations to get the metric tensor that determines the curvature of spacetime. We can measure the curvature, so we can measure the mass/energy density, in absolute term. This article gives an example of how that would work: https://en.wikipedia.org/wiki/Gravitational_binding_energy#N.. https://en.wikipedia.org/wiki/Gravitational_binding_energy#N....
        crdrost 6y ago
        So, uh, general relativity makes all of this stuff really complicated. First off, the gravitational energy is not a direct component in the stress-energy tensor that serves as a "source" for gravitational curvature. So it is formally incorrect to say that the gravitational potential energy is a source for gravity. However, the Einstein equations are not linear the way the Maxwell equations are, and you can maybe interpret these nonlinearities as a self-interaction of the gravitational field, causing gravity to be a source of gravity. But the analogy does not seem to be as easy as just saying "here is the gravitational potential energy and by E = mc² ..." [1] Secondly, I cannot reiterate enough that general relativity does not in its usual formulations conserve energy. General relativity does not generally give you a good way over a non-infinitesimal part of spacetime to define a volume and sum up all of the energy in that volume, so it makes it very hard to define energy. And then when you try to define it in some obvious ways, you run into some weird paradoxes -- for example due to the expansion of space light from distant stars is redshifted when it arrives at us; one can either do a complicated argument from classical electromagnetic theory or a weaker but more straightforward argument from quantum theory (just track a collection of photons of definite number!) to conclude that this redshift must correspond to a loss of energy. This broader idea that the entire universe has zero net "energy" for some definition of energy does exist in some sparse literature, notably Krauss's pop-sci book A Universe from Nothing, but generally those treatments have not made a very robust case for this[2] and I don't think if I asked professional cosmologists that they would regard it as an established fact of modern cosmology, moreso than maybe just a way of speculating that maybe the universe is closed or asymptotically flat or so. (Asymptotical flatness is one way to try to give yourself an "out" so that you can again define mass, either by the ADM or Bondi methods[3].) Finally, there is indeed a correlation given by thermodynamics and stellar evolution that due to the virial theorem, "the total internal energy of the star is simply −(1/2) of its gravitational binding energy." [4] [1] Baez says a lot of this more eloquently at http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_g... [2] e.g. https://arxiv.org/abs/1405.6091 https://arxiv.org/abs/1405.6091 as critical of that particular book, some further history at https://en.wikipedia.org/wiki/Zero-energy_universe https://en.wikipedia.org/wiki/Zero-energy_universe . Note that the wikipedia article's claim has a very nice formulation of "could a quantum fluctuation create a star given that it apparently has no net energy?" and like I would view the lack of experimental evidence for this as perhaps experimental disconfirmation, but perhaps there is a more detailed calculation which confirms it as rare or some other excuse that means that you can't do it with a star but only a whole universe. [3] This topic is adequately mentioned at https://en.wikipedia.org/wiki/Mass_in_general_relativity https://en.wikipedia.org/wiki/Mass_in_general_relativity . [4] PDF warning, https://websites.pmc.ucsc.edu/~glatz/astr_112/lectures/notes3.pdf https://websites.pmc.ucsc.edu/~glatz/astr_112/lectures/notes...
BariumBlue 6y ago
When you fall, that is potential energy stored via gravity turning into kinetic energy. When you land, that kinetic energy is then turned into thermal energy. If you're interested, this would be physics. Physics 1 (usually kinematics) isn't bad if you like math, though I think physics 2 and beyond start to be demanding
fennecfoxen 6y ago
When you send a rocket into space, you burn chemicals (releasing energy located in electrons in the chemical bonds holding the atoms together) gaining thermal energy (the energy of moving atoms in hot rocket exhaust) which is then redirected into kinetic energy (still moving atoms, but they're all moving in the same direction, and more orderly about it). When the spaceship moves upwards relative to a large mass (Earth), this kinetic energy is traded for gravitational potential energy. It's still associated with the atoms and other particles, just like the kinetic energy was. This does mean that objects in distant space have more of this energy than objects at the bottom of a gravity well, and are therefore more massive (because of mass-energy equivalence). Falling into an intense gravity well is actually one of the more efficient ways to convert mass into energy, which is why the area around black holes is often particularly energetic, with the matter falling in emitting energetic X-rays. This is why you can see quasars even though they are billions of light-years away (the radiation heats the surrounding material and it is very bright.) Besides general physics reading, I can recommend the book "Einstein's Universe" to help build a general understanding of both special and general relativity, with particular regard to matters such as mass-energy equivalence.
- saagarjha 6y ago
  > Falling into an intense gravity well is actually one of the more efficient ways to convert mass into energy, which is why the area around black holes is often particularly energetic, with the matter falling in emitting energetic X-rays. The heat comes from gravitational compression and friction with other material in the accretion disk, not because objects at the bottom of a gravity well have less mass.
  - fennecfoxen 6y ago
    It's true that object in freefall being attracted towards a gravity well does not have less mass as it approaches deeper parts in the well. However, it will have less rest mass from its potential energy, and more mass due to its kinetic energy. If the object continues on a trajectory past the gravity well and out to space beyond, it will lose that kinetic energy, and regain the potential energy. So in that sense, it's true: the overall mass doesn't change. But if the object smacks into something, like an accretion disk around the black hole, or the atmosphere of a planet, then the kinetic energy escapes the object, and is lost to the surrounding system (through mechanisms like friction and compression, as you've mentioned). Per Einstein, this lost energy really does mean the object has lost mass.
opless 6y ago
The heavier an object is, the deeper its ‘gravity well’ is. You’ve seen the space-time is a rubber sheet metaphor I hope? Imagine it like dropping a stone in a literal well. The potential energy is higher on the ground than two meters down the well. Since we are in the gravity well of a planet (earth) the energy needed to reach escape velocity is a certain value, because we have to climb the walls of the gravity well. A denser planet has a deeper gradient so requires more energy to escape. Falling down the gravity well requires no energy, but the difference of potential energy in “outer space” and “on the planet” is what you’d experience. There’s no energy stored per se, when you pump water up into a dam or when you enter orbit, as you expend the energy into work, resulting in a difference of potential energy in two rest states.
- k__ 6y ago
  Why are they called gravity wells and not gravity sinks?
  - opless 6y ago
    I’d just say that English is a obscure and difficult language. :-)
  - cdirkx 6y ago
    Because you can get out of them again. A sink, to me, has the connotation of something being removed from a system.
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auntienomen 6y ago
The energy is stored in the gravitational field. Gravitational fields can carry energy and momentum. The same is true for electromagnetic fields, as anyone sitting in warm sunlight has noticed.
jiggawatts 6y ago
I was going to write up a long-winded spiel about rubber sheets and various ways to look at gravity, including my own unique perspective, then I realised that XKCD has already covered this in "Gravity Wells": https://xkcd.com/681/ https://xkcd.com/681/ You're just describing the "landscape" of a field. Any field works this way, including the electromagnetic field, it's just that it tends to cancel out over long distances. Your scenario is like two people living in shallow valleys atop a high mesa discussing the "free energy" that can be gained by going down to the ocean.
acqq 6y ago
> But when we talk about sending say, a spaceship into space, it may fall on a different planet, with a different gravity (a function of the planet's mass), changing the amount of energy yielded by the fall. I guess you imagine there "failing" as something we experience on the surface of the Earth when we drop an object. It's not what is happening when "sending a spaceship into space." It's actually very hard (as in, one really has to spend a lot of energy) to get a spaceship to some star or planet. One can't imagine it as "just let it fall". I suggest reading the following article for the start: https://www.forbes.com/sites/startswithabang/2016/10/01/ask-ethan-why-dont-we-shoot-earths-garbage-into-the-sun/ https://www.forbes.com/sites/startswithabang/2016/10/01/ask-... "Ask Ethan: Why Don't We Shoot Earth's Garbage Into The Sun?" Much less energy is needed to shoot something out of the Solar system than is needed to send something into the Sun, even if you want "just" to send it there to burn. Nothing from Earth can "just fall" there, huge amounts of energy from the outside are needed. What is actually happening is what already Newton figured out: when we're on Earth, we aren't aware of it but the whole Earth is constantly free failing towards the Sun, and yet we're not leaving the orbit around the Sun. > Where does the extra energy, if the ship falls into a heavier planet, come from? A lot of the energy needed for the ship to reach the planet has to be spent by the ship, for the current ships it is launched as the fuel with the ship and later "spent" by the ship (in reality "traded"). The ship can also "spend" the energy of other objects and that energy is really then "spent" (as in bookkeeping, debited to other account, that's what the conservation of energy and momentum is, a kind of bookkeeping) -- the object (e.g. a planet) is indeed slowing down a little, but the change is conveniently small enough: https://en.wikipedia.org/wiki/Gravity_assist https://en.wikipedia.org/wiki/Gravity_assist "This explanation might seem to violate the conservation of energy and momentum, apparently adding velocity to the spacecraft out of nothing, but the spacecraft's effects on the planet must also be taken into consideration to provide a complete picture of the mechanics involved. The linear momentum gained by the spaceship is equal in magnitude to that lost by the planet, so the spacecraft gains velocity and the planet loses velocity. However, the planet's enormous mass compared to the spacecraft makes the resulting change in its speed negligibly small even when strictly compared to the orbital perturbations planets undergo due to interactions with other celestial bodies on astronomically short timescales." In short, there's nothing "extra" happening -- all the energies and momentums are completely accounted for for everything that happens on the scales reachable to our spaceships. The only energy we can't account to anything else is the one we call "dark" but it is observable just in the movements of the whole galaxies, on the scales practically unreachable to our spaceships.
zelphirkalt 6y ago
Just a lay person here. I would say, that that "potential energy" is not really energy, at least not in the sense of "the energy is inside this object" or "this object has so-and-so much energy". I would say you move an object into a different environment and the environment is all that is needed to make the object act in a certain way. If that environment is a heavier planet, then it means, that the object will behave differently than on Earth. The energy you spent to bring the object away from Earth if you want, is stored in the fact, that this object is now in a different environment than before you spent that energy. If the object were to return to its place, it would make use of energy the other way around. Looking at the object alone, you cannot see any of that energy, because it is not really in there, but stored in the fact, that this object is in a different environment. That is what gives it potential.
- cygx 6y ago
  Potential energy is energy associated with a system's configuration. For example, arrangements of charges can be more or less energetic, as in, you need to perform or can extract work when you go from one configuration to another. This is a rather abstract notion. However, as energy (or rather, stress-energy-momentum) is the source of gravity, you actually do need to know precisely where that energy is located if you want to do general relativity. In case of the example above, the energy will be stored in the electromagnetic field. The gravitational field itself is an exception to this: While it can be used to store energy, in the general case, it's impossible to locate it. I'd argue this is a consequence of general relativity's unification of gravity and inertia.
Tade0 6y ago
The question is: Where is the energy stored in the meantime? Where does the extra energy, if the ship falls into a heavier planet, come from? From a classical physics perspective everything that has mass is immersed in the gravitational field of everything else, it's just that most of that potential energy turns into kinetic energy only when you get very close to something. Basically your potential energy is this mind-bogglingly large number which would only approach zero if the whole universe were condensed into a single point.
martin-adams 6y ago
Here's my lay person thinking on this. By lifting objects up or down, you're altering the gravity wells of each object. A spaceship leaving Earth would make Earth's well a bit smaller. The question you pose makes me think the following: if it requires no energy to fall down a gravity well, but you can use that falling to generate energy (hydroelectric dam for example), then where does that energy come from? So this makes me think that as the following: 1. Increasing gravity wells (object falling) creates energy debt to the well (the debt is larger), in exchange for harnessed energy by things like hydroelectric dams. 2. Decreasing gravity wells (object being lifted) pays off the energy debt from the well by you using energy in the lifting action. So for me, there's no energy stored, it's just a debt in the form of the size of a gravity well.
- pdonis 6y ago
  > A spaceship leaving Earth would make Earth's well a bit smaller. While this is technically true, the effect is much too small to measure, and is not what is responsible for the usual phenomena we associate with gravitational potential energy.
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Koshkin 6y ago
> if the ship falls into a heavier planet I would be a genuine paradox if this other planet appeared out of nowhere at some point in the process. That's the key: both the source and the destination mass points have been there all along, and so you must, as usual, take into account all the bodies involved from the very beginning.
noobermin 6y ago
So there are cases in which you could consider energy being "stored" somewhere. The canonical example from EM is energy being stored in the EM field itself like in radiation. In this case for classical static gravitation fields, the potential energy has more to due the motion of the orbiting mass and the nature of the force itself being conservative and so it isn't really stored somewhere in that sense. What I mean by that is a mass that moves against that gravitation force will lose motion due to that force (specifically motion with a component away from the planet) and from math you can then calculate that kinetic energy that would be lost from the orbiting body if it were to get a certain distance away from the mass for example. This process is reversible, that is you can trade distance for kinetic energy, which is a familiar situation like that of a ball being thrown straight up in the air, slowing and then stopping at some height, then falling back down. The reversibility of this trade off then allows you to associate an energy, potential energy, with given heights off the ground which makes calculations easier by considering it as a trade off between the motion and these greater distances. This however isn't the case for all forces. For air resistance this isn't the case, so you can't associate air resistance with a potential energy as a counterexample. There are however cases in which energy is real in a sense and isn't just a convenience, and for the most part this is when that energy is associated with motion (momentum specifically) of something. Also, that something would be localized and thus could be considered being stored somewhere. For some examples, of course kinetic energy, but for EM energy, you could consider it being the momentum carried by photons after QED.
visarga 6y ago
> Where does the extra energy, if the ship falls into a heavier planet, come from? It comes from the heavier planet. It also falls onto the spaceship, just slightly changing its trajectory.
ksec 6y ago
I thought it is actually a very interesting question. Hopefully this stays on the front page for another few hours when HN get most of its traffics. Then a physicist will likely pop in with an answer.
lazyjones 6y ago
> It seems that the energy is being stored somewhere, and that matter itself allows us to tap into that, but is not the place where it is stored. It's not stored anywhere, it's always present since the gravitational masses are always present, it's just that a part of it turns into kinetic energy when you are close to large masses (and back once you leave them).
smdz 6y ago
My thoughts on this - The energy used in sending a craft up is already lost when the craft leaves the planet. And potential energy can only be induced into the object from a larger mass. That is based on the object's mass and the large mass like a planet pulling the object. And the "value" (read: usefulness/utility) of that potential energy is based on where it exists P.S. I am not a physics guy and could be completely wrong here.
brandmeyer 6y ago
There's a bunch of half-answers in this thread. > Generally speaking, when we lift matter high above, we get some of the used energy back when it falls back down. > Where is the energy stored in the meantime? Potential energy in a gravity well is negative. The "zero" point is the limit when you get far away from all gravitational bodies. The better question is: Where does the kinetic energy come from when you descend down the gravity well in the first place? The answer is mass. All potential energy is ultimately realized as a change in mass, equal to the famous `mass_difference * speed_of_light^2`. As the body falls down the gravity well, it gains kinetic energy and loses mass. When atoms bind together in an exothermic reaction, their final state is bound more tightly than their initial state. They released some energy in the form of molecular-scale kinetic energy (ie, heat), and they have lost mass. When atoms change configuration in an endothermic reaction, their final state is bound less tightly than their initial state. Some energy was required to push them up to that higher-energy state (sometimes in the form of heat), and they have gained mass. When light nuclei fuse together, their final state is more tightly bound than their initial state. They released some energy in the form of atomic-scale kinetic energy (ie, heat) and a photon, and they have lost mass. Ironically, when very heavy nuclei have fissioned, their final state is more tightly bound than their initial state. They released some energy in the form of atomic-scale kinetic energy (ie, heat) and photons, and they have lost mass. When a spring is compressed, the molecules within the elastic material are stretched apart from one another such that their final state is less tightly bound than their initial state, and the spring has gained mass.
- saagarjha 6y ago
  > Potential energy in a gravity well is negative. The "zero" point is the limit when you get far away from all gravitational bodies. Like all potential energies, its definition is relative and its zero point convention. > As the body falls down the gravity well, it gains kinetic energy and loses mass. So you’re a saying that hydrogen lower in a gravity well will fuse worse?
  - brandmeyer 6y ago
    > Like all potential energies, its definition is relative and its zero point convention. Turns out that this one does have a unique definition because the limit exists, and because the sum of kinetic energy and rest mass determines the gravitational influence of a body. > > As the body falls down the gravity well, it gains kinetic energy and loses mass. > So you’re a saying that hydrogen lower in a gravity well will fuse worse? No. The processes are orthogonal to each other. Forget for the moment that its the release of kinetic energy by dropping down the gravity well that makes the subsequent fusion possible at all. Whether you fused first and then dropped down the well, or dropped down the well first and then fused, the final state is the same.
thanatropism 6y ago
I'm not a physicist, but my master's thesis was on numerical ODEs preserving physical (symplectic structure). I think the best way of looking at these problems is thinking of space already equipped with a vector field (like streamline plots with little arrows) and objects just being carried by them, as if laying down small rocks on a busy river. Of course, this vector field is not constant and the object's mass alters it, but this will be small in the spaceship vs. planets setting. This is more of a mathematical intuition than history-of-the-universe answer, but I think it should help. Ed: This is very simple to reason about when talking about independent particles in one dimension; the phase space arises out of a simple variational problem. It's also easy to reason about in 3 dimensions, but the problem is that the "river" in one dimension has two coordinates, momentum and position. So to think of two dimensional space you have to imagine four dimensions, etc. But thanks to Darboux's theorem these dimensions are "coupled in pairs" (this is what symplectic means), and it's not that difficult to visualize four dimensions. Cf. this illustration of the "symplectic camel theorem": https://encrypted-tbn0.gstatic.com/images?q=tbn%3AANd9GcRtOKkv_wL9wnSkr7rkCxm11ZOmgKXjNBUsVGIkE2BVYlryctvd&usqp=CAU https://encrypted-tbn0.gstatic.com/images?q=tbn%3AANd9GcRtOK...
pdonis 6y ago
> when we talk about sending say, a spaceship into space, it may fall on a different planet, with a different gravity (a function of the planet's mass), changing the amount of energy yielded by the fall. In a situation like this, you have to consider the total energy of the system, including both massive bodies and the potential energy due to them. The total energy remains constant, but how it is distributed will change: the potential energy of the spaceship will be different at the end than it was at the start. > It seems that the energy is being stored somewhere It is stored in the geometry of spacetime; at least, that's the only way I know of to conceptualize it.
- acqq 6y ago
  > It is stored in the geometry of spacetime And because changing the relative position of objects changes the geometry and the conservation of energy and momentum exists, can't we say that we don't even have to think about the geometry as the "store", but that it's enough to know that it is a property of the whole system we observe and that when it changes the conservation is not violated?
  - pdonis 6y ago
    > changing the relative position of objects changes the geometry Yes, although the change due to the spaceship moving is far too small to measure. (Technically, the global spacetime geometry doesn't "change", since spacetime is a 4-dimensional geometry that already contains all the information about "changes" in it, so spacetime itself doesn't change at all. But if we think of the local spacetime geometry changing along the worldline of an object like the spaceship, that concept makes sense.) > conservation of energy and momentum exists You have to be careful with this. There are actually three different senses in which this can be taken in GR, a local one and two global ones. The local sense is that the covariant divergence of the stress-energy tensor is zero. This basically says that matter and energy in tangible form (which does not include "energy stored in the gravitational field" or "energy stored in spacetime geometry") can't be created or destroyed. The first global sense is that in certain kinds of spacetimes, we can do integrals that evaluate the "total energy" of the spacetime, and these integrals will obey certain conservation laws. But those laws don't necessarily correspond to what we normally think of as "conservation of energy" in a global sense. The second global sense is that, again in certain kinds of spacetimes, there is a constant of free-fall motion that can be interpreted as the kinetic plus potential energy of the free-falling object, such as a spaceship on a free-fall trajectory between two planets. This is the sense that has implicitly been used in this discussion. But this constant of free-fall motion only exists in stationary spacetimes, and in stationary spacetimes, the relative positions of gravitating masses cannot change with time. If the relative positions of gravitating masses do change with time, the spacetime is not stationary and the constant of free-fall motion that is interpreted as kinetic plus potential energy does not exist. So there is no such thing as a spacetime which has changing relative positions of objects and conservation of energy in the sense we have been using that term in this discussion.
bcgraham 6y ago
I always thought it was stored in the “angle” of your geodesic. When you come out of the gravity well, your travel through spacetime is angled to be traversing more time relative to space, making you look like you’re moving slower (when considering only spatial dimensions). As you fall into a gravity well, spacetime curvature changes your geodesic such that you traverse more space relative to time, meaning your travel looks faster in the spatial dimensions.
Rury 6y ago
>It seems that the energy is being stored somewhere, and that matter itself allows us to tap into that Like others have mentioned, you need to look at the two planets and the spaceship as a system together. The gravitational potential energy you gain from raising a spaceship out of one planet's gravity is equal to the energy you spend getting it out of that planet's gravity well (neglecting efficiencies). While the trip from Planet A to Planet B might seemingly "gain" you extra potential energy because of the differences in gravity wells. The reverse trip of going from Planet B to Planet A will reverse that "extra" potential energy you gained as it's more costly to get out of Planet B's gravity well. As so, there's no "extra" or created energy within the system. As for where energy is stored, it's simply stored as gravitational potential energy.
- gizmo686 6y ago
  > it's simply stored as gravitational potential energy This answers the "how", but not the "where". Since energy itself has mass, we would expect the potential energy to carry mass as well; and that that mass/energy should be concentrated in some region of space.
  - Rury 6y ago
    >This answers the "how", but not the "where Potential energy is a property of a system rather than an actual physical thing. It's not energy an object has, it's energy an object potentially has relatively speaking. And it's relative to whatever you use as your reference point in your system. >Since energy itself has mass Energy doesn't have mass. It is equivalent to mass.
    - kwk1 6y ago
      This is a good point. Potential energy is relative to a datum, which is arbitrary. So you could say a box on a table has no potential gravitational energy relative to the table, but it does have some relative to the floor. But there was no actual change in energy between the two measurements.
    - cygx 6y ago
ozy 6y ago
Bending your spacetime frame of reference costs energy because you have to leverage something and change its frame. And you can use your bend spacetime frame to extract energy by bringing it in contact with another frame. But the bending itself, or maintaining it, or changing it, doesn't cost or yield energy.
bluGill 6y ago
It doesn't go away. Newtons law of gravity still applies F = Gm1m2/r^2. When you are on earth the gravity of Mars is still affecting you, but r is such a large number (and then we square it) that you can basically ignore it, but it is still there even if we can't measure it. When you get in the rocket you increase the r for how earth's gravity affects you, eventually getting to the point where it is insignificant and then where Mars gravity dominates everything else and so we ignore it, but earth is still affecting things even though it couldn't be measured. The above applies to the moon, sun, even every atom in the universe. Fortunately most of them are far enough away that we can ignore them: we don't actually know how to calculate all the math if we wanted to account for them all.
- saagarjha 6y ago
  This answers a different question, which is “where does the gravitational force go when I can’t feel it”.
d_tr 6y ago
> The question is: Where is the energy stored in the meantime? Where does the extra energy, if the ship falls into a heavier planet, come from? You can say it came from the fact that there existed a point in space with a lower gravitational potential than the one at the spaceship's point of departure. Each point p in space is assigned a "gravitational potential" value V(p) measured in energy / mass. The function V(p) is determined by the distribution of mass in the space. A body of mass m moving from point a to point b will gain "gravitational potential energy" equal to (V(b) - V(a)) * m. In your case, the spaceship moved from a higher to a lower gravitational potential, thus losing gravitational potential energy (and gaining kinetic energy). Note that you are only interested in the differences of values of V at different points, so you can add or subtract any constant to the function V.
mercer 6y ago
I'm currently reading The Information (I highly recommend it!) and while I feel a vague sense of an answer, I'll defer to others more knowledgeable. I just wanted to say that I think this is a wonderful question! Thanks!
SuoDuanDao 6y ago
The best description I've heard is that energy is a measure of relationship. The spaceship on earth and in space has a relationship with the other planet the whole time, there's just very little that will happen as a result of that relationship until the distance between the two gets smaller.
archgoon 6y ago
So the term you want to search for is "Conservative Forces". I'll give a brief explanation here so you can find a better one more easily. :) Let's take a few steps back and talk about forces, and work. Gravity isn't really a key part of your question. Suppose you had a uniform downward force field like so: | | | | | | | | v v v v v v v v | | | | | | | | v v v v v v v v | | | | | | | | v v v v v v v v Those are vectors pointing downwards. Now suppose you built a miniature roller coaster. Your first one is super boring | | | | | | | | v v v v v v v v car |---\ <--------------------> | | | | | | | | v v v v v v v v It just goes back and forth, perpendicular to the field. Now, if you really grease up the tracks, and have really good bumpers on the end (noiseless, don't heat up, a perfect elastic collision), the car will just bounce back and forth for a very long time, and not speed up or slow down much at all (and ideally, none at all). There is a concept of 'work' in physics. When a force points in the direction of motion, we say that a force does work. The total work done along a path is simply the sum (integral) of all the parts of the path, multiplied by their length, multiplied by the amount of force in the direction of the path. This definition chosen so that the amount of kinetic energy gained, or lost, is equal to the work done. Since the force is perpendicular for the above path, the 'work' done is said to be zero. Now let's have a more interesting path. Let's give the car a kick. car <- /---| /--------<-------\ | | v ^ | | \------->--------/ Here we have two parts, top and bottom, where the force does no work. However, we have a part going down, and a part going up. The car will accelerate going down, and decelerate going up. If we grease the tracks up, then the car will always be traveling at the same speed on the top track. It will also always be moving faster on the bottom track. We say that the force does 'work' when the car is on the tracks going down and going up. Now, the total amount of work that the force does, for any loop, ends up being zero. This is clear here, since the distance the force is applied downwards, is the same as the distance of the force applied upwards. What is slightly less obvious is that any looped track we could build here would have the same property. If we had something like /---\ | | \ | \ | \| (and ignore the mechanics of those super sharp turns) we would find that the work done was still zero. The downward diagonal bits would end up still adding up the same amount of work as the upward straight bits (bit of vector math shows this). The fact that the work done on any loop is zero makes this particular force a 'conservative force'. Not all force fields need to be conservative! Suppose our field was only present on the left side. /---| /--------------------\ | | | | v | v | | | | | | | v | v | | | | | | | v | v | \--------------------/ This would only do work going down. This would mean that the track car would keep getting accelerated faster and faster. Potential energy only makes sense as a concept in the presence of conservative fields. You define a reference point, sometimes chosen to be at infinity, and define the potential energy at each point to be the amount of work done to bring the object to that test point. However, you don't need to just be a constant field to be conservative. Gravity, as expressed by Newton, is a conservative field. Regardless of how you arrange a bunch of objects or planets, looped paths will always have zero net work done. This is because the mathematical form of Gravity can be epxressed as the gradient of a scalar (takes a position, gives a number) function. You seem happy with the idea that locally on earth, energy is conserved because you get the energy back when you travel in a loop. This doesn't change when you travel in a more complicated path from one planet to another. The force field does work on the object to accelerate it when it moves in the direction of the force, and decelerates it when you pull it back. The big takeaway point here is that energy, as a concept, is built on top of (at least classically) the more primitive concept of 'force'.
agumonkey 6y ago
It's funny indeed that a field is not constant and decreases over space so after some point the potential goes "down" ?
raverbashing 6y ago
You're confused because basically when we think of potential energy on the surface of the Earth that's an approximation In case of a spaceship going from one planet to another you need to consider the potential energy function for both planets in relation to each other (and the Sun's potential energy as well) Potential energy on the surface is more like going up and down a ramp in a line. Potential energy in the case of multiple planets is more like a curvy skate park, where you can go from one place to another but have a small change in potential energy in the end. Also, potential energy is always relative to something. So if you go to another planet you shouldn't be calculating the potential energy from its surface, but from where you came from.
steve76 6y ago
> Where is the energy stored in the meantime? Where does the extra energy, if the ship falls into a heavier planet, come from? Take your reference from the Sun, not the planets.
gmantg 6y ago
Kinetic energy is fake. We may as well say that it's planet falling on the ship and thus it's energy is the mass of the planet multiplied by v^2/2 and where did that energy come from?
JProthero 6y ago
>It seems that the energy is being stored somewhere, and that matter itself allows us to tap into that, but is not the place where it is stored. Is there anything I can read to better understand what we know about this? Here are some quick recommendations for views on potential and gravitational energy: This [1] is a short video from a presentation given by Alan Guth — an early proponent of inflationary cosmology — in which he outlines a thought experiment designed to illustrate how gravitational fields are associated with negative energy. This thought experiment is a standard feature of Guth's introductory talks on inflation; recordings of his academically-oriented talks with more rigorous treatments of the subject are available on YouTube. The logic of the thought experiment that Guth describes in the presentation is not controversial among physicists, but there are different perspectives on how to describe it. Guth's former MIT colleague, Sean Carroll, prefers to avoid descriptions involving negative energy and says instead that energy is not conserved in General Relativity. He explains his perspective here [2]. The cosmologists Luke Barnes and Geraint Lewis discuss different perspectives on fields and energy conservation in this video [3]. In this video [4], the theoretical physicist Sabine Hossenfelder briefly discusses the physical reality of potentials in the context of the Aharonov-Bohm effect. If you have several hours, Sean Carroll has a video series about the Biggest Ideas in the Universe [5]. I haven't been through the playlist yet myself, but I imagine he might go into some detail on the questions you raise, particularly in his discussion of Conservation (1), Force, Energy and Action (3), Spacetime (6) and Fields (9). The Q&A videos following each instalment might also be useful, because his audience frequently asks questions like yours. [1] https://www.youtube.com/watch?v=15IGPyRHOaY https://www.youtube.com/watch?v=15IGPyRHOaY [2] http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/ http://www.preposterousuniverse.com/blog/2010/02/22/energy-i... [3] https://youtu.be/bcE5RQ7A7Ys?t=423 https://youtu.be/bcE5RQ7A7Ys?t=423 [4] https://youtu.be/0GCHzbmMZf0?t=144 https://youtu.be/0GCHzbmMZf0?t=144 [5] https://www.youtube.com/playlist?list=PLrxfgDEc2NxZJcWcrxH3jyjUUrJlnoyzX https://www.youtube.com/playlist?list=PLrxfgDEc2NxZJcWcrxH3j...
gpsx 6y ago
It is useful to think about all the energy in the problem. There is a case where the only energy in the problem is the kinetic energy of the spaceship and the potential energy from gravity. That is the case of firing the spaceship with a giant catapult so that it goes and hits the other planet. In this case, it will be going really fast when it leaves earth. It will slow down as it moves away from earth, leaving its gravity. Then, it accelerates as it falls towards the heavier planet. Then splat, even faster than it was moving when it left earth. At any point, the change in kinentic energy will be the negative of the change in potential energy. A more common scenario, at least in our imagination, is that a thruster on the spacecraft is adding and removing lots of energy from the spacecraft. The math will still work out. And this way there is hope that the spacehip ends up at rest, not destroyed, on the larger planet. But this time the thruster did lots of work and has to be included in the energy balance.

by phn 6y ago

simonh 6y ago

That turns out to be a really interesting question. In physics the gravitational energy of a massive body is considered to be negative. That takes a bit of getting used to. If you consider the universe after the Big Bang, it consisted of extremely thinly dispersed clouds of hydrogen and helium atoms. Over billions of years these collapsed into galaxies and stars. During the process of that collapse these atoms accelerated towards each other, impacted and heated up considerably due to the release of kinetic energy. This kinetic energy has to come from somewhere, so we consider that the energy of the gravitational field of a star or planet is negative energy. It's an energy deficit. As an side, it is estimated that if you add up the energy in star, including the matter it's composed of, and deduct the negative energy of it's gravitational field they cancel out. It seems likely that the net energy of the universe may well be very close to zero. https://en.wikipedia.org/wiki/Zero-energy_universe https://en.wikipedia.org/wiki/Zero-energy_universe

goldenkey 6y ago
I am happy that you provided the only realistic answer here. Gravity is the negative to light...radiation pressure pushes and gravity pulls. We live in a yin yang universe. Like Alan Watts would say, the universe is in the middle of cosmic dance, and eventually all positives and negatives will neutralize and we will be back where we started. This bides well for cyclic theories of cosmology. https://en.m.wikipedia.org/wiki/Radiation_pressure https://en.m.wikipedia.org/wiki/Radiation_pressure Radiation pressure - Wikipedia
- pestaa 6y ago
  I listen to Alan Watts quite regularly, and always fascinated by the broad range of sciences he can tap into. Even more interesting given the criticism he has towards academia.
  - dmix 6y ago
    aka Hippies feigning science for some pseudo-religious ends.
    - pestaa 6y ago
      Are you sure he misrepresented science? On the contrary, I get the impression that Alan took scientific discoveries and matched those onto his philosophical stances. He objected to the personal beliefs of the stereotypical scientist, and put forward an argument that the science works even if you stop viewing the universe as a purely materialistic mechanism. Which is not to say it then automatically possesses meaning, an intelligent designer or purpose.
- saagarjha 6y ago
  > This bides well for cyclic theories of cosmology. But the second law of thermodynamics really doesn’t. I should note the radiation pressure does not have much to do with gravity, either, unless the only thing you’re focusing on is whether a star is going to collapse.
- kleer001 6y ago
  I'd love to hear a refutation to this (however simple) rather than an anonymous and empty downvote.
  - simonh 6y ago
    Gravity and light are unrelated phenomena, one being a curvature of space and the other a fluctuation in electromagnetic fields. They are in no way the opposite of each other. It’s like saying apples are the opposite of oranges. It’s not even intelligibly refutable.
pdonis 6y ago
> it is estimated that if you add up the energy in star, including the matter it's composed of, and deduct the negative energy of it's gravitational field they cancel out No, this is not true. If it were, the star's mass would be zero. What "gravitational potential energy is negative" means is that the mass of the star is less than the sum of the mass of all of its constituents. Or, to put it another way, if we make a star out of a highly diffuse cloud of atoms as you describe, the process will have to release energy (by emitting radiation into space that escapes the star system). But the energy left over is far from zero: the mass of the star is energy. > It seems likely that the net energy of the universe may well be very close to zero. This is a speculative hypothesis that is not the same as the usual concept of "gravitational potential energy" being negative. The fact that it is often framed in similar terminology is misleading.
- simonh 6y ago
  >No, this is not true. If it were, the star's mass would be zero. All it means is that if we were to take all the constituent particles in a star and spread them out at 'infinity' from each other (effectively all across the universe), pushing them apart working against the gravity of the star holding it together, the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc). The energy cost of disassembling a star equals the energy in the star, so the net energy of the star is zero. It's just a matter of doing the relevant calculations to see this, which was first done by Pascual Jordan in the 1940s. I realise it's a ridiculously counter-intuitive result. Apparently when Einstein was told this, he stopped dead in his tracks while crossing a busy road.
  - saagarjha 6y ago
    It doesn’t actually make much sense to do that for each individual particle, because the strong force will prevent some of them from being separated due to color confinement.
  - pdonis 6y ago
    > f we were to take all the constituent particles in a star and spread them out at 'infinity' from each other (effectively all across the universe), pushing them apart working against the gravity of the star holding it together, the energy required would be equal to the energy in those particles (from their mass, nuclear forces, etc). This would require that all of the particles in the end state of the disassembly were massless (so all of the rest mass goes away in the disassembly process); but if they are massless, they can't also have zero kinetic energy (the only kind of energy a massless particle can have in the absence of gravity) or they don't exist at all. So I don't think the scenario you are implicitly relying on here is possible. > It's just a matter of doing the relevant calculations to see this, which was first done by Pascual Jordan in the 1940s. Do you have a reference? (And no, I don't mean the pop science references in the "Zero energy universe" Wikipedia article, I mean an actual reference to the published paper by Jordan where he makes the calculations you refer to. Or a more recent paper where someone else makes similar calculations.)
gpderetta 6y ago
Apart from the zero-energy universe, which is quite an interesting conjecture, as far as I understand, considering gravitational energy to be negative it is only a convenience. There is really no absolute energy value, the only thing that matter is energy deltas and you can pick a zero point arbitrarily. IANAP of course.
- lambdatronics 6y ago
  It's a little different in general relativity, because the mass-energy equivalence implies there really is a zero point -- the gravitational energy of the planet would itself add to the apparent mass of the planet, in the same way that the binding energy of nuclei make the nuclei have measurably different masses compared to the sum of the constituents.
  - gpderetta 6y ago
    Can you elaborate on that? I understand that there is a self interaction component in general relativity, but how does that lead to an absolute zero point? Note that 'because the math says so' can be an acceptable answer as I wouldn't be able to understand the math.
    - lambdatronics 6y ago
      I'm not an expert on GR, but basically: energy density and mass density are summed up in the Einstein field equations to get the metric tensor that determines the curvature of spacetime. We can measure the curvature, so we can measure the mass/energy density, in absolute term. This article gives an example of how that would work: https://en.wikipedia.org/wiki/Gravitational_binding_energy#N.. https://en.wikipedia.org/wiki/Gravitational_binding_energy#N....
      - crdrost 6y ago
        So, uh, general relativity makes all of this stuff really complicated. First off, the gravitational energy is not a direct component in the stress-energy tensor that serves as a "source" for gravitational curvature. So it is formally incorrect to say that the gravitational potential energy is a source for gravity. However, the Einstein equations are not linear the way the Maxwell equations are, and you can maybe interpret these nonlinearities as a self-interaction of the gravitational field, causing gravity to be a source of gravity. But the analogy does not seem to be as easy as just saying "here is the gravitational potential energy and by E = mc² ..." [1] Secondly, I cannot reiterate enough that general relativity does not in its usual formulations conserve energy. General relativity does not generally give you a good way over a non-infinitesimal part of spacetime to define a volume and sum up all of the energy in that volume, so it makes it very hard to define energy. And then when you try to define it in some obvious ways, you run into some weird paradoxes -- for example due to the expansion of space light from distant stars is redshifted when it arrives at us; one can either do a complicated argument from classical electromagnetic theory or a weaker but more straightforward argument from quantum theory (just track a collection of photons of definite number!) to conclude that this redshift must correspond to a loss of energy. This broader idea that the entire universe has zero net "energy" for some definition of energy does exist in some sparse literature, notably Krauss's pop-sci book A Universe from Nothing, but generally those treatments have not made a very robust case for this[2] and I don't think if I asked professional cosmologists that they would regard it as an established fact of modern cosmology, moreso than maybe just a way of speculating that maybe the universe is closed or asymptotically flat or so. (Asymptotical flatness is one way to try to give yourself an "out" so that you can again define mass, either by the ADM or Bondi methods[3].) Finally, there is indeed a correlation given by thermodynamics and stellar evolution that due to the virial theorem, "the total internal energy of the star is simply −(1/2) of its gravitational binding energy." [4] [1] Baez says a lot of this more eloquently at http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_gr.html http://math.ucr.edu/home/baez/physics/Relativity/GR/energy_g... [2] e.g. https://arxiv.org/abs/1405.6091 https://arxiv.org/abs/1405.6091 as critical of that particular book, some further history at https://en.wikipedia.org/wiki/Zero-energy_universe https://en.wikipedia.org/wiki/Zero-energy_universe . Note that the wikipedia article's claim has a very nice formulation of "could a quantum fluctuation create a star given that it apparently has no net energy?" and like I would view the lack of experimental evidence for this as perhaps experimental disconfirmation, but perhaps there is a more detailed calculation which confirms it as rare or some other excuse that means that you can't do it with a star but only a whole universe. [3] This topic is adequately mentioned at https://en.wikipedia.org/wiki/Mass_in_general_relativity https://en.wikipedia.org/wiki/Mass_in_general_relativity . [4] PDF warning, https://websites.pmc.ucsc.edu/~glatz/astr_112/lectures/notes3.pdf https://websites.pmc.ucsc.edu/~glatz/astr_112/lectures/notes...

BariumBlue 6y ago

When you fall, that is potential energy stored via gravity turning into kinetic energy. When you land, that kinetic energy is then turned into thermal energy. If you're interested, this would be physics. Physics 1 (usually kinematics) isn't bad if you like math, though I think physics 2 and beyond start to be demanding

fennecfoxen 6y ago

When you send a rocket into space, you burn chemicals (releasing energy located in electrons in the chemical bonds holding the atoms together) gaining thermal energy (the energy of moving atoms in hot rocket exhaust) which is then redirected into kinetic energy (still moving atoms, but they're all moving in the same direction, and more orderly about it). When the spaceship moves upwards relative to a large mass (Earth), this kinetic energy is traded for gravitational potential energy. It's still associated with the atoms and other particles, just like the kinetic energy was. This does mean that objects in distant space have more of this energy than objects at the bottom of a gravity well, and are therefore more massive (because of mass-energy equivalence). Falling into an intense gravity well is actually one of the more efficient ways to convert mass into energy, which is why the area around black holes is often particularly energetic, with the matter falling in emitting energetic X-rays. This is why you can see quasars even though they are billions of light-years away (the radiation heats the surrounding material and it is very bright.) Besides general physics reading, I can recommend the book "Einstein's Universe" to help build a general understanding of both special and general relativity, with particular regard to matters such as mass-energy equivalence.

saagarjha 6y ago
> Falling into an intense gravity well is actually one of the more efficient ways to convert mass into energy, which is why the area around black holes is often particularly energetic, with the matter falling in emitting energetic X-rays. The heat comes from gravitational compression and friction with other material in the accretion disk, not because objects at the bottom of a gravity well have less mass.
- fennecfoxen 6y ago
  It's true that object in freefall being attracted towards a gravity well does not have less mass as it approaches deeper parts in the well. However, it will have less rest mass from its potential energy, and more mass due to its kinetic energy. If the object continues on a trajectory past the gravity well and out to space beyond, it will lose that kinetic energy, and regain the potential energy. So in that sense, it's true: the overall mass doesn't change. But if the object smacks into something, like an accretion disk around the black hole, or the atmosphere of a planet, then the kinetic energy escapes the object, and is lost to the surrounding system (through mechanisms like friction and compression, as you've mentioned). Per Einstein, this lost energy really does mean the object has lost mass.

opless 6y ago

The heavier an object is, the deeper its ‘gravity well’ is. You’ve seen the space-time is a rubber sheet metaphor I hope? Imagine it like dropping a stone in a literal well. The potential energy is higher on the ground than two meters down the well. Since we are in the gravity well of a planet (earth) the energy needed to reach escape velocity is a certain value, because we have to climb the walls of the gravity well. A denser planet has a deeper gradient so requires more energy to escape. Falling down the gravity well requires no energy, but the difference of potential energy in “outer space” and “on the planet” is what you’d experience. There’s no energy stored per se, when you pump water up into a dam or when you enter orbit, as you expend the energy into work, resulting in a difference of potential energy in two rest states.

k__ 6y ago
Why are they called gravity wells and not gravity sinks?
- opless 6y ago
  I’d just say that English is a obscure and difficult language. :-)
- cdirkx 6y ago
  Because you can get out of them again. A sink, to me, has the connotation of something being removed from a system.

deleted 6y ago

[deleted]

auntienomen 6y ago

The energy is stored in the gravitational field. Gravitational fields can carry energy and momentum. The same is true for electromagnetic fields, as anyone sitting in warm sunlight has noticed.

jiggawatts 6y ago

I was going to write up a long-winded spiel about rubber sheets and various ways to look at gravity, including my own unique perspective, then I realised that XKCD has already covered this in "Gravity Wells": https://xkcd.com/681/ https://xkcd.com/681/ You're just describing the "landscape" of a field. Any field works this way, including the electromagnetic field, it's just that it tends to cancel out over long distances. Your scenario is like two people living in shallow valleys atop a high mesa discussing the "free energy" that can be gained by going down to the ocean.

acqq 6y ago

> But when we talk about sending say, a spaceship into space, it may fall on a different planet, with a different gravity (a function of the planet's mass), changing the amount of energy yielded by the fall. I guess you imagine there "failing" as something we experience on the surface of the Earth when we drop an object. It's not what is happening when "sending a spaceship into space." It's actually very hard (as in, one really has to spend a lot of energy) to get a spaceship to some star or planet. One can't imagine it as "just let it fall". I suggest reading the following article for the start: https://www.forbes.com/sites/startswithabang/2016/10/01/ask-ethan-why-dont-we-shoot-earths-garbage-into-the-sun/ https://www.forbes.com/sites/startswithabang/2016/10/01/ask-... "Ask Ethan: Why Don't We Shoot Earth's Garbage Into The Sun?" Much less energy is needed to shoot something out of the Solar system than is needed to send something into the Sun, even if you want "just" to send it there to burn. Nothing from Earth can "just fall" there, huge amounts of energy from the outside are needed. What is actually happening is what already Newton figured out: when we're on Earth, we aren't aware of it but the whole Earth is constantly free failing towards the Sun, and yet we're not leaving the orbit around the Sun. > Where does the extra energy, if the ship falls into a heavier planet, come from? A lot of the energy needed for the ship to reach the planet has to be spent by the ship, for the current ships it is launched as the fuel with the ship and later "spent" by the ship (in reality "traded"). The ship can also "spend" the energy of other objects and that energy is really then "spent" (as in bookkeeping, debited to other account, that's what the conservation of energy and momentum is, a kind of bookkeeping) -- the object (e.g. a planet) is indeed slowing down a little, but the change is conveniently small enough: https://en.wikipedia.org/wiki/Gravity_assist https://en.wikipedia.org/wiki/Gravity_assist "This explanation might seem to violate the conservation of energy and momentum, apparently adding velocity to the spacecraft out of nothing, but the spacecraft's effects on the planet must also be taken into consideration to provide a complete picture of the mechanics involved. The linear momentum gained by the spaceship is equal in magnitude to that lost by the planet, so the spacecraft gains velocity and the planet loses velocity. However, the planet's enormous mass compared to the spacecraft makes the resulting change in its speed negligibly small even when strictly compared to the orbital perturbations planets undergo due to interactions with other celestial bodies on astronomically short timescales." In short, there's nothing "extra" happening -- all the energies and momentums are completely accounted for for everything that happens on the scales reachable to our spaceships. The only energy we can't account to anything else is the one we call "dark" but it is observable just in the movements of the whole galaxies, on the scales practically unreachable to our spaceships.

zelphirkalt 6y ago

Just a lay person here. I would say, that that "potential energy" is not really energy, at least not in the sense of "the energy is inside this object" or "this object has so-and-so much energy". I would say you move an object into a different environment and the environment is all that is needed to make the object act in a certain way. If that environment is a heavier planet, then it means, that the object will behave differently than on Earth. The energy you spent to bring the object away from Earth if you want, is stored in the fact, that this object is now in a different environment than before you spent that energy. If the object were to return to its place, it would make use of energy the other way around. Looking at the object alone, you cannot see any of that energy, because it is not really in there, but stored in the fact, that this object is in a different environment. That is what gives it potential.

cygx 6y ago
Potential energy is energy associated with a system's configuration. For example, arrangements of charges can be more or less energetic, as in, you need to perform or can extract work when you go from one configuration to another. This is a rather abstract notion. However, as energy (or rather, stress-energy-momentum) is the source of gravity, you actually do need to know precisely where that energy is located if you want to do general relativity. In case of the example above, the energy will be stored in the electromagnetic field. The gravitational field itself is an exception to this: While it can be used to store energy, in the general case, it's impossible to locate it. I'd argue this is a consequence of general relativity's unification of gravity and inertia.

Tade0 6y ago

The question is: Where is the energy stored in the meantime? Where does the extra energy, if the ship falls into a heavier planet, come from? From a classical physics perspective everything that has mass is immersed in the gravitational field of everything else, it's just that most of that potential energy turns into kinetic energy only when you get very close to something. Basically your potential energy is this mind-bogglingly large number which would only approach zero if the whole universe were condensed into a single point.

martin-adams 6y ago

Here's my lay person thinking on this. By lifting objects up or down, you're altering the gravity wells of each object. A spaceship leaving Earth would make Earth's well a bit smaller. The question you pose makes me think the following: if it requires no energy to fall down a gravity well, but you can use that falling to generate energy (hydroelectric dam for example), then where does that energy come from? So this makes me think that as the following: 1. Increasing gravity wells (object falling) creates energy debt to the well (the debt is larger), in exchange for harnessed energy by things like hydroelectric dams. 2. Decreasing gravity wells (object being lifted) pays off the energy debt from the well by you using energy in the lifting action. So for me, there's no energy stored, it's just a debt in the form of the size of a gravity well.

pdonis 6y ago
> A spaceship leaving Earth would make Earth's well a bit smaller. While this is technically true, the effect is much too small to measure, and is not what is responsible for the usual phenomena we associate with gravitational potential energy.

deleted 6y ago

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Koshkin 6y ago

> if the ship falls into a heavier planet I would be a genuine paradox if this other planet appeared out of nowhere at some point in the process. That's the key: both the source and the destination mass points have been there all along, and so you must, as usual, take into account all the bodies involved from the very beginning.

noobermin 6y ago

So there are cases in which you could consider energy being "stored" somewhere. The canonical example from EM is energy being stored in the EM field itself like in radiation. In this case for classical static gravitation fields, the potential energy has more to due the motion of the orbiting mass and the nature of the force itself being conservative and so it isn't really stored somewhere in that sense. What I mean by that is a mass that moves against that gravitation force will lose motion due to that force (specifically motion with a component away from the planet) and from math you can then calculate that kinetic energy that would be lost from the orbiting body if it were to get a certain distance away from the mass for example. This process is reversible, that is you can trade distance for kinetic energy, which is a familiar situation like that of a ball being thrown straight up in the air, slowing and then stopping at some height, then falling back down. The reversibility of this trade off then allows you to associate an energy, potential energy, with given heights off the ground which makes calculations easier by considering it as a trade off between the motion and these greater distances. This however isn't the case for all forces. For air resistance this isn't the case, so you can't associate air resistance with a potential energy as a counterexample. There are however cases in which energy is real in a sense and isn't just a convenience, and for the most part this is when that energy is associated with motion (momentum specifically) of something. Also, that something would be localized and thus could be considered being stored somewhere. For some examples, of course kinetic energy, but for EM energy, you could consider it being the momentum carried by photons after QED.

visarga 6y ago

> Where does the extra energy, if the ship falls into a heavier planet, come from? It comes from the heavier planet. It also falls onto the spaceship, just slightly changing its trajectory.

ksec 6y ago

I thought it is actually a very interesting question. Hopefully this stays on the front page for another few hours when HN get most of its traffics. Then a physicist will likely pop in with an answer.

lazyjones 6y ago

> It seems that the energy is being stored somewhere, and that matter itself allows us to tap into that, but is not the place where it is stored. It's not stored anywhere, it's always present since the gravitational masses are always present, it's just that a part of it turns into kinetic energy when you are close to large masses (and back once you leave them).

smdz 6y ago

My thoughts on this - The energy used in sending a craft up is already lost when the craft leaves the planet. And potential energy can only be induced into the object from a larger mass. That is based on the object's mass and the large mass like a planet pulling the object. And the "value" (read: usefulness/utility) of that potential energy is based on where it exists P.S. I am not a physics guy and could be completely wrong here.

brandmeyer 6y ago

There's a bunch of half-answers in this thread. > Generally speaking, when we lift matter high above, we get some of the used energy back when it falls back down. > Where is the energy stored in the meantime? Potential energy in a gravity well is negative. The "zero" point is the limit when you get far away from all gravitational bodies. The better question is: Where does the kinetic energy come from when you descend down the gravity well in the first place? The answer is mass. All potential energy is ultimately realized as a change in mass, equal to the famous `mass_difference * speed_of_light^2`. As the body falls down the gravity well, it gains kinetic energy and loses mass. When atoms bind together in an exothermic reaction, their final state is bound more tightly than their initial state. They released some energy in the form of molecular-scale kinetic energy (ie, heat), and they have lost mass. When atoms change configuration in an endothermic reaction, their final state is bound less tightly than their initial state. Some energy was required to push them up to that higher-energy state (sometimes in the form of heat), and they have gained mass. When light nuclei fuse together, their final state is more tightly bound than their initial state. They released some energy in the form of atomic-scale kinetic energy (ie, heat) and a photon, and they have lost mass. Ironically, when very heavy nuclei have fissioned, their final state is more tightly bound than their initial state. They released some energy in the form of atomic-scale kinetic energy (ie, heat) and photons, and they have lost mass. When a spring is compressed, the molecules within the elastic material are stretched apart from one another such that their final state is less tightly bound than their initial state, and the spring has gained mass.

saagarjha 6y ago
> Potential energy in a gravity well is negative. The "zero" point is the limit when you get far away from all gravitational bodies. Like all potential energies, its definition is relative and its zero point convention. > As the body falls down the gravity well, it gains kinetic energy and loses mass. So you’re a saying that hydrogen lower in a gravity well will fuse worse?
- brandmeyer 6y ago
  > Like all potential energies, its definition is relative and its zero point convention. Turns out that this one does have a unique definition because the limit exists, and because the sum of kinetic energy and rest mass determines the gravitational influence of a body. > > As the body falls down the gravity well, it gains kinetic energy and loses mass. > So you’re a saying that hydrogen lower in a gravity well will fuse worse? No. The processes are orthogonal to each other. Forget for the moment that its the release of kinetic energy by dropping down the gravity well that makes the subsequent fusion possible at all. Whether you fused first and then dropped down the well, or dropped down the well first and then fused, the final state is the same.

thanatropism 6y ago

I'm not a physicist, but my master's thesis was on numerical ODEs preserving physical (symplectic structure). I think the best way of looking at these problems is thinking of space already equipped with a vector field (like streamline plots with little arrows) and objects just being carried by them, as if laying down small rocks on a busy river. Of course, this vector field is not constant and the object's mass alters it, but this will be small in the spaceship vs. planets setting. This is more of a mathematical intuition than history-of-the-universe answer, but I think it should help. Ed: This is very simple to reason about when talking about independent particles in one dimension; the phase space arises out of a simple variational problem. It's also easy to reason about in 3 dimensions, but the problem is that the "river" in one dimension has two coordinates, momentum and position. So to think of two dimensional space you have to imagine four dimensions, etc. But thanks to Darboux's theorem these dimensions are "coupled in pairs" (this is what symplectic means), and it's not that difficult to visualize four dimensions. Cf. this illustration of the "symplectic camel theorem": https://encrypted-tbn0.gstatic.com/images?q=tbn%3AANd9GcRtOKkv_wL9wnSkr7rkCxm11ZOmgKXjNBUsVGIkE2BVYlryctvd&usqp=CAU https://encrypted-tbn0.gstatic.com/images?q=tbn%3AANd9GcRtOK...

pdonis 6y ago

> when we talk about sending say, a spaceship into space, it may fall on a different planet, with a different gravity (a function of the planet's mass), changing the amount of energy yielded by the fall. In a situation like this, you have to consider the total energy of the system, including both massive bodies and the potential energy due to them. The total energy remains constant, but how it is distributed will change: the potential energy of the spaceship will be different at the end than it was at the start. > It seems that the energy is being stored somewhere It is stored in the geometry of spacetime; at least, that's the only way I know of to conceptualize it.

acqq 6y ago
> It is stored in the geometry of spacetime And because changing the relative position of objects changes the geometry and the conservation of energy and momentum exists, can't we say that we don't even have to think about the geometry as the "store", but that it's enough to know that it is a property of the whole system we observe and that when it changes the conservation is not violated?
- pdonis 6y ago
  > changing the relative position of objects changes the geometry Yes, although the change due to the spaceship moving is far too small to measure. (Technically, the global spacetime geometry doesn't "change", since spacetime is a 4-dimensional geometry that already contains all the information about "changes" in it, so spacetime itself doesn't change at all. But if we think of the local spacetime geometry changing along the worldline of an object like the spaceship, that concept makes sense.) > conservation of energy and momentum exists You have to be careful with this. There are actually three different senses in which this can be taken in GR, a local one and two global ones. The local sense is that the covariant divergence of the stress-energy tensor is zero. This basically says that matter and energy in tangible form (which does not include "energy stored in the gravitational field" or "energy stored in spacetime geometry") can't be created or destroyed. The first global sense is that in certain kinds of spacetimes, we can do integrals that evaluate the "total energy" of the spacetime, and these integrals will obey certain conservation laws. But those laws don't necessarily correspond to what we normally think of as "conservation of energy" in a global sense. The second global sense is that, again in certain kinds of spacetimes, there is a constant of free-fall motion that can be interpreted as the kinetic plus potential energy of the free-falling object, such as a spaceship on a free-fall trajectory between two planets. This is the sense that has implicitly been used in this discussion. But this constant of free-fall motion only exists in stationary spacetimes, and in stationary spacetimes, the relative positions of gravitating masses cannot change with time. If the relative positions of gravitating masses do change with time, the spacetime is not stationary and the constant of free-fall motion that is interpreted as kinetic plus potential energy does not exist. So there is no such thing as a spacetime which has changing relative positions of objects and conservation of energy in the sense we have been using that term in this discussion.

bcgraham 6y ago

I always thought it was stored in the “angle” of your geodesic. When you come out of the gravity well, your travel through spacetime is angled to be traversing more time relative to space, making you look like you’re moving slower (when considering only spatial dimensions). As you fall into a gravity well, spacetime curvature changes your geodesic such that you traverse more space relative to time, meaning your travel looks faster in the spatial dimensions.

Rury 6y ago

>It seems that the energy is being stored somewhere, and that matter itself allows us to tap into that Like others have mentioned, you need to look at the two planets and the spaceship as a system together. The gravitational potential energy you gain from raising a spaceship out of one planet's gravity is equal to the energy you spend getting it out of that planet's gravity well (neglecting efficiencies). While the trip from Planet A to Planet B might seemingly "gain" you extra potential energy because of the differences in gravity wells. The reverse trip of going from Planet B to Planet A will reverse that "extra" potential energy you gained as it's more costly to get out of Planet B's gravity well. As so, there's no "extra" or created energy within the system. As for where energy is stored, it's simply stored as gravitational potential energy.

gizmo686 6y ago
> it's simply stored as gravitational potential energy This answers the "how", but not the "where". Since energy itself has mass, we would expect the potential energy to carry mass as well; and that that mass/energy should be concentrated in some region of space.
- Rury 6y ago
  >This answers the "how", but not the "where Potential energy is a property of a system rather than an actual physical thing. It's not energy an object has, it's energy an object potentially has relatively speaking. And it's relative to whatever you use as your reference point in your system. >Since energy itself has mass Energy doesn't have mass. It is equivalent to mass.
  - kwk1 6y ago
    This is a good point. Potential energy is relative to a datum, which is arbitrary. So you could say a box on a table has no potential gravitational energy relative to the table, but it does have some relative to the floor. But there was no actual change in energy between the two measurements.
  - cygx 6y ago

ozy 6y ago

Bending your spacetime frame of reference costs energy because you have to leverage something and change its frame. And you can use your bend spacetime frame to extract energy by bringing it in contact with another frame. But the bending itself, or maintaining it, or changing it, doesn't cost or yield energy.

bluGill 6y ago

It doesn't go away. Newtons law of gravity still applies F = Gm1m2/r^2. When you are on earth the gravity of Mars is still affecting you, but r is such a large number (and then we square it) that you can basically ignore it, but it is still there even if we can't measure it. When you get in the rocket you increase the r for how earth's gravity affects you, eventually getting to the point where it is insignificant and then where Mars gravity dominates everything else and so we ignore it, but earth is still affecting things even though it couldn't be measured. The above applies to the moon, sun, even every atom in the universe. Fortunately most of them are far enough away that we can ignore them: we don't actually know how to calculate all the math if we wanted to account for them all.

saagarjha 6y ago
This answers a different question, which is “where does the gravitational force go when I can’t feel it”.

d_tr 6y ago

> The question is: Where is the energy stored in the meantime? Where does the extra energy, if the ship falls into a heavier planet, come from? You can say it came from the fact that there existed a point in space with a lower gravitational potential than the one at the spaceship's point of departure. Each point p in space is assigned a "gravitational potential" value V(p) measured in energy / mass. The function V(p) is determined by the distribution of mass in the space. A body of mass m moving from point a to point b will gain "gravitational potential energy" equal to (V(b) - V(a)) * m. In your case, the spaceship moved from a higher to a lower gravitational potential, thus losing gravitational potential energy (and gaining kinetic energy). Note that you are only interested in the differences of values of V at different points, so you can add or subtract any constant to the function V.

mercer 6y ago

I'm currently reading The Information (I highly recommend it!) and while I feel a vague sense of an answer, I'll defer to others more knowledgeable. I just wanted to say that I think this is a wonderful question! Thanks!

SuoDuanDao 6y ago

The best description I've heard is that energy is a measure of relationship. The spaceship on earth and in space has a relationship with the other planet the whole time, there's just very little that will happen as a result of that relationship until the distance between the two gets smaller.

archgoon 6y ago

So the term you want to search for is "Conservative Forces". I'll give a brief explanation here so you can find a better one more easily. :) Let's take a few steps back and talk about forces, and work. Gravity isn't really a key part of your question. Suppose you had a uniform downward force field like so: | | | | | | | | v v v v v v v v | | | | | | | | v v v v v v v v | | | | | | | | v v v v v v v v Those are vectors pointing downwards. Now suppose you built a miniature roller coaster. Your first one is super boring | | | | | | | | v v v v v v v v car |---\ <--------------------> | | | | | | | | v v v v v v v v It just goes back and forth, perpendicular to the field. Now, if you really grease up the tracks, and have really good bumpers on the end (noiseless, don't heat up, a perfect elastic collision), the car will just bounce back and forth for a very long time, and not speed up or slow down much at all (and ideally, none at all). There is a concept of 'work' in physics. When a force points in the direction of motion, we say that a force does work. The total work done along a path is simply the sum (integral) of all the parts of the path, multiplied by their length, multiplied by the amount of force in the direction of the path. This definition chosen so that the amount of kinetic energy gained, or lost, is equal to the work done. Since the force is perpendicular for the above path, the 'work' done is said to be zero. Now let's have a more interesting path. Let's give the car a kick. car <- /---| /--------<-------\ | | v ^ | | \------->--------/ Here we have two parts, top and bottom, where the force does no work. However, we have a part going down, and a part going up. The car will accelerate going down, and decelerate going up. If we grease the tracks up, then the car will always be traveling at the same speed on the top track. It will also always be moving faster on the bottom track. We say that the force does 'work' when the car is on the tracks going down and going up. Now, the total amount of work that the force does, for any loop, ends up being zero. This is clear here, since the distance the force is applied downwards, is the same as the distance of the force applied upwards. What is slightly less obvious is that any looped track we could build here would have the same property. If we had something like /---\ | | \ | \ | \| (and ignore the mechanics of those super sharp turns) we would find that the work done was still zero. The downward diagonal bits would end up still adding up the same amount of work as the upward straight bits (bit of vector math shows this). The fact that the work done on any loop is zero makes this particular force a 'conservative force'. Not all force fields need to be conservative! Suppose our field was only present on the left side. /---| /--------------------\ | | | | v | v | | | | | | | v | v | | | | | | | v | v | \--------------------/ This would only do work going down. This would mean that the track car would keep getting accelerated faster and faster. Potential energy only makes sense as a concept in the presence of conservative fields. You define a reference point, sometimes chosen to be at infinity, and define the potential energy at each point to be the amount of work done to bring the object to that test point. However, you don't need to just be a constant field to be conservative. Gravity, as expressed by Newton, is a conservative field. Regardless of how you arrange a bunch of objects or planets, looped paths will always have zero net work done. This is because the mathematical form of Gravity can be epxressed as the gradient of a scalar (takes a position, gives a number) function. You seem happy with the idea that locally on earth, energy is conserved because you get the energy back when you travel in a loop. This doesn't change when you travel in a more complicated path from one planet to another. The force field does work on the object to accelerate it when it moves in the direction of the force, and decelerates it when you pull it back. The big takeaway point here is that energy, as a concept, is built on top of (at least classically) the more primitive concept of 'force'.

agumonkey 6y ago

It's funny indeed that a field is not constant and decreases over space so after some point the potential goes "down" ?

raverbashing 6y ago

You're confused because basically when we think of potential energy on the surface of the Earth that's an approximation In case of a spaceship going from one planet to another you need to consider the potential energy function for both planets in relation to each other (and the Sun's potential energy as well) Potential energy on the surface is more like going up and down a ramp in a line. Potential energy in the case of multiple planets is more like a curvy skate park, where you can go from one place to another but have a small change in potential energy in the end. Also, potential energy is always relative to something. So if you go to another planet you shouldn't be calculating the potential energy from its surface, but from where you came from.

steve76 6y ago

> Where is the energy stored in the meantime? Where does the extra energy, if the ship falls into a heavier planet, come from? Take your reference from the Sun, not the planets.

gmantg 6y ago

Kinetic energy is fake. We may as well say that it's planet falling on the ship and thus it's energy is the mass of the planet multiplied by v^2/2 and where did that energy come from?

JProthero 6y ago

>It seems that the energy is being stored somewhere, and that matter itself allows us to tap into that, but is not the place where it is stored. Is there anything I can read to better understand what we know about this? Here are some quick recommendations for views on potential and gravitational energy: This [1] is a short video from a presentation given by Alan Guth — an early proponent of inflationary cosmology — in which he outlines a thought experiment designed to illustrate how gravitational fields are associated with negative energy. This thought experiment is a standard feature of Guth's introductory talks on inflation; recordings of his academically-oriented talks with more rigorous treatments of the subject are available on YouTube. The logic of the thought experiment that Guth describes in the presentation is not controversial among physicists, but there are different perspectives on how to describe it. Guth's former MIT colleague, Sean Carroll, prefers to avoid descriptions involving negative energy and says instead that energy is not conserved in General Relativity. He explains his perspective here [2]. The cosmologists Luke Barnes and Geraint Lewis discuss different perspectives on fields and energy conservation in this video [3]. In this video [4], the theoretical physicist Sabine Hossenfelder briefly discusses the physical reality of potentials in the context of the Aharonov-Bohm effect. If you have several hours, Sean Carroll has a video series about the Biggest Ideas in the Universe [5]. I haven't been through the playlist yet myself, but I imagine he might go into some detail on the questions you raise, particularly in his discussion of Conservation (1), Force, Energy and Action (3), Spacetime (6) and Fields (9). The Q&A videos following each instalment might also be useful, because his audience frequently asks questions like yours. [1] https://www.youtube.com/watch?v=15IGPyRHOaY https://www.youtube.com/watch?v=15IGPyRHOaY [2] http://www.preposterousuniverse.com/blog/2010/02/22/energy-is-not-conserved/ http://www.preposterousuniverse.com/blog/2010/02/22/energy-i... [3] https://youtu.be/bcE5RQ7A7Ys?t=423 https://youtu.be/bcE5RQ7A7Ys?t=423 [4] https://youtu.be/0GCHzbmMZf0?t=144 https://youtu.be/0GCHzbmMZf0?t=144 [5] https://www.youtube.com/playlist?list=PLrxfgDEc2NxZJcWcrxH3jyjUUrJlnoyzX https://www.youtube.com/playlist?list=PLrxfgDEc2NxZJcWcrxH3j...

gpsx 6y ago

It is useful to think about all the energy in the problem. There is a case where the only energy in the problem is the kinetic energy of the spaceship and the potential energy from gravity. That is the case of firing the spaceship with a giant catapult so that it goes and hits the other planet. In this case, it will be going really fast when it leaves earth. It will slow down as it moves away from earth, leaving its gravity. Then, it accelerates as it falls towards the heavier planet. Then splat, even faster than it was moving when it left earth. At any point, the change in kinentic energy will be the negative of the change in potential energy. A more common scenario, at least in our imagination, is that a thruster on the spacecraft is adding and removing lots of energy from the spacecraft. The math will still work out. And this way there is hope that the spacehip ends up at rest, not destroyed, on the larger planet. But this time the thruster did lots of work and has to be included in the energy balance.