Why is Gravitational Potential Energy Negative? (Gravity, Physics)

[@johnyfausman959]

This was the best explanation I have seen for this. Thanks, we need more people like you.

[@alohahoward1]

The negative sign has always bothered me, now it make sense. When I took physics in school I found it difficult but fascinating. I think the problem I had was that the concepts were not adequately explained. I could work the problems because I memorized the formulas but I still was often confused. Thanks for the great explanations.

[@GodJesusChristlovesyou_knows_u]

Brothers and Sisters, God loves us so much that He sent His Son Jesus Christ for us to save us from our sins, and he bled and died on a cross for us to redeem us from death, and to gain life everlasting, for those who put their trust in him. And what's more, he has risen, and is willing to call you to repentance (correction) and as his witness, by his grace indeed. "For it is by grace you are saved, through faith". Jesus loves you, God cares for you! Therefore, repent and believe the gospel.

[@dominusdone5023]

@@GodJesusChristlovesyou_knows_u sttfu buddy its a physics video

[@a_0vi]

This channel is a great resource for physics enthusiasts as your explanations are very intuitive. Keep up the great work!

[@PhysicsMadeEasy]

I will! Thank you for your kind words.

[@frankdimeglio8216]

@@PhysicsMadeEasy Time DILATION ultimately proves ON BALANCE that E=mc2 is F=ma, AS ELECTROMAGNETISM/ENERGY IS GRAVITY. The Earth AND the Sun are CLEARLY E=mc2 as F=ma. The stars AND PLANETS are POINTS in the night sky. TIME is NECESSARILY possible/potential AND actual IN BALANCE, AS ELECTROMAGNETISM/energy is gravity; AS E=MC2 is F=ma. Very importantly, outer "space" involves full inertia; AND it is fully invisible AND black. The sky is blue, AND the Earth is ALSO BLUE. It ALL CLEARLY makes perfect sense. E=mc2 IS F=ma. Gravity IS ELECTROMAGNETISM/energy. By Frank DiMeglio

[@eksa-5213]

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[@eksa-5213]

@@frankdimeglio8216 I also have an objection let's figure it out together

[@eksa-5213]

@@frankdimeglio8216 your comment looks kind of interesting but I really didn't understand what you mean by can you explain it more intutively

[@rajkumars6086]

Amazingly lucid ! The conclusion with atrractive and repulsive forces was stunning. Thank you Sir. Please continue your good work. It is so rare to find a teacher who gives conceptual clarity.

[@PhysicsMadeEasy]

Thank you for your kind words, Raj. I am glad this video clarifies a few things!

[@rajkumars6086]

@@PhysicsMadeEasy I hope someday you will do 'Math in Physics - a simplified overview'

[@raindrops6342]

youtube is a blessing, so is this man

[@PhysicsMadeEasy]

Thanks raindrop! KZbin is a great tool for learning. I wished it had been there when I was in my teens and early 20s ;-)!

[@Ii-fo8pq]

Please never stop posting videos.. I've been searching this kind of video for months now I finally understood why negative energy particle falls into the black hole and positive energy particle escapes.. you're a great teacher I must say👏👏

[@PhysicsMadeEasy]

Hi, please tell me what you understood here... Hawkings radiation does not really refer to a positive energy or negative energy particle (just to a particle and its antimatter counterpart that originally should be virtual, but because of their proximity to the event horizon of a BH become real because irremediably separated). In reference to the video, once the particles became real, the particle that falls into the BH loses gravitational PE energy (its GPE becomes more negative), and the one that escapes gains some (its GPE becomes less negative). Overall, what one lost, the other gained. Thank you for your kind words, btw :-)

[@yan-amar]

@@PhysicsMadeEasy Is this explanation, involving the position of a virtual particle and antiparticle pair near the event horizon really true, or is it a simplification? I read the explanation involves field theory and allows particles to form quite far from the event horizon (outside of the BH). But this is all well beyond my understanding.

[@thegoldenllama8787]

I have genuinely spent the entire day looking for a good explanation of GPE and your video has made it all crystal clear - subscribed!

[@PhysicsMadeEasy]

Thank you, I am glad it helped!,

[@amritrajdash663]

It's a shame this channel isn't more popular. Loved the explanation!

[@PhysicsMadeEasy]

Thank you Amritraj, I am glad you enjoy my work . Channel not popular enough? Talk about it around you ;-)!

[@syedmuhammadfaheem7458]

after all, my confusion gets clear. thank you very much sir. so clear and calculus based explanation.

[@PhysicsMadeEasy]

Thank you, I am glad I could help

[@ndwani5724]

I see this KZbin is really a fantastic social media which is giving us the golden opportunity to get our clarifications of every aspects but only drawback is we find these golden videos and golden creators rare and finding them is very hard. After watching 10 videos 2/10 are found fantastic videos.

[@PhysicsMadeEasy]

Yes, if one steps back and stops taking things for granted, services like KZbin (and now chatgpt) are truly amazing. I wish I had access to this when I was a high school student in the late 80s ... My grades would have been very different (in the good sense haha). I hope that one of the 2/10 videos were on my channel :-)! PS: for chat gpt, I do not recommend it for physics, yet... I tested it with a few problems, the answers were very poor, and sometimes completely wrong, but for other things, it's amazing.

[@yan-amar]

@@PhysicsMadeEasy In the face of the proliferation of low effort science channels, that spits out AI generated content that is often over simplified or simply false, I can only fear these tools. Right now one useful trick for telling these channels apart from the good ones is to watch only those where the host shows their face. But soon we will not be able to distinguish a real human from a computer generated one.

[@PhysicsMadeEasy]

@@yan-amar yeah I know... check the latest fake ads in crypto currencies. You see the CEOs of Microstrategy or the CEO Solana talking, so you kind of trust what is being said. But it's all a scam! The video is made by AI. I wonder how youtube allows that!

[@Brainiac51]

Hello At time 5:24 you mention work also has to be done if you want to bring a mass m closer to mass M, in this case is the work done negative as the potential energy is decreasing and so is the distance between the two masses ?

[@PhysicsMadeEasy]

The applied force (in the positive direction) is preventing the object to accelerate due to the gravitational force (in the negative direction). We do so to make the mass m move at a constant speed so that kinetic energy remains constant, meaning that all the work done would become potential energy. The force vector is in the positive direction, and the displacement vector in the negative direction, so work done on m is negative (energy is taken out from the system). Work being a transfer of energy, that means that the PE of the system decreases (it becomes more negative). You can see it like this if you want : You attach yourself to the mass m : it pulls you because it is attracted by M. So m is doing (positive) work on you, thus loses energy.

[@Brainiac51]

@@PhysicsMadeEasythankyou

[@Tzaounis]

The best explanation I have watched so far. With mathematical and logical clarity.

[@PhysicsMadeEasy]

Thank you Panagis, I am glad you enjoyed the video :-)

[@The_Green_Man_OAP]

2:53 The energy that was _transmitted to_ the moon _increased_ the _gravitational potential energy level_ , which _decreased_ the _gravitational potential binding energy_ for the moon/planet system at that distance from it's _barycenter._ For a conservative applied forces, potential gradient exist such that the work done on masses can be calculated for displacements over equipotential lines instead of using along the exact trajectories. The work done is then path independent.

[@The_Green_Man_OAP]

∆Energy=∫F•ds , ds along path. U=∆Kinetic=∆K=∫dp•v , dp along trajectory. W=∆Potential=∆V=-∫F•dr , dr along equipotential levels. Work done by any applied force, in terms of kinetic energy=U(K,K')=orbit K gain to mass. Work done by conservative force, in terms of potential energy= -W(V,V')=orbit K loss.

[@PhysicsMadeEasy]

Yeah, you see, I do have a totally different approach to that. I find the use of conservative force quite intellectually elitist because it is not needed for understanding the fundamentals and correctly solving questions and problems. Additionally, it can seriously confuse some students that otherwise would be fine. For me it is like using the concept of centrifugal force in circular motion (thankfully, the latter has nearly disappeared from physics programs). Work is a transfer (or if you prefer, transmission) of total mechanical energy. A single object cannot have PE. A system needs at least two objects in interaction for it to have PE. Within this context , a gravitational force linking two objects cannot do work on that system, because it is an internal force to that system, the PE only exists for a system, not for one of its objects. That’s where the idea of gravity being a conservative force shows up in order to apply things like W = ∆Kinetic. Maybe practical in some limited contexts, but the approximation is not worth it imho. Unnecessary and Elitist. In the video, when an external force pulls the moon away from the planet, that force is doing work that increases the PE of the system. I have been teaching IB, and the concept of conservation forces are discussed just in a short paragraph, probably just in case a student sees it somewhere out of the scope of the curriculum… I teach also A-Levels, and I see this concept being mentioned but minored also. Recently, I have been requested by one of my students to teach AP Physics C, and I discovered that they really put this concept of conservative force quite at the heart of their problem solving. Sad… Why make things more complicated by dividing situations into categories while they can be treated within a more general and simpler paradigm? hopefully, this will evolve like it did in other curricula.

[@mathu_black5782]

An underated channel 😢❤

[@PhysicsMadeEasy]

maybe, maybe not ;-) all depend on our many physics students or physics lovers there are out there :-) thanks for your words mathu

[@ello-isa]

I would like to say, thank you. This video really helped me a lot to understand the concepts

[@PhysicsMadeEasy]

Cheers Elo, I am glad this video helped. Feel free to explore the channel for more...

[@aryanurs2499]

Sir, during integration in the last step, the direction of external force applied by you and the displacement dx are in opposite directions, wouldn't that give it a negative sign to the terms?

[@PhysicsMadeEasy]

See my response to Akshat's comment: the negative sign is "hidden" in the integral's limits. It shows up later in the derivation.

[@sweetumms5308]

This was reeaallyy helpful! Thank you so much❤

[@lexanris]

Thanks for making this amazing video. Explanations very concise! But i still have 1 query: What I you understood is that Since GPE = -W by gravitational force to bring from infinity to r & W done by gravitational force in this case must >0 as the displacement is in same direction as force, GPE must be

[@PhysicsMadeEasy]

Hi Lexa, It is not the the gravitational force that is being integrated, but the force applied so that the kinetic energy of the test mass m remains constant (remember this proof is a thought experiment). Naturally that applied force will have the same magnitude as the gravitational force but be of opposite sign… thus of opposite sign of that of the displacement. I mention this in the video, but you might have missed it: See the video starting from around 5:30, it should help clarify 😊

[@AdhamMGhaly]

A brilliant explanation. Glad to see Phil Collins explaining physics :)

[@PhysicsMadeEasy]

haha! You are not the first one making me this kind of compliment (I am a musician too :-). Some also mentioned Bruce Willis lol!

[@dhurbaraj6355]

sir your lectures are far better than our college lectures .. Sir where are you from ? loved the video

[@PhysicsMadeEasy]

Hello Dhurba, thank you very much , I am glad my work helps you. My lessons are aimed at giving a fundamental understanding on which a student can build: After watching them, go back to your notes from class, and try to make better sense of your notes at the light of what you understood from my videos.

[@ajunbabu5547]

Thank you very much

[@r.i.pnandy2854]

This channel really makes physics easy❤️❤️❤️❤️❤️👌🏿👌🏾👌🏽👌🏻

[@azeeartie2468]

Great content, Sir.🤩 I would be grateful if you could explain on Electomagnetism and on the internal energy of an ideal gas system. Thank you, Sir.😊

[@PhysicsMadeEasy]

Hi. Electromagnetism on the way... A big subject! For Internal energy of an ideal gas, it's on the list, but you might need to wait... A quick explanation: It's actually very simple. Internal energy of a system is the sum of PE and KE of all the particles that make the system. An ideal gas is a model built so that the gas has no PE, therefore the internal energy of an ideal gas is the sum of the KE of all the particles of the gas. U = sum(KE) = N * 1.5 kT (1.5kT is the average KE of the particles of the system and N the number of particles).

[@azeeartie2468]

@@PhysicsMadeEasy Loads of thanks Sir.🤩

[@gloryikuku8621]

This is the best explanation I have had on why GPE is always negative. Thank you very much sir

[@PhysicsMadeEasy]

You are warmly welcome. I am glad my work was helpful to you. :-)

[@nabeeln.a4968]

Good work sir. I have a doubt. When you calculated the work done for bringing mass m from infinity to R, you took the force positive, but dx should be taken as negative, right? Because the displacement is against the applied force in this case. Then total PE is positive. Please help me with it

[@PhysicsMadeEasy]

Hi I took the force positive because I considered an applied force counteracting gravity, which is in the positive direction of the axis (to the right). The displacement on the other hand is negative (the test mass is moving to the left, towards mass M). The work done by the applied force is therefore negative (dW = Vector F x Vector dx). So the PE decreases (becomes more negative). Maybe you are confused because you don’t see the negative sign : it’s in the integral bounds. As shown in the video, the negative shows up when you carry out the integration.

[@InsocialDev]

Sir you give us the best explanations

[@PhysicsMadeEasy]

Thank you Warmess! You know what present I got myself for Christmas (A RTX 3060 for my PC... I am 51, but a gamer too (PC and VR) ;-).

[@gamertothecore-borngamer2347]

What is this a coincidence? I got this doubt now on 18th October 2021, and you uploaded this on 17th October, 2020. Haha, nice helping me on the first anniversary of this video! Amazing explanation btw :)

[@PhysicsMadeEasy]

Hi Fanatic Gamer! The advantage of making videos about the basics of Physics is that they are super evergreen! this video helped you 1 year and 1 day after I published it, maybe another person will be helped in 10 years and 1 day, and maybe 50 years and 1 day! Anyway, thank you , your comment is very encouraging to me!

[@gamertothecore-borngamer2347]

@@PhysicsMadeEasy exactly! These basic fundamentals will forever be required to propel someone into the ocean of physics! and your hard work will keep paying off for decades! And I really appreciate your effort to reply to the comments.

Thanks man you helped a lot, changed my point of view about potential energy

[@PhysicsMadeEasy]

I am glad I was able to give you a new perspective on this really important notion in Physics. Now, for you, the fun can start !

[@BokangLikhoeli]

that makes a lot more sense now, thank you for sharing this

[@PhysicsMadeEasy]

I am glad my work clarified some things for you! Thank for letting me know!

[@The_Green_Man_OAP]

3:14 The problem with that is that the best estimate for the size of the Universe is at about 93 billion light years, which is a far cry from infinity, therefore we need a better definition of potential that doesn't invoke infinities. Btw, another way to get zero gravitational potential is at Lagrange points of reference, where gravitational fields cancel out. So, instead of invoking infinities, it may be better to map Lagrange points out throughout the galaxy and estimate them throughout the Universe.

[@PhysicsMadeEasy]

The use of an infinity here is just a mathematical thought experiment. If that makes you uncomfortable, you can instead consider the gravitational Force as approximated to zero at very large distances (because proportional to 1/x^2). Btw, The Universe is not 93 billion light years across. This is the size of the observable universe. The curvature of the universe is measured at 0 with an error of roughly +/-1%. If it is 0 or negative it means it is infinite. If it is slightly positive, then it implies that the size of the universe remains very large compared to that of the observable universe… At Lagrange points, there are still gravitational interactions. The net force is zero because the various forces acting on the object placed there are balanced. So the gravitational field strength is zero. The gravitational potential at that point is not zero but has an absolute value which is minimum, i.e. the top of a hill of a potential position graph, (as the potential is always negative, it is locally the closest to zero). (reminder: In classical field theory, the field strength is the derivative with position of potential: when it is zero, the potential position graph reaches a maximum or minimum).

[@The_Green_Man_OAP]

​@@PhysicsMadeEasy To clarify, I pointed out Lagrange points _because_ the field strength is zero there and no work can be done by fields to masses with zero net force, so they are logical reference points to use for potentials. However, any slight motion away from them would mean work was being done, so it makes sense that the field potentials just around those points are non zero. So, you admit that the field potential is zero nowhere and the infinity potential is a fiction. That's ultimately all I wanted. Thank You for your response.

[@The_Green_Man_OAP]

Of course, to model this would require adding an outside contribution, even to a supposedly isolated system. Therefore it's impossible to have any true isolation, even in theory. If work is done on a mass m from one orbit at initial point "O" (which is rº from CM of main mass M=μ/G), to the nearest lagrange point at "a" ( which is rª > rº along a radial line that connects them), then W=μm(1/rª - 1/rº) is the work contribution on m due to M in the interval [rº,rª]= [rº, rª-dr]+(rª-dr,rª], with the rhs interval having its field strength contribution cancelled out by exterior fields. If an additional field is present from an outside source then it can be described as coming from the total field strength of all masses exterior to the system, acting from the CM of that exterior system. This exterior field will vary from point to point along O→a, so will be a function of position along this line, and of time. If the extra force from exterior masses at points that are distances {r'} from {r: along O→a } is μ'm.Σ{(r-r')/|r-r'|³}, then μ'=μ'(r,r',t)=GM' , where M' is a function of time due to redistribution of masses as they orbit others, & the extra work done is: ∫μ'm.Σ{(r-r')•dr/|r-r'|³}:=mΦ, where displacements {dr} are along the line: O→a. So, total work on m due to main field from M and all exterior fields is: Σ "Work on m, due to field from M' variable" =W+ΣW'=μm(1/rª - 1/rº+(μ'/μ)Φ), over interval [rº,rª]= [rº, rª-dr]+(rª-dr,rª]. We can assume that rhs interval is where μ'Φ dominates, possibly cancelling out with μm/rª or redirecting some motion tangentially at that point.

[@The_Green_Man_OAP]

However, its only an assumption that G is constant throughout the Universe! It may vary over long distances for all we know. G units: [G]= [(ML/T²).(R/M)²]=[(R²/T)/(M/L)] could be associated with areal velocities of various orbital contributions to m from distant masses (roughly proportional to π(R')²/T, R'~distance variable from external mass to point along line O→a), and with the exterior forces in the L' directions which are pointed slightly ahead of the R' directions if you thought of m orbiting each M'. As L' is along a force line from a distant mass, M'/L' would be like the mass per unit length in a tense string. In a vibrating string, M/L is proportional to tension/v² , where v=wave velocity. So, possibly: G~ (Areal Speed).v² / tension (let Å:="Areal speed", V'="potential"=μ'/R') We can assume that v=c, and we already know that c²~Σ{μ'/R'}, therefore: G~Å.(ΣV')/m(dV/dR')~k.Å.∆R'/m for some unitless k, and L'~∆R'.

[@star9710]

3:45 but wasn't work negative of potential energy? Is it because when the work is being done on it (the system), it's like the system is doing negative work, leading to positive delta potential energy?

[@PhysicsMadeEasy]

In order to move away two objects that attract each other (for example two magnets), one needs to work on them. Seen from the magnets, the work is positive, because the energy is transferred to them: that's why their PE increases. Seen from the person pulling the magnets apart, work is negative (the person loses energy) When A works on B, A loses energy (for A the work is negative), and B gains energy (For B, work is positive) . I hope this helps!

[@kaushik-sarkar-droid]

Wonderfully explained!!! Its so under-subscribed channel...

[@Ri-bc5ff]

very good explanation. Kindly also make a video explaining why total energy of the universe is zero,i.e. positive expansion energy is balanced by negative gravitational energy.

[@PhysicsMadeEasy]

Hi Ri, it is a good idea, yet I would need to go a little beyond High School Physics for that. But I'll keep it in mind. Remember though, that this is an hypothesis, and the consequent flatness of the Universe which we observe is only based on the observation of the observable universe... We do not know (and probably if the flatness is confirmed, will never know) the real size of the Universe, or even if it has a size...

[@OluwabunmiFayomi]

Quick question, did the gravitational force become zero because the object m is no longer in the gravitational field formed by the other object M?

[@PhysicsMadeEasy]

Yes, that's correct.

[@gaad45]

Best explanation , no doubt . thank you so much .love from Sri Lanka

[@PhysicsMadeEasy]

Thank you Gaad :-)

[@smartphoneguitar6780]

Great explanation. Inmediately suscribed!!

[@PhysicsMadeEasy]

Thank you, and welcome to Physics Made Easy. I hope you will enjoy the voyage :-)

[@hasankapoor9141]

@6:30 Sir but Work applied and displacement are on opposite directions so it should be negative?

[@PhysicsMadeEasy]

Hi Hasan, the negative sign is hidden in the boundaries of the integral...

[@hx756]

Hi sir, can you please help me with this? I have some confusions regarding gravitational potential energy. Lets say I have two masses, m1 of y kg, and m2 of 1kg, I am going to fix the position of m1 and vary the position of m2 and study the gravitational force and gravitational potential energy. So lets say m1 is fixed at position O, m2 is at position A for now, and m1 and m2 are R distance apart from each other. Then the gravitational potential energy of m2 at A, can be thought of as the work done to bring m2 from infinity to A right? Then i also thought that because at A, or any point within the gravitational field, gravitational force is still acting on m2. This must mean that the gravitational potential energy of m2 at A, can also be thought of as the work done needed to overcome the gravitational force of attraction to stop m2 from going any closer than a distance of R away from m1. Because without this gravitational potential energy, m2 would go closer to m1 due to the gravitational force of attraction. But these 2 ways of thinking contradict and i don’t know what is wrong. Because the work done is the area under the gravitational force vs distance x graph. And the area under graph between position x=0 and x=A, just won’t happen to be equal to the area under graph between position x=A and x= infinity?

[@PhysicsMadeEasy]

Hi HX, I believe there is a confusion in your reasoning : You appear to see Gravitational potential energy as a concept equivalent to that of gravitational force… In your example, you state « This must mean that the gravitational potential energy of m2 at A, can also be thought of as the work done needed to overcome the gravitational force of attraction to stop m2 from going any closer than a distance of R away from m1 » You are writing that an energy can counter a force… No, it is a force that counters another force… To visualize the difference, try to think about it this way. You have m2 at rest at A, and you do not want the gravitational force Fg to make it fall towards M1 and see its GPE converted to KE. So you apply another force, Fapp, of same magnitude than Fg and opposite direction. M2 remains thus at rest. What is the work done by Fapp ? Work = Fapp x displacement = Zero. (Work = energy transfer) The force applied didn’t modify the energy of m2, it just prevented it to convert to KE. You also wrote this which reinforce my diagnostic on your confusion: « Because without this gravitational potential energy, m2 would go closer to m1 due to the gravitational force of attraction. » Gravitational potential energy is not what prevents an object to fall. It is the work the object has the capacity to provide (think about the damage to m1 if m2 crashes on it). By applying a counter force, you are just preventing the energy to be released, you are not changing that energy. Your confusion seems to lie deep. And these are important concepts. Review the the definitions, and take time to think about them. Have you seen my video « what is energy « ? If not, the first section could be useful as a starting point.

[@hx756]

@@PhysicsMadeEasy hi sir. Thank you so much for helping me clear my confusion! My understanding is clearer now and i will think through it again. Yes I’ve watched your video on what is energy previously but i will rewatch it again, i must’ve missed out some important infos. Thank you once again for your explanation! I really appreciate it

[@tune490]

Great Job explaining this

[@simongross3122]

Great explanation, yet again. Thank you. Once again I think that my confusion about this topic stems from the word "potential". Potential energy is not energy, it is the difference in energy between two places. Or have I misunderstood yet again? I feel that my potential understanding can only be positive.

[@PhysicsMadeEasy]

Hi Simon, Thank you. Potential is not a difference of energy between two points... Electric Potential is a quantity associated with a position. It represents the potential energy that a charge placed at that position would have PER UNIT CHARGE. I bolded the 'per Unit Charge', because this is the part to really understand. Example: You park in a red parking zone where it costs 5 Euros/hour to park. You park there for 2 hours. You pay 5 E/h x 2h = 10 E. At a position with 5V ( = 5 Joules / Coulomb - a volt is a joule per coulomb), you place a charge of 2 Coulomb, the charge gets a potential energy of 10 Joules. I recommend you check this video, where I explain what a potential is in detail: kzbin.info/www/bejne/oGSqo4KBp8qSlZI, that should increase your potential for understanding ;-) !

[@lawrencelam2333]

Very clear and lucid lecture! can explain on picture at 7.48 ..why did'nt you choose the lower limit as R and upper limt as infinity just to make it same direction as dx and Fa ( from left origin at centre of mass M to the right towards mass m? Thanks

[@PhysicsMadeEasy]

That would work too Lawrence . The objective is to derive the PE of m at R. I chose this order for limits because in the thought experiment, I was moving the mass m from infinity (initial position, initial energy = 0) to R (final position, final energy that I can obtain directly by calculating the integral). Your suggestion does work also, but here you would be calculating the amount of work needed to be provided to go from R to infinity. You would find W = +GMm/R. And because at infinity the PE is 0, you then would deduce that the PE at R is -GMm/r. You see, same result. :-)

[@lawrencelam2333]

Many Thanks! Your Teaching is awesome!

[@yogeshchoudhary6336]

Sir, why are we considering only work done by external agent in the picture as on 2:46. Please explain by considering work done by system.

[@PhysicsMadeEasy]

Hello Yogesh. In the situation at 2:46, The force is separating the two masses, leading to an increase of the energy of the system: Work is done on it by the external agent applying the force. The system does not do work, work is being done on it. But if you insist considering the system as your reference to the work done, it would be acceptable to say, that the system is carrying out negative work on the external agent… But I find it more complicated to think that way. In the derivation I propose for the gravitational potential energy formula around 6:00, the situation is reversed. the two masses are let approaching each other: the work done by the external agent (which is aimed at keeping the speed constant in that situation), is negative (Here, the system is doing work on the external agent - the external agent being pulled in).

[@gtfantafizzy1295]

@@PhysicsMadeEasy So in another metaphor would it be correct to say it’s like a person with negative 15 dollars and a person works by giving the other 15+ dollars which would equal to him being at zero. Meaning the only person that can do work is the one with 15+ dollars and to say that the other is doing negative work by taking 15 dollars wouldn’t make sense, since he has the potential dollars of zero. ?

[@farhanabbasi486]

The best explanation like ever.

[@PhysicsMadeEasy]

Thank you Farhan

[@teacherbrad9534]

Lovely explanation.

[@nehakolaparthi5280]

Tysm bro ur the dude who finally made me understand this :)

[@MichaelOyinkolawa]

Very good Explanation❤

[@PhysicsMadeEasy]

Thank you

[@lightwave2334]

If the gravitational potential energy is negative, does that means it emits negative gravity? If so, can we measure it? Have we? And where would the effect of this negative energy be strongest? Inside the respective objects, or between them?

[@PhysicsMadeEasy]

Energy is a relative quantity. The fact that it is negative is just that we set it up at zero while it has capacity to do work on its environment. Actually any force triggered by a charge that has an attractive quality, will lead to a negative value of the corresponding system's energy. In other words, the negative sign results from a human convention that was set to allow to perform calculations easily.

[@poolofanxiety885]

how is the kinetic energy zero and all the work done is used in increasing PE only? if a body is moving then it has KE right? whether the body is accelerating or moving with constant velocity, it is in motion i didnt understood this

[@PhysicsMadeEasy]

Hi, As you understood, this calculation is based on a thought experiment: To calculate the PE of a point mass when placed at a certain point, you need to make sure that the work you do on that mass goes entirely to the PE: that means that the KE stays constant. The KE does not need to be zero, indeed, you need a little bit for the mass to move, but the important thing is that the KE (thus the velocity) stays constant during the full process of bringing the mass from infinity to the final point.

[@tand9854]

I think what Sir meant was change in KE was zero NOT KE is zero

[@jakirhossain-vn5ou]

Will you please make a video about the cause of the equal and opposite reaction of an action,according to the newtons third law

[@PhysicsMadeEasy]

This is a good suggestion! thank you. The real cause of the 3rd law is never mentionned in text books. You can find it in classical field theory by expressing the force of a charge (or a mass) on another, and vice versa. In all cases, you realise that the two forces have to be equal in magnitude and opposite in direction.

[@arhamapon9282]

Sir why did you intigrate if the gravitational force is needed to be equal to the applied force?

[@PhysicsMadeEasy]

Hello Arham, The work done to bring the mass from infinite to a point without changing its kinetic energy provides the gravitational potential energy (GPE) of that mass. So to calculate the GPE, you need to calculate the work done when doing so. Work is force by displacement, right? But the gravitational force changes with the position (F=GMm/x^2)… Ouch! What value for force should you use when calculating the work? … Well, you have to calculate the work done for infinitely small intervals of displacements (where you can approximate the force constant), and add all these works together. That will give you the total work done on the mass, thus its GPE. This operation is called integration ;-)

[@shine_simone]

Thank you sir, this really helped alot

[@stefanopogany3767]

Very, very, very good! Now I can finally sleep and stop thinking about this!

[@PhysicsMadeEasy]

@@stefanopogany3767 hi Stefano, physics shouldn't give you insomnia! I am glad my videos help you in finding the serenity for good night sleeps 😊.

[@ifrazali3052]

Is the work done to balance the attractive force external or internal work?

[@PhysicsMadeEasy]

Hi Ifraz, To know if work is internal or external, you just need to define your perspective on the object doing the work (applying a force) relatively to the object receiving the work (subjected to that force): If the system you are considering contains both objects, then the work is internal, otherwise, it is external. Another way to see it is just to look at the change in total (internal) energy of a system: If it changes, then the work is external. In the video, the system I consider is actually the two masses interacting gravitationally. The force applied to balance gravitational force (and prevent any change in ‘internal’ KE) is external: the work I apply on the system is thus external: The system’s potential (thus total) energy change.

[@ifrazali3052]

@@PhysicsMadeEasy Thank you so much

[@jonhicks8999]

Looks to me like the positive/negative on potential energy, in this circumstance, is fairly arbitrary. Negative was chosen and agreed upon by physicists to describe attractive forces and vice versa. It doesn't seem like the math would be that different if you flipped it to focus on the force applied at any given point. From which you would get a positive. I feel like this is more because this all started with a question of why things go down when dropped.

[@PhysicsMadeEasy]

You are totally correct. Choosing a relevant point for the zero is actually what happens in mechanics where we consider the zero at ground level (PE = mgh). Concerning the sign of the forces, well, the convension probably originates from Coulomb's law F=kQq/d^2, where both situations occur and are dictated by the sign of the electrio charges (Q and q same sign, repulsive, and it happens that F ends up being >0, Q and q1 of opposite signs, and it happens that F ends up being

[@jonhicks8999]

@@PhysicsMadeEasy I love it when those nuances reveal themselves. I had never considered the possibility of any kind of standardization from law to law. Really cool!

[@vishtrinity]

Why did we not take dx negative in the integral..the force applied on the system is in direction opposite to displacement so it should ideally be negative

[@PhysicsMadeEasy]

Hello Vishtrinity. The negative sign is there, it is just hidden in the order I chose for the integral boundaries (x final is R a x initial is infinity) :-)

[@alexandermrkich8734]

Good explanation. Thank you.

[@sakshamshrivastava4221]

Thank you sir you helped a lot by making this video ❤

[@PhysicsMadeEasy]

You are welcome Saksham. I am glad if it was usesul to you.

[@zakirhussain-js9ku]

What happens to gravitational potential energy of electron and positron after annihilation as photons so created have no mass and therefore no gravity.How gravitational field emerges back after pair production?

[@PhysicsMadeEasy]

Einstein: what we perceive as gravity is the curvature of space time. The shape of space time is affected by mass AND energy... So a photon of energy equivalent to the rest mass of an electron, should interact gravitationally the same way than an electron. (note that for particles, gravity is extremely weak, so week that I believe we do not really understand it at this level, where QFT should be considered instead).

[@zakirhussain-js9ku]

@@PhysicsMadeEasy Thank you for reply. Electric, magnetic & gravitational fields arise from interaction b/w electric charges, magnetic poles & masses. Does gravitational field b/w photon and other mass prove that photon may not be massless. Electron & positron have mass, electric charge & magnetic moment. If energy of photon is equal to rest mass energy of electron or positron then where is electric & magnetic energy of these particles? I think mass is carrier of KE, electric & magnetic charges carriers of EM energy.

[@NiharikaAChirayil]

Thanks for the video! It helped.

[@PhysicsMadeEasy]

Hello Niharica, I am glad it did. Thank you for letting me know!

[@victoriacorcimaru1731]

This video is amazing!!! Special thanks for using calculus!!!!

[@PhysicsMadeEasy]

Thank you Victoria

[@Editer978]

I have a doubt that is troubling my mind for long time ,the question is that "I had calculated that during solar eclipse sun exerts larger force then earth , so why not moon go away from the earth orbit during solar eclipse "

[@PhysicsMadeEasy]

It does, but you are forgetting an important fact: these objects are in motion...: The Moon orbits the Earth, but also orbits the sun with a wobbly perturbation that is the effect of the Earth (that we consider the orbit around the Earth). You cannot just place 3 static masses, and observe what happens: If so, both the moon and the Earth would fall into the sun... Consider the motion of the masses, and things will start making sense :-)

[@mahirbalayev5835]

And does it mean that the potential energy of charged capacitor is negative? And how this negative potential energy has to interact with spacetime? And what will happen to normal particles in the field of negative energy?

[@PhysicsMadeEasy]

Hello Mahir, Energy is a relative quantity. Something has energy because it can lose some (by performing an action). The fact that GPE is always negative is just a question of where we place the zero. It is natural to place it at infinity because this is where an object has no capacity to do work (not being impacted by any external gravitational force). If you placed a partially charged capacitor A with a voltage V=Q/C, and you placed it in parallel with an identical capacitor B with a higher charge (for the same capacitance), then you could say that the energy contained in A is negative RELATIVELY to the energy of Capacitor B (B would start charging A). "And what will happen to normal particles in the field of negative energy?" Well they just fall further: they have now negative potential energy that will become even more negative as it convert to KE… and you would need to work to get them out of there (i.e. render their PE equal to zero) See my video « What is Energy », and check the second part where I discuss how energy is relative. Be well

[@mahirbalayev5835]

@@PhysicsMadeEasy thanks a lot for your prompt response and explanation. I have asked this question, because I want to know if these capacitors - A and B will will operate as a warp drive? Or "negative energy" for warp drive is another kind of energy?

[@mahirbalayev5835]

@@PhysicsMadeEasy math on physics can predict some processes. If GPE at the infinite point is zero, then in the center of planet or something massive it will be negative infinite. And Universe is expanding, and GPE is coming closer to zero. And could we assume that the reason of expanding of universe is black holes? :)

[@zakirhussain-js9ku]

Does the gravitational field borrow its energy from space. Space is only thing around or b/w mass bearing objects & could be the only source of gravitational potential energy.

[@PhysicsMadeEasy]

Hi Zakir. it depends on what perspective you are taking (relativity or quantum mechanics). We'll consider relativity, because this is the theoretical realm of gravity... Yes you could kind of say that... But remember, first, you needed to place the massive object in that space... that requires work... So your question is maybe, what is the origin of energy as a whole. At the big bang, a pot of energy was given to the universe, it was then just distributed among its elements, and is exchanged between them... Note though that gravitational potential energy needs a second object to really exist... If you have an universe with only one mass, therefore a single gravitational field, can one consider that this gravitational field has energy if it cannot work on anything?

[@Rus-gomez]

Thanks a lot sir.... Your explanations are far better than my teachers in school... Gotta show them your videos 😤😤... Love from India🇮🇳❤️

[@jahnvigupta1678]

Amazing explanation sir but I have a doubt in the first case when we are moving mass m from r to infinity we are doing work on the system losing energy in the process therefore the PE of the system increases now in the second case when we are moving mass m from infinity to r can we say we are working against the system that's why the PE decreases?

[@PhysicsMadeEasy]

Hello Jahn, I’ll try to clarify this for you The situation: a mass M at the origin of an axis and a test mass m on the positive side of the axis at a distance r from M. M applies a gravitational force on m directed towards M (negative direction) Case 1 : m is moved from r to infinity. To do so, we need to pull m away with a force F directed towards the positive direction. The work done on the mass m (F * displacement) is positive, so the energy of the system increases. The GPE becomes less negative. Cade 2 : m is moved from infinity to r. if we do not act on m, the mass m will naturally accelerate towards M, thus gain kinetic energy (therefore the GPE decreases because of conservation of energy of the M-m system). In the video, I apply a force in the positive direction equal to the gravitational force so that the velocity (and the KE) remains constant . Still the mass moves towards M (in the negative direction), so the work done on the mass m is negative : the GPE decreases (becomes more negative). "in the second case when we are moving mass m from infinity to r can we say we are working against the system that's why the PE decreases?" : you could says that the system is working on you (you are waterskiing!), or that you are performing negative work on the system (=taking some of its energy)...

[@johnyfausman959]

I have a question. If work= change in potential energy, then moving an object from a -3J GPE to -5J GPE would result in the work being -2J which is negative. How can work be negative? because I'm taught in my high school that work can not be negative. I understand the transfer of energy thing going on here but this contradiction is unsettling to me. Thanks for your time :)

[@PhysicsMadeEasy]

Hi Johny, Work is always positive if you adjust your perspective… Suppose you have an object with an energy Eo and you have an energy Ep. When you work on an object, you transfer energy to that object. Suppose you work an amount W. The energy of the object would now be Eo + W. But what about you? this energy came from you, you lost the energy. So your energy is Ep - W. But what if it is the object pulling you? You could consider the same: Object : Eo + W , You: Ep - W, in that case the work would be negative. In order to keep the work positive, you need to change perspective, and say that it is the object working on you…In that case, the energy of the object is now Eo - W, and your energy Ep + W. So if you wish to keep the work positive in the example you propose, you must consider the work done from the perspective of the object that is losing energy. The object that went from -3J to -5J worked on you +2J, and you gained + 2J (it pulled you down with it!). I hope it helps

[@johnyfausman959]

@@PhysicsMadeEasy Thanks for your detailed reply, this makes sense.

[@lifeandmore_as]

IT HELPED ME A LOT ,TNX

[@AK56fire]

At 2:50 you wrote W = Delta PE. But W = - Delta PE. You missed the minus sign there.

[@PhysicsMadeEasy]

Hi there, it is a positive sign... When you provide energy to an object when working, it increases its energy.... W = Work done by me on the object of interest (i.e. received by that object of interest) PE = Potential Energy of the object of interest note: if I were referring to myself applying a force over a distance as the system., in that case, yes, there would be a negative sign...

[@faizah6043]

Very useful sir! One question though, i don’t really understand why the mass has to travel with constant velocity

[@PhysicsMadeEasy]

Hi Faizah. I suppose you are referring to when we are deriving the equation for gravitational potential energy. To do so, we calculate the work needed to bring a mass m from infinity (where GPE = 0) to a distance R from the mass M. If the speed of the motion of m would not be constant, the work would also contribute to the change of kinetic energy of the mass m. The GPE would not be equal to the work applied on mass m. To summarise. W = DeltaGPE + deltaKE. So to have W = GPE, in addition to have GPE initial = 0, we need deltaKE to be 0 (speed = contant).

[@faizah6043]

@@PhysicsMadeEasy ahhh i see, i get it now. thankyou!:-)

[@sudhanshurazz4344]

But Sir why the GPE has to be equal to the external work done; why it can't be less than the work done by external force ..?? If this is silly question I m sorry.

[@קטיהלסמן]

Thank you great video. Could you please explain why you integrated?

[@PhysicsMadeEasy]

Hi, In the video, I want to calculate the work (done by a force F) on a test mass to bring it at a distance R from the mass creating the gravitational field. I can’t use W = FdeltaX, because the force F depends on the position (I chose F opposite but equal in magnitude to the gravitational force so not to introduce any change in the kinetic energy of the point mass…) . This is why I have to integrate: I sum all the works done on infinitely small intervals dx for which the force can be considered as constant. I hope this helps, Greetings, E.

[@קטיהלסמן]

@@PhysicsMadeEasy thank you! This really helps. Your videos are great!!!

[@shravanichawathe1358]

But what about mgh formula which was for small height why is it positive and the general formula has a negative sign . From the video i got to completely acknowledge the fact that epwhy it has nehative but mgh still bothers me.😢

[@PhysicsMadeEasy]

Hi Shravani. mgh is a formula for a potential energy difference: it is actually mgDELTAh, and for convenience we set the height of the ground to zero (hence mgh). Another way to see it: If the real formula (-GMm/R) was used, when bringing an object up, the value would for PE would become less negative (absolute value would decrease). If you consider starting from zero, than it appears positive. Note that mgh can only be used when g is approximated independent from position (i.e. the Grav field is considered uniform)

[@dipsankhanal017]

Thank you sir😃

[@Ii-fo8pq]

You're doing a great job really

[@jonathanstevens1917]

Brilliant !! Thank you

[@PhysicsMadeEasy]

Thank you Jonathan :-)

[@narrowbtw3970]

At the start you stated w > 0, and then ended up with w = -G(mM)/r. How?

[@PhysicsMadeEasy]

In a few words, W>0 means that the object being worked on is receiving energy. Now, in order to bring a system of 2 objects interacting gravitationally infinitely away from each other (that is far enough so that the gravitational interaction is negligible), you need to apply a force counteracting the gravitational force: You are doing work on that system, in other words the energy of the system is increasing. When the two objects are infinitely far away from each other, there is no more gravitational interaction, their gravitational energy is zero. This implies automatically that the gravitational energy they had when interacting gravitationally was negative.

[@narrowbtw3970]

@@PhysicsMadeEasy wow ok now i got thanks so much!!!

[@naseemessa9092]

Thank you so much for this

[@AbhraTalksaLot]

Amazing! Thanks a lot

[@sudarshann7194]

Is this definition of 'gravitational potential energy at a point' is correct? 👉 " it is the work done to bring a unit mass from infinity to that point against!! gravity " . If "no" why , if "yes" why . (If anyone know answer , please answer it ). Will you please Answer this sir

[@sudarshann7194]

There's against gravity in that definition . My teacher taught me and I'm not able to know if it's right . So I want the answer to be more intuitive . (And I love your videos like hell )❤️

[@PhysicsMadeEasy]

What you wrote is the definition of a gravitational potential because you are talking about a unit mass. The definition of gravitational potential energy is: The Gravitational Potential Energy of a test mass located at a given position is the work required to bring that test mass from infinity to that position. To say “against gravity” is not incorrect but not absolutely necessary either. Your teacher probably wishes to insist that the force doing the such work must be opposite to the gravitational force and of the same magnitude at all times. This is so that the work done does not contribute to an increase of the kinetic energy of the object, which would invalidate the definition. However by saying ‘work required’ or ‘work needed’ instead of just ‘work done’ in the definition, you eliminate the idea that some of the work will go to do other things like increasing the kinetic energy. In that case, in my opinion, you do not need to say ‘against gravity’. All this is just wording… It is important, but what I find more important is that a student understands clearly the difference between gravitational Potential and gravitational potential energy! I hope my answer helps!

[@sudarshann7194]

@@PhysicsMadeEasy 🙏 thank you sir . That explanation jus made me like 'shut my mouth' , it was more detailed . It helped me. And I love my teacher to say so that defination bcz It made me think more on it . I wish someday I want to be your student . ❤️

[@actioncamsports]

The singularity at r=0 is unsettling.

[@PhysicsMadeEasy]

It is, but not so much :-) How could you position yourself at r = 0... You need an object with a mass to generate a G field. That object has to have a size, thus the source of gravity cannot be punctual, so r=0 has no meaning... It has a meaning when the density become infinite (like theoretically at the center of a black hole). But we haven't been in a black hole to check out if matter really collapses in a singularity at its center. There could be a pressure of some kind that prevents this to occur (Quark stars, Strange stars, and even if this is not enough, we get limited by the Plank length). So to get unsettled, we need to discover a theory of quantum gravity, and validate it. Today, that would be the Saint Graal in Physics! Give it a shot! Succeed, and you get the Nobel prize ten years in a row :-)

[@theunpredictable1354]

Awesome❤❤

[@akashhera]

This is the most intuitive explanation I have ever seen of this concept! Thanks a lot sir

[@PhysicsMadeEasy]

Thank you Akash. I am glad my approach on explaining GPE helped!

[@saadkarim3516]

Sir , can you please explain Absolute potential energy ? , Nothing in the universe is absolute then , why is it called absolute potential energy..?? Please explain

[@PhysicsMadeEasy]

Hello Saad, You are correct in saying that potential energy is never absolute (See my video ‘What is Energy’). The subtlety is that the reference position, where PE = 0, is set at infinity. I do not remember mentioning an ‘absolute’ PE in the video, but it’s been a long time since I shot it, so if I did, it was probably to differentiate from when setting PE = 0 at the level of Earth’s ground…

[@zizerokub6799]

every mass in solar system or universe have Gravitation to each other much or less? so there should be no reference point? or every mass in earth should have reference point inside center of the earth not earth surface?

[@PhysicsMadeEasy]

Hey Zizero, A reference point is arbitrary. You can choose it where you want (more precisely, where it is most convenient) depending on what quantity you are considering (energy, gravity field stength etc.). For example, when dealing with the GPE of things on the surface of the Earth, where gravity field strength g is assumed constant (GPE = mgh), it is convenient to set the reference point at the sea level (= setting a height of zero). For two objects interacting in space (like the Earth and the moon), it is more convenient to set the reference point (or more precisely the distance between objects) at infinity (GPE = -GMm/d) Example: "every mass in earth should have reference point inside center of the earth not earth surface?" You can calculate a problem in both ways: either you consider the height, that is distance between ground and object . The distance between center of Earth and ground has already been factored (in g). leading to PE = mgh (condition to do so, g is assumed constant) either you consider the distance between the object and the center of the earth PE=-GMm/d I hope this helps

[@zizerokub6799]

​@@PhysicsMadeEasy Light is 2D object in this 3D universe?

[@saidhanif5887]

Please explain total energy of electron on this basis

[@PhysicsMadeEasy]

Hi Said, can you reformulate your question? What do you mean by "total energy of electron on this basis". Maybe you are talking about electrons in atoms? If so, at such scale, gravity is so weak compared to the electric force, that it is neglected (at this scale, gravity is something like 10^40 times weaker than the electric interaction).

[@or5026]

What are qualities of protons of themselves alone? Are any of them in cosmos in pure form without negative charges around? Maybe proton stars?

[@PhysicsMadeEasy]

Of course! actually most of protons in the universe are by themselves: in space most of the hydrogen gas is ionised by the UV light from stars... You get protons by themselves. And in stars, the temperature is too high for electrons to stay kindly around the protons. So your 'proton stars' are just stars ;-)

[@cookiecains3045]

Great video!

[@chitral5665]

Why we take integration of force for work

[@PhysicsMadeEasy]

Because if you just write W = F * displacement... What will you put for F? F changes with the position (inversely proportional to the square of the distance)... So you need to consider each infinetely small displacement dx, for which the force can be considered constant. Apply the formula dW = F * dx (where dW is the corresponding infinitely small amount of work), and sum the whole thing to get the total work...

[@vimalathithand917]

sir what is the significance of gravitational potential?and what is meant by negative energy? how can a body have negative energy?

[@PhysicsMadeEasy]

Gravitational potential is like gravitational potential energy but per Coulomb of charge. You can check my video on electric potential, where I also explain this concept. kzbin.info/www/bejne/oGSqo4KBp8qSlZI. An energy is a relative quantity. So depending where/what you consider for the origin, the energy can be considered as negative. Think about GPE in mechanics (where g is considered constant). Bring up a mass m at a height h from the ground, it gains a PE = mgh. Now, before you let the object fall, you dig a hole of 1 meter deep. You let if fall: it loses all its PE for KE. How much did it lose? mg(h+1), and now that it is in the hole, its PE is -mg1... negative. Just because you took the ground as reference... So, you see a negative PE does not have a special physical meaning... I hope this helps

[@vimalathithand917]

@@PhysicsMadeEasy thank you sir .it helps me a lot

[@erick.gudino]

Thannk you!

[@sidhanaravind925]

Sir the applied force does negative work right which is gpe.but gravity also does some work right because the force and displacement are at same side. in which form that energy is stored.if that work done is also equal to negative of gpe thus positive.then the body has no energy 😵‍💫😵‍💫😵‍💫

[@PhysicsMadeEasy]

Hello Sidhana, I know where your confusion comes from: be careful at what system you are considering. The system considered in the video is the two masses. The applied force is external to that system so does work on it. The gravitational force is internal to that system, so does not do work on the system. If you consider the test mass as being the system. What applies is the work-kinetic energy theorem. (the concept of potential energy does not apply on a system made of one single entity).

[@sidhanaravind925]

@@PhysicsMadeEasy Thank you sir

[@tanvibadadare2622]

you are very talented!!

[@PhysicsMadeEasy]

Thank you Tanvi

[@ahmedatifabrar7698]

What a great soul you are!

[@PhysicsMadeEasy]

Thank you Ahmed :-)

[@maciej12345678]

hmm what is Grawitation kinetic energy and is ther eqaution like Ec = Ek + Ep analog to shordineger eqaution. Maybe imagynary mass can be obser like bending space -time oposit to positive energy-mass

[@PhysicsMadeEasy]

Hi Neo Vision, There are two types of mechanical energies: Kinetic Energy (KE) and Potential Energy (PE). KE is the energy of a body related to its motion, and PE is the energy related to its position. The adjective you place in front of PE refers to the cause of the energy (Electrical, elastic, gravitational). We need that because the calculation of that energy is directly related to the cause. For KE, the formula is always the same (1/2 * m * v^2), so there is no need to mention the cause. For example, an object in motion because it has been under the influence of a gravitational force could be said to have gravitational kinetic energy, but the fact that it is gravity that made it move is of no consequence: if the force at the origin of the motion had been electrical, with the same magnitude, direction and time of application, the motion of the object would be the same. Negative mass is a concept that is explored. Check this for more info: en.wikipedia.org/wiki/Negative_mass

[@ff_gaming8179]

Thank you for you help I can understand it very easily

[@PhysicsMadeEasy]

Thanks FF, I am glad my work helped you do so :-)

[@ExtraterrestrialIntelligence]

what about negative mass?

[@PhysicsMadeEasy]

Matter with such property has never been seen observed, only speculated about. Although one alternative could be... antimatter: Negative mass would react in a gravitational field created by a normal mass by feeling a force in the opposite direction as normal mass would. Experiments have been carried out to evaluate how antimatter reacts in Earth' s gravitational field: Does it fall downwards like normal mass, or does it fall "up"? Unfortunately, the uncertainty on such experiments are two high to conclude either way. So we need to wait until measurement techniques improve significantly to have an answer... And as for the subject of the video, if in the future we discover that two particles can indeed experience a repulsive force of gravitational origin, then positive gravitational potential energy would be possible (like it is with electrical potential energy between two charges of opposite signs)

[@namelesss8226]

Why doesn't the centripetal force cause objects to collapse into the center? And if centripetal force acts on a car when rounding a bend why does the passenger get pushed to the other direction?

[@PhysicsMadeEasy]

"Why doesn't the centripetal force cause objects to collapse into the center?" Hi, think about free fall. When a ball is travelling horizontally, the only force is gravity acting vertically, so the ball is accelerated vertically downwards. But because it still has horizontal velocity, the ball doesn't fall vertically, but as a parabolic motion. It is a similar idea with circular motion: If there is a centripetal force, and the object affected by this force has velocity (in another direction than the center), it will not fall directly towards the center because of its velocity. “if centripetal force acts on a car when rounding a bend why does the passenger get pushed to the other direction?” Because the Force is acting on the car not the passenger: the passenger has velocity forwards, unless a force acts on him, he will continue going in a straight line (1st law of Newton)… The passenger will therefore not have the same motion as the seat or the inside part of the car door. But because he is in contact with those things, friction at the (seat - passenger's ass) interface will occur and if the passenger gets in contact with the door, there will be a reaction force of the door on the passenger too. In the end, these forces will push the passenger to follow the path of the car.

[@namelesss8226]

@@PhysicsMadeEasy omg, thank you so so much, the analogy of the free fall made circular motion much clearer. you made it so simple!!

[@shivamithra1651]

What is C,P,T? Where the time going?

[@PhysicsMadeEasy]

I suppose you are referring to the CPT (Charge, Parity and Time) symmetries. Way too complex to talk about this in a comment, but I wrote an article about it a few years ago: you can consult here: steemit.com/steemstem/@muphy/the-mystery-of-the-missing-antimatter-an-introduction-to-cpt-symmetry-particle-physics-series-episode-4c

[@shivamithra1651]

@@PhysicsMadeEasy Tqs for reply

[@AnilJangraOM]

Just wonderful 👍

[@ndwani5724]

Many unworthy ones are getting millions of views whereas many worthy content creators get less views and subscribers conpared , I also did not get my worth in my another channel. If God created the system that everyone gets what he deserves, it would be good. But to reality, "one gets what he want not what he deserves". We have to constantly chase, that is one kind of have greed to get something, the wanted thing rather needs more attention than worth of the gainer.

[@PhysicsMadeEasy]

Yes, but if what you want is simplicity? Even if I deserved to a be a super famous teacher star on (inter)national television, and earn millions (with all the negative consequences, stress, not being able to live a simple life, little time to spend with my friends and loved ones)... Well, I don't want that... I'd rather be the way I am now ;-)

[@samahashah9826]

Can energy be negative?

[@PhysicsMadeEasy]

Energy per say is not negative because it is a relative quantity. In the video the explanation is that 2 objects that do not see each other (therefore that cannot work on each other), are considered as having no energy. For two objects attracting each other, you need to bring energy to them to separate them fully (in order for them to reach an energy of zero). Therefore, originally, their potential energy had a negative value. Yet, when saying "energy" you have to consider the sum of the Kinetic energy and potential energy. When something falls on something else, the potential energy of the system becomes more negative, but the Kinetic energy becomes more positive, balancing things out... So energy as a whole for an isolated system energy is not negative. Consider the universe as a system. it had an initial amount of (positive) energy. Assumed it is isolated (it does not interact on something else). The distribution of this energy among its components changes with time, but the system as a whole still has the same amount now (positive).