Hey, I just wanted to say thank you for making this video. Often in physics, I often feel frustrated because I feel that I don't understand equations that are given to, I don't understand the explanation behind them, how they were formed, and how the derivation leads to the equation doing what its supposed to do. This video really helped me understand ths work of gravity euation/ concept, and I now feel my frustration has been relieved. You earned a new subscriber
Hi there, just wanted to say great video! What you have on this channel is really special, and the nice visuals really help with understanding too. Like Parv already said keep at it, it will all pay of I am sure of it :)
Oh that’s a nice question and it’s because the external force (whatever it is) must exactly overcome the gravitational force but not give the object any kinetic energy. So the real definition is the work done to move the masses together from infinity at a roughly constant low velocity.
also can i say that -gMm/r is also the work done to put that mass from that point to ground. And that means that the energy spent to take the object from infinity to that point would be equal to the energy from that point to earth under only earths gravitational field.
Well of course that isn’t technically true since the ground is well above the centre of the earth (even more technically, Earth’s centre of mass) but I see what you mean. It’s an interesting point
@suspended3785 Oh yeah sorry for getting back to you on this late and it’s useful cause it allows me to be a little clearer than I was a year ago. The external force needs to be there in both definitions so that the KE of the object doesn’t change as it comes in from infinity to its position (otherwise finding that work done which we know is energy transferred wouldn’t tell us straight away that all the W went into changing the PE, it would have also changed the KE). What I meant, I think, is that you could say the GPE is the work done by the external force or the negative of the work done by gravity (both would give the same answer, so I see no reason you couldn’t define it like that)
I understand the math behind the derivation, I just don't understand why conceptually PE is defined as the work done by some external force in bringing 2 bodies from infinitely far away to the distance they are now. Why is potential energy defined this way? Also, is this a definition that can be used to derive other forms of potential energy, or does is this only how we define GPE?
Great questions! Your second question is easier to answer: other forms of potential energy are defined this way such as electrical potential energy. Your first question is interesting, I think it can basically be attributed to the fact that potential energy is really just a bookkeeping device, so having a number that instantly tell you how much energy is needed to pull them apart is nice. Plus, conservation of energy wouldn’t work without this definition. But both of those seem like answers a mathematician would give, not a physicist, so I hope that’s helpful anyways but let me know if you want a better explanation.
Here's the thing For all conservative forces the change in potential energy of an object equals negative of the work done by that force on that object now, a lot of people just to make things easier, write that negative work done by the force as a the positive work done by another external agent to balance the original force out I'm myself struggling with how the directions work tho honestly, i can't figure out what the direction of dr is
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
Sorry about the late response but the important point is the Work done we’re talking about is the work done by an external force to make sure that on the whole journey from infinity to r you don’t gain any KE. The gravitational force is pulling you inwards (negative sign in radial coordinates) so this external force has to push you equally outwards (positive sign in radial coordinates). Hopefully that clears it up
@@TheBrainFiller I understand everything, except since Potential energy is negative, the more infinite you go the greater r is. Since it is negative, the P.E fraction gets smaller(which means it gets stronger since -1 is greater than -2). So if it’s invite shouldn’t the P.E be absolutely huge, not zero. That’s the only problem I have. It makes sense with force, but not with P.E.
“The origin of momentum creation is founded on the application of energy. Momentum serves as the initial cause of pushing and pulling. The energy, emanating from the internal core of a planet, is what gives rise to momentum, commonly referred to as gravity, and not due to the planet’s size, mass, or space-time curvature. For example, when energy is applied, it generates the necessary momentum for the actions of pushing and pulling. Without the application of energy, momentum cannot be established. Consequently, force cannot be exerted for pushing and pulling, as it lacks a physical existence. This, in turn, clarifies why force is not gravity.” ~Guadalupe Guerra
you made a mistake here: 3:45 Firstly, let's derive the work kinetic energy theorem: W=\int_Xi^Xs F dx W=\int_Xi^Xs ma dx W=\int_Xi^Xs m.dv/dt dx W=\int_Xi^Xs m(dv/dx)(dx/dt) dx dx/dt=u du=d/dx(dx/dt)dx du=(dv/dx)dx d(Xi)/dt=Vi d(Xs)/dt=Vs W=\int_Vi^Vs m(du/dx)(u) dx W=\int_Vi^Vs m.u(du/dx) dx W=\int_Vi^Vs m.u du W=(1/2)m.(Vs)^2-(Vi)^2 W=ΔKE...(1) and we know the conservation of energy KE+PE=PE'+PE' -ΔPE=ΔKE If we substitute equation number 1 -ΔPE=W so you was say ''PE=W''
It’s all about thinking who is doing the work. That changes the negative sign. I defined the potential energy as the work done by an external force going from infinity to the present location. Now think about your derivation a bit more carefully (the maths is correct but think about the interpretation). Specifically, what is the change of the kinetic energy of my object across the whole process? It started at infinity at rest and it ends at its current location again at rest so the change in the kinetic energy is 0! Wait so what’s happening was work done? Yes but the net work done is 0. So this external force did some work and it is exactly the negative of the work done by the gravitational field. That’s why I don’t need the negative sign. Have a look at this stack exchange post for some clarification: physics.stackexchange.com/questions/761716/potential-energy-and-work-done-by-external-force Thanks for watching and I appreciate that you’re thinking about the material deeply this is tricky stuff.
Also thinking about a possible issue remember work is actually the integral of the force vector dotted with the dr vector. When we’re moving from infinity to r the gravitational force is pointed radially inwards so in spherical polar coordinates there is a negative sign on that force vector and then we dot it with a displacement dr which is positive. Soo….overall F•dr is negative so then if we were doing an integral from r to infinity that minus sign would still be present. Luckily I swapped the limits of integration and when from infinity to r so that cancels out the negative sign on F•dr. So honestly an argument could be made I was not clear enough on where I was getting that exact integral expression from. An important point is potential energy can be defined in terms of the work done by an external force or also equivalently by the work done by the gravitational field but the negative sign swaps in the two definitions. My final answer is certainly the correct standard result used and the literal maths is not wrong but I should have been clearer on the physics.
@@TheBrainFiller Actually, what I don't understand is why we are trying to apply extra force and make the kinetic energy 0. Is this really necessary? The decrease in potential energy is equal to the work done by the force applied by the source of potential energy, and at the same time the work done is equal to the change in kinetic energy, but the change in kinetic energy does not have to be 0 for us to establish a relationship between potential energy and work. Well W=-∆U=∆K What I don't understand is why extra force is needed.
@spaceghost00 Because if you want some kind of definition that you can write down and think about independently of KE you need this kind of definition. Physicists like energy as a bookkeeping device where they can add up some independent sources of energy and have a total like E = KE + PE and they want this definition of PE to be separate to KE. So if you’re trying to quantify what is the energy associated with a certain configuration of your system that is just due to the position alone then in your definition you need to cancel out any accumulating KE as you construct your system (bringing everything from infinity to its current setup) so that the final number you get from your definition is just about position in space and the presence of a gravitational field. This definition also emphasizes the path independence of this potential energy quantity whereas defining it in terms of the change in KE kind of obscures that. And we want a nice way to directly connect potential energy to the forces so that we can more naturally think about fields and stuff. Honestly it’s a good question worth looking into a bit more. Let me know if you find a nice intuitive explanation somewhere.
Hi there, I'm confused about something. In our example of an external force doing work on our objects to bring them from infinity to r, shouldn't this work be positive since the force and displacement vectors are in the same direction? This would be strange, since the derived formula is negative. Thanks for your help!
The external force is actually opposing the gravitational force because part of the definition is bringing the object in at a constant speed, if the external force was in the same direction as the gravitational force then the object would be accelerating as it came in and so also gaining kinetic energy but we want to isolate the potential energy. This was a tricky point for me to get too, so hopefully that clears it up but let me know
Hey, I'm hitting quite a roadblock when trying to derive the work done by gravity in bringing an object from infinity to r this work done should in theory be equal to +GMm/r but here's the derivation both the gravitational force and dr (direction of instantaneous movement) are in the same direction so it evaluates to the integral from infinity to r -> (GMm/r^2)dr = -GMm/r this is so counter intuitive its crazy. like my instantaneous work done was positive and everything made sense but the limits just completely changed the game and made the result -ve Is there any obvious mistake you can catch? it would be a ton of help
Yeah this is tricky so I get you. First just to be clear for anyone else who reads this to avoid confusion, rando_guy is calculating the work done by gravity, I was calculating the work done by the external force. Of course these have to exactly cancel out because when the object finally arrives at r from infinity it has 0 kinetic energy. So the change in the kinetic energy is 0 therefore the net work done has to be 0 too. Anyways onto your question…the point is I’m controlling the direction of the dr by choosing limits infinity to r rather than r to infinity. So flipping the sign of the dr but then also using my funky limits is doing the same thing twice which cancel each other out
Assuming you’re familiar with line integrals which I’m getting the sense you are W= int (F dot dr) from r_1 to r_2 yes but it’s always easier to think about line integrals by parameterizing the chosen path with a variable t and then doing W= int(F dot dr/dt dt) from t_1 to t_2 in which we can interpret t as being a time and dr/dt being the velocity vector of our particle along the path. Now let’s choose a path where the velocity vector is radially inwards (same direction as the gravitational force) the whole trip and is a constant speed v. So F dot dr/dt = (GMm/(r(t))^2)*v and r(t) = (X - v*t) (where we’ll take the limit X goes to infinity at the end cause at t=0, our t_1, the mass is at infinity and at t = (X+r_2)/v the mass is at location r_2). Then just do that integral with those correct time limits and finally take the limit as X goes to infinity and you should get the answer you’re looking for. Alternatively, if you just want the direction of vec(dr) its conventionally considered radially outwards and by flipping the limits of my integration I’m sort of flipping the direction of the dr to be radially inwards. But it is probably much clearer to define a velocity vector dr/dt and do the integral dt for no ambiguity
@@TheBrainFiller The velocity approach is interesting, I've never really come across it before. I pondered a bit more on this doubt, it's like saying if (dx î)/dt = -v î, then direction of dx is along -î. for which, yeah the small displacement is along that direction but we don't write that on the LHS while denoting the velocity right (that wouldn't make sense). If we multiply that equation with dt we would get the instantaneous motion along dx î only, regardless of what v vector is. Maybe not completely concrete but this is a way to think about it. I think it's in a way similar to the logic where your approach contributes
I highly recommend you do look into line integrals in that case cause yeah you’re getting the right idea and it’s a really interesting topic. dx/dt is a vector and so I’d write in -v r hat (for the radial unit vector in spherical coordinates) and then you dot product that with the Force vector which points in the -r hat direction too so F dot v will be positive the whole journey and then you’ll get a positive answer when you integrate from t=0 to t= time at which you reach the final position
hi just want to ask... why the graph in 2:58 shows infinity is the upper limit while r is the lower but in 3:04, u put the upper limit as r while the lower limit is infinity?
Right so the graph just shows a 1/r^2 relationship and of course that takes on values from 0 to infinity (I suppose I could have shown it from infinity to 0 but that’s not really conventional). I’m then suggesting that the object begins its trajectory infinitely far away and then moves in slowly until it arrives at r
thank you for the great explanation, although the video was pretty straight forward i may have missed something isn’t the work equal to Ep+Ek why did we ignore the kinetic energy of the planet? because the velocity of it was very low towards the begining? but as we approach to close distances that wont be the case any more so we cant ignore it. also didn’t we integrate from infinity to r, but the when he showed the graph of Force function it looked like we were integrating from r to infinity. i’ve been trying to derive this for 3 days and i tried to compute it like this: 0 to r(radius of earth) + r to some R (distance between moon and earth for example) im only a highschool student so im very new to this stuff i would be really happy if some one could tell me where was i wrong.
The external force applied throughout the trip from infinity to r basically works to keep the velocity of the object constant. It does this by exactly matching the force of gravity therefore, by Newton’s first and second law, the net force on the object is 0 so it continues moving at constant velocity. As a result the kinetic energy of the object is constant the whole time (and importantly very low…to the point it can be ignored). I can’t exactly remember what I did in the video but when I integrate from infinity to r, I should have got an expression like this: [-GMm/r] evaluated from infinity to r which means (-GMm/r)-(-GMm/(infinity)) (because you do the final state minus the initial state). There are two ways of defining a potential field: in terms of an external force or the force of the field. The latter definition is that Ep= - integral F dx and the one in terms of the external force is Ep= integral Fext dx. But the external force is obviously just -F (it faces the opposite direction) so both of these definitions are equivalent (as they should be).
To respond to your own attempt at a derivation. When you’re integrating from 0 to r note that the force of gravity isn’t just changing because of the distance from 0 but also the amount of mass below you. Look up Gauss’s law for gravity if you’re interested. The main thing to realise is that the treatment of gravity you’re learning applies to point objects only and Newton showed that spherical objects can be treated as point-like beyond their radius so it’s completely ok for us to think about orbits and such in terms of point particles.
@@TheBrainFiller thank you for yet another great explanation i will check your other videos as well and better my understanding of physics before the university. and if something bothers my mind i reckon i can ask them to you via comments.
As far as I know, Newton derived his law using the relationship between the period of an orbit and its radius (T^2=kr^3 which is Kepler’s third law) and centripetal force. I actually use the equation for gravitational force to derive Kepler’s third law so it would be weird to make a video where I go the other way round lol. If you want to see that video I’ve linked it here: Orbital Mechanics - Orbital Speed, Total Energy and Kepler’s 3rd Law ruclips.net/video/SP41C33FBT0/видео.html
i was AHHHAAAAAAing so hard at this you wouldant believe
That’s awesome to hear. Thanks for watching
Hey, I just wanted to say thank you for making this video. Often in physics, I often feel frustrated because I feel that I don't understand equations that are given to, I don't understand the explanation behind them, how they were formed, and how the derivation leads to the equation doing what its supposed to do. This video really helped me understand ths work of gravity euation/ concept, and I now feel my frustration has been relieved. You earned a new subscriber
Glad it was helpful. Thanks for watching and subscribing!
Hi there, just wanted to say great video! What you have on this channel is really special, and the nice visuals really help with understanding too. Like Parv already said keep at it, it will all pay of I am sure of it :)
Wow thanks so much
hey buddy. i just wanna say. dont ever think to stop. because believe me your videos are great. you will succeed
Lol I appreciate that. I’m glad you found it helpful.
You said it is the work done by an external force so why when integrate you put the force of gravity
Oh that’s a nice question and it’s because the external force (whatever it is) must exactly overcome the gravitational force but not give the object any kinetic energy. So the real definition is the work done to move the masses together from infinity at a roughly constant low velocity.
also can i say that -gMm/r is also the work done to put that mass from that point to ground. And that means that the energy spent to take the object from infinity to that point would be equal to the energy from that point to earth under only earths gravitational field.
Well of course that isn’t technically true since the ground is well above the centre of the earth (even more technically, Earth’s centre of mass) but I see what you mean. It’s an interesting point
Thank you!!!
this is the masterpiece that I've been looking for for a decade ♥
Very kind, thanks for watching
@@TheBrainFiller wish U the best
Well explained brother.👍
may i ask why an externla force has to show up in the definition of GPE? why do we not consider the work done by gravity itself instead?
Oh you absolutely can. In fact there are two equivalent definitions of GPE: one involving an external force and one not.
@@TheBrainFillerwhat's the other one
@suspended3785 Oh yeah sorry for getting back to you on this late and it’s useful cause it allows me to be a little clearer than I was a year ago. The external force needs to be there in both definitions so that the KE of the object doesn’t change as it comes in from infinity to its position (otherwise finding that work done which we know is energy transferred wouldn’t tell us straight away that all the W went into changing the PE, it would have also changed the KE). What I meant, I think, is that you could say the GPE is the work done by the external force or the negative of the work done by gravity (both would give the same answer, so I see no reason you couldn’t define it like that)
Bro. You have the greatest visualization vid. Dont stop !
Thanks!
I understand the math behind the derivation, I just don't understand why conceptually PE is defined as the work done by some external force in bringing 2 bodies from infinitely far away to the distance they are now. Why is potential energy defined this way? Also, is this a definition that can be used to derive other forms of potential energy, or does is this only how we define GPE?
Great questions! Your second question is easier to answer: other forms of potential energy are defined this way such as electrical potential energy. Your first question is interesting, I think it can basically be attributed to the fact that potential energy is really just a bookkeeping device, so having a number that instantly tell you how much energy is needed to pull them apart is nice. Plus, conservation of energy wouldn’t work without this definition. But both of those seem like answers a mathematician would give, not a physicist, so I hope that’s helpful anyways but let me know if you want a better explanation.
@@TheBrainFiller That's helpful, thanks!
Here's the thing
For all conservative forces the change in potential energy of an object equals negative of the work done by that force on that object
now, a lot of people just to make things easier, write that negative work done by the force as a the positive work done by another external agent to balance the original force out
I'm myself struggling with how the directions work tho honestly, i can't figure out what the direction of dr is
Very helpful and well explained video bro thank you. And great visual effects to show the physics concepts!
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
Sorry about the late response but the important point is the Work done we’re talking about is the work done by an external force to make sure that on the whole journey from infinity to r you don’t gain any KE. The gravitational force is pulling you inwards (negative sign in radial coordinates) so this external force has to push you equally outwards (positive sign in radial coordinates). Hopefully that clears it up
This video was amazing and you explained it very well! Keep making such awesome videos.
Thank you for the video
W=Fdxcos180 ;Fext.^dx =180 w=work done by external force but you didn’t write "cos180", why?
Basically for simplicity’s sake lol. Thanks for watching
@@TheBrainFiller I understand everything, except since Potential energy is negative, the more infinite you go the greater r is. Since it is negative, the P.E fraction gets smaller(which means it gets stronger since -1 is greater than -2). So if it’s invite shouldn’t the P.E be absolutely huge, not zero. That’s the only problem I have. It makes sense with force, but not with P.E.
“The origin of momentum creation is founded on the application of energy. Momentum serves as the initial cause of pushing and pulling. The energy, emanating from the internal core of a planet, is what gives rise to momentum, commonly referred to as gravity, and not due to the planet’s size, mass, or space-time curvature. For example, when energy is applied, it generates the necessary momentum for the actions of pushing and pulling. Without the application of energy, momentum cannot be established. Consequently, force cannot be exerted for pushing and pulling, as it lacks a physical existence. This, in turn, clarifies why force is not gravity.” ~Guadalupe Guerra
you made a mistake here: 3:45
Firstly, let's derive the work kinetic energy theorem:
W=\int_Xi^Xs F dx
W=\int_Xi^Xs ma dx
W=\int_Xi^Xs m.dv/dt dx
W=\int_Xi^Xs m(dv/dx)(dx/dt) dx
dx/dt=u
du=d/dx(dx/dt)dx
du=(dv/dx)dx
d(Xi)/dt=Vi
d(Xs)/dt=Vs
W=\int_Vi^Vs m(du/dx)(u) dx
W=\int_Vi^Vs m.u(du/dx) dx
W=\int_Vi^Vs m.u du
W=(1/2)m.(Vs)^2-(Vi)^2
W=ΔKE...(1)
and we know the conservation of energy
KE+PE=PE'+PE'
-ΔPE=ΔKE
If we substitute equation number 1
-ΔPE=W
so you was say ''PE=W''
It’s all about thinking who is doing the work. That changes the negative sign. I defined the potential energy as the work done by an external force going from infinity to the present location. Now think about your derivation a bit more carefully (the maths is correct but think about the interpretation). Specifically, what is the change of the kinetic energy of my object across the whole process? It started at infinity at rest and it ends at its current location again at rest so the change in the kinetic energy is 0! Wait so what’s happening was work done? Yes but the net work done is 0. So this external force did some work and it is exactly the negative of the work done by the gravitational field. That’s why I don’t need the negative sign. Have a look at this stack exchange post for some clarification: physics.stackexchange.com/questions/761716/potential-energy-and-work-done-by-external-force
Thanks for watching and I appreciate that you’re thinking about the material deeply this is tricky stuff.
Also thinking about a possible issue remember work is actually the integral of the force vector dotted with the dr vector. When we’re moving from infinity to r the gravitational force is pointed radially inwards so in spherical polar coordinates there is a negative sign on that force vector and then we dot it with a displacement dr which is positive. Soo….overall F•dr is negative so then if we were doing an integral from r to infinity that minus sign would still be present. Luckily I swapped the limits of integration and when from infinity to r so that cancels out the negative sign on F•dr.
So honestly an argument could be made I was not clear enough on where I was getting that exact integral expression from. An important point is potential energy can be defined in terms of the work done by an external force or also equivalently by the work done by the gravitational field but the negative sign swaps in the two definitions.
My final answer is certainly the correct standard result used and the literal maths is not wrong but I should have been clearer on the physics.
@@TheBrainFiller Actually, what I don't understand is why we are trying to apply extra force and make the kinetic energy 0. Is this really necessary? The decrease in potential energy is equal to the work done by the force applied by the source of potential energy, and at the same time the work done is equal to the change in kinetic energy, but the change in kinetic energy does not have to be 0 for us to establish a relationship between potential energy and work. Well
W=-∆U=∆K
What I don't understand is why extra force is needed.
@spaceghost00 Because if you want some kind of definition that you can write down and think about independently of KE you need this kind of definition. Physicists like energy as a bookkeeping device where they can add up some independent sources of energy and have a total like E = KE + PE and they want this definition of PE to be separate to KE. So if you’re trying to quantify what is the energy associated with a certain configuration of your system that is just due to the position alone then in your definition you need to cancel out any accumulating KE as you construct your system (bringing everything from infinity to its current setup) so that the final number you get from your definition is just about position in space and the presence of a gravitational field. This definition also emphasizes the path independence of this potential energy quantity whereas defining it in terms of the change in KE kind of obscures that. And we want a nice way to directly connect potential energy to the forces so that we can more naturally think about fields and stuff.
Honestly it’s a good question worth looking into a bit more. Let me know if you find a nice intuitive explanation somewhere.
Hi there, I'm confused about something. In our example of an external force doing work on our objects to bring them from infinity to r, shouldn't this work be positive since the force and displacement vectors are in the same direction? This would be strange, since the derived formula is negative. Thanks for your help!
The external force is actually opposing the gravitational force because part of the definition is bringing the object in at a constant speed, if the external force was in the same direction as the gravitational force then the object would be accelerating as it came in and so also gaining kinetic energy but we want to isolate the potential energy. This was a tricky point for me to get too, so hopefully that clears it up but let me know
Hey, I'm hitting quite a roadblock when trying to derive the work done by gravity in bringing an object from infinity to r
this work done should in theory be equal to +GMm/r
but here's the derivation
both the gravitational force and dr (direction of instantaneous movement) are in the same direction
so it evaluates to the integral from infinity to r -> (GMm/r^2)dr = -GMm/r
this is so counter intuitive its crazy. like my instantaneous work done was positive and everything made sense but the limits just completely changed the game and made the result -ve
Is there any obvious mistake you can catch? it would be a ton of help
Yeah this is tricky so I get you. First just to be clear for anyone else who reads this to avoid confusion, rando_guy is calculating the work done by gravity, I was calculating the work done by the external force. Of course these have to exactly cancel out because when the object finally arrives at r from infinity it has 0 kinetic energy. So the change in the kinetic energy is 0 therefore the net work done has to be 0 too. Anyways onto your question…the point is I’m controlling the direction of the dr by choosing limits infinity to r rather than r to infinity. So flipping the sign of the dr but then also using my funky limits is doing the same thing twice which cancel each other out
@@TheBrainFiller Thanks for the quick response! The point that the limits control the direction seems to be what I was missing. Pretty clear now :)
Assuming you’re familiar with line integrals which I’m getting the sense you are W= int (F dot dr) from r_1 to r_2 yes but it’s always easier to think about line integrals by parameterizing the chosen path with a variable t and then doing W= int(F dot dr/dt dt) from t_1 to t_2 in which we can interpret t as being a time and dr/dt being the velocity vector of our particle along the path. Now let’s choose a path where the velocity vector is radially inwards (same direction as the gravitational force) the whole trip and is a constant speed v. So F dot dr/dt = (GMm/(r(t))^2)*v and r(t) = (X - v*t) (where we’ll take the limit X goes to infinity at the end cause at t=0, our t_1, the mass is at infinity and at t = (X+r_2)/v the mass is at location r_2). Then just do that integral with those correct time limits and finally take the limit as X goes to infinity and you should get the answer you’re looking for.
Alternatively, if you just want the direction of vec(dr) its conventionally considered radially outwards and by flipping the limits of my integration I’m sort of flipping the direction of the dr to be radially inwards. But it is probably much clearer to define a velocity vector dr/dt and do the integral dt for no ambiguity
@@TheBrainFiller The velocity approach is interesting, I've never really come across it before. I pondered a bit more on this doubt, it's like saying if (dx î)/dt = -v î, then direction of dx is along -î. for which, yeah the small displacement is along that direction but we don't write that on the LHS while denoting the velocity right (that wouldn't make sense). If we multiply that equation with dt we would get the instantaneous motion along dx î only, regardless of what v vector is. Maybe not completely concrete but this is a way to think about it. I think it's in a way similar to the logic where your approach contributes
I highly recommend you do look into line integrals in that case cause yeah you’re getting the right idea and it’s a really interesting topic. dx/dt is a vector and so I’d write in -v r hat (for the radial unit vector in spherical coordinates) and then you dot product that with the Force vector which points in the -r hat direction too so F dot v will be positive the whole journey and then you’ll get a positive answer when you integrate from t=0 to t= time at which you reach the final position
Job well done
Thanks for watching!
hi just want to ask... why the graph in 2:58 shows infinity is the upper limit while r is the lower but in 3:04, u put the upper limit as r while the lower limit is infinity?
Right so the graph just shows a 1/r^2 relationship and of course that takes on values from 0 to infinity (I suppose I could have shown it from infinity to 0 but that’s not really conventional). I’m then suggesting that the object begins its trajectory infinitely far away and then moves in slowly until it arrives at r
Thankyouuuuuu❤️
Thankyouuuuuuu for watching!
2:53
It’s not really the same it’s more of a continuous some of each single tiny step in height
Right, the connection I was making is that an area is being found in both cases but you are of course correct
thank you for the great explanation, although the video was pretty straight forward i may have missed something isn’t the work equal to Ep+Ek why did we ignore the kinetic energy of the planet? because the velocity of it was very low towards the begining? but as we approach to close distances that wont be the case any more so we cant ignore it. also didn’t we integrate from infinity to r, but the when he showed the graph of Force function it looked like we were integrating from r to infinity. i’ve been trying to derive this for 3 days and i tried to compute it like this: 0 to r(radius of earth) + r to some R (distance between moon and earth for example) im only a highschool student so im very new to this stuff i would be really happy if some one could tell me where was i wrong.
The external force applied throughout the trip from infinity to r basically works to keep the velocity of the object constant. It does this by exactly matching the force of gravity therefore, by Newton’s first and second law, the net force on the object is 0 so it continues moving at constant velocity. As a result the kinetic energy of the object is constant the whole time (and importantly very low…to the point it can be ignored). I can’t exactly remember what I did in the video but when I integrate from infinity to r, I should have got an expression like this: [-GMm/r] evaluated from infinity to r which means (-GMm/r)-(-GMm/(infinity)) (because you do the final state minus the initial state). There are two ways of defining a potential field: in terms of an external force or the force of the field.
The latter definition is that Ep= - integral F dx and the one in terms of the external force is Ep= integral Fext dx. But the external force is obviously just -F (it faces the opposite direction) so both of these definitions are equivalent (as they should be).
To respond to your own attempt at a derivation. When you’re integrating from 0 to r note that the force of gravity isn’t just changing because of the distance from 0 but also the amount of mass below you. Look up Gauss’s law for gravity if you’re interested. The main thing to realise is that the treatment of gravity you’re learning applies to point objects only and Newton showed that spherical objects can be treated as point-like beyond their radius so it’s completely ok for us to think about orbits and such in terms of point particles.
@@TheBrainFiller thank you for yet another great explanation i will check your other videos as well and better my understanding of physics before the university. and if something bothers my mind i reckon i can ask them to you via comments.
Great video. Perhaps you can derive GMm/r^2 as well?
As far as I know, Newton derived his law using the relationship between the period of an orbit and its radius (T^2=kr^3 which is Kepler’s third law) and centripetal force. I actually use the equation for gravitational force to derive Kepler’s third law so it would be weird to make a video where I go the other way round lol. If you want to see that video I’ve linked it here:
Orbital Mechanics - Orbital Speed, Total Energy and Kepler’s 3rd Law
ruclips.net/video/SP41C33FBT0/видео.html