Lecture 20

[Music] hello and welcome to another lecture session on classical motion of a single particle in this lecture we simply continue our previous discussion on the motion of a single particle under an inverse Square law of force and also to be very precise here we will be mostly dealing with the attractive inverse Square law of force and this is because I mean the main attractive inverse Square law of force can be seen and we actually experience every day and we can even claim that we live for that that is the attraction uh force between two Mass particles that is the gravitational attraction huh and if you just uh like remember Newton's law of gravitation this is simply nothing but an attractive in Square law of force okay so now coming back to the analytical part so as I said in previous lectures we have discussed several points actually already uh on inverse Square uh law of force so first we said that again so this is a central Force law so I can write generally an inverse Square law by this expression minus K by r² * R cap where K is a positive quantity that gives me attractive inverse Square law and then the corresponding potential will be equal to minus K by R so the potential or the potential energy of okay will be a negative quantity because K is positive and R is also positive so this is a negative quantity then I said that the total energy can be written now in this way where we know of course that these two terms actually are coming from the kinetic energy term this is coming from the radial part of the velocity and this is coming from the cross radial part of the velocity and this is the true potential energy and the sum of the two remains constant because this is a conservative force field now this quantity and uh is called the effective uh kinetic energy and this quantity and this quantity together they are called the effective potential energy why this effective terms because if you write in this way then you it looks that the motion is effectively one dimension and the only space variable under consideration is R okay now you can also write this in this way that half Mr do square plus some V effective you can call the sum as V effective so we already discussed how V effective behaves as a function of r and we had a plot like this and if you remember the nature of the plot actually you will see there is a very interesting uh point to note that under such a condition if you somehow so here you see for very small R it starts from a very high value then it decreases and being positive okay that's why V effective has a positive value for very small R and then with uh increasing R it decreases and then it crosses R is I mean V effective is equal to zero and then it reach is some local minimum value again it increases and finally it asymptotically touches the r from negative side okay so in a sense if you now think the motion under such an effective potential you will simply can classify this by saying this is such an potential well the effective potential well is such that or the potential I mean energy uh structure is such that if you give the system any energy which is slightly greater than zero huh here for example here or here or here or just but anything positive the total energy is if it is positive then always you see that the system will simply have one turning point and the other turning point is missing that mean it is at Infinity so this is an unbounded motion this is true for inverse Square law of force which is attractive in nature and if you give the system anything negative okay which is even slightly less than zero here or here and even here okay until this okay you will always have a bounded motion because you will then see that always the particle will have two turning points okay what happens exactly at zero since it is asymptotically approaching zero then even at zero we can think that the system is having only one Turning Point thereby making the motion unbounded okay so if I write that for E greater than equal to Z the motion is unbounded and E less than zero the motion is bounded okay so that we already can conclude from this graphical analysis and that is the power of such analysis now you know that whenever you have uh such an attractive inverse Square law Force we have already using uh B equation we showed that the orbit of such a particle okay would be given by a conic section which is uh simply R is equal to P by 1 + Epsilon or E cos Theta Epsilon and E will be I mean somehow used indistinguishably in the course sorry for that H so this is known as the semil latus rectum see you just uh search on internet what they are and I think most of you know from your coordinate geometry class in the high schools and this is the eccentricity okay and so this is a and this is positive because this is equal to this m h² by K which is positive and this is a quantity this is a dimensionless quantity this Epsilon this is an eccentricity and this is given by this quantity MH s c by K okay so these two actually defines the nature of the conic section okay and uh so we will we will actually discuss that whenever the conic section so uh maybe all of you know that in nature we have uh four common commonly encountered conic sections as far as my knowledge is concerned huh so one is the ellipse another is the parabola the third one is the hyperbola and the fourth one is a special case of ellipse which is known as a circle okay so I mean so broadly classifying they will be of three kinds elliptic Parabola and hyperbola so an elliptic orbit is basically giving you uh I mean a closed trajectory basically and whereas a parabola and a hyperbola huh can be giving you the possibility of a open trajectory okay because a hyperbola and a parabolic trajectory they are not closed that we know from our geometric knowledge okay now you try to associate this knowledge to bartran theorem bartran theorem already said that if you are dealing with an inverse Square Force law any bounded motion will assertively give you a closed trajectory that means that here for this here we can easily use this knowledge by saying that if I have the total energy of the system to be positive or equal to zero then I will have an unbounded motion and I cannot expect assertively a closed trajectory okay so my trajectory can be be open as well although I cannot always say that the trajectory can be open or closed but again you can use your common sense just by thinking that if your motion is unbounded then there is by Common Sense there is no possibility of a closed trajectory okay on the other hand if your motion is bounded then your trajectory can be closed and open so if your motion is unbounded already you can you can analyze and conclude that your motion is unbounded for this then only open trajectories are possible and then you can already think that only possibilities can be parabolic orbits or hyperbolic orbits whereas if you go to this case e less than zero that is the total mechanical energy of the particle is negative then basically you will have a bounded system now here there is there is a possibility of open versus closed trajectory and butm said for inverse Square law we don't have to think anything we don't have have to be anxious about the nature of the trajectory it will be a closed one and here we know this is this will be a closed conic section so it will be either ellipse or Circle so I can now write from this point that if I have a positive or equal to zero total energy then I will have either parabolic or hyperbolic trajectory whereas if we start from negative energy I mean if we if we deal with negative energy negative total mechanical energy then it will be either ellipse or simply Circle and actually you can say in general it's an ellipse okay so you see already just by doing this much of analysis we can conclude about the one to one correspondence to the nature of energy whether it is positive or negative the total mechanical energy and the nature of the orbit okay here we will somehow proceed in the same direction much more formally and we will see that how this the features of the conic section that means the semil latus rectum and the eccentricity are really associated with the dynamical variables of the system for that what we will simply do we will just start with this solution and you can say that okay this Theta 0 can be taken to be zero so this will be my solution u u is nothing but 1 / R okay so I am writing in case you want a recapitulation u is nothing but 1 / R so I can now so U is equal to I know K by m h² + C cos Theta where C was a constant okay and then I can easily write du D Theta is equal to so this is a constant quantity this is a function of theta so C minus C sin Theta okay and now we again write the whole energy conservation now the energy conservation gives half m r dot square plus l s by 2 m r² minus K by R is equal to e for an inverse Square law for an inverse Square law of course attractive for our case for an inverse Square Force I write like that okay then if I can write this I also know that I know r dot is equal to minus L by m du D Theta how would I know that this is very pretty easy to understand huh r dot is Dr DT so this is our DOT and this is nothing but Dr D Theta D Theta DT so this is a mathematical manipulation you have to use everywhere in the central force and then this D Theta DT is Theta Dot and then Dr D Theta is nothing but DD Theta of 1 / U and sorry and then this becomes this part becomes Min - 1 / u sare duu d Theta times Theta Dot and Theta Dot is nothing but L by m r² or H by R square is as you know and from that I can actually wrap this bracket off yeah okay and you can actually have so r sare and u sare they just make one so it will be minus L by m du D Theta which is exactly what you have over here okay if you know that then simply you can write the total conservation of energy is equal to half m l ² by m² du D Theta whole Square you can also write plus l s by 2 m now R sare you can write as 1 / U sare so it will be here and then you can write minus K by R as minus k u and that is your constant energy okay okay and from that what you get is a fantastic relation that is almost like B uh type of equation but for the conservation of energy you can write du D Theta whole squar plus u² is equal to 2 m by l s e minus V okay where we know that V is nothing but minus k KU or k by R okay that we have say minus K by R okay so this is the uh expression which we will use now huh and uh so what we will do that we will simply put our solution U is equal to this and du D Theta is equal to this in this one and that's why you will finally see that finally we will have something very very interesting so du D Theta whole Square will be c² sin Square Theta okay so I write c² sin Square Theta again U Square will be this whole thing whole Square so it will be K by m h² + C cos Theta squar is equal to 2 m by l s e minus V clear so if you if you if you see this is the final I mean this is the expression what you have then you can do a little bit of more simplification and you can come c² plus just using that sin Square Theta + cos sare Theta is equal to 1 you will have k s by m² H to 4 + 2 K C cos Theta by MH s okay and on the other hand you write simply this is equal to so if you just see e minus V okay so V is again just you remember I just write 2 m by l s e minus V is Ku so it will be e+ KU and again instead of U you just write this value and then you will have 2 e by m h² + 2 k s m² h to the 4 plus uh yes I have to multiply everything just a minute yes by everything like this yeah plus uh 2 K C cos Theta by MH s you see here again instead of youu you wrote this expression and throughout this whole expression you have wrote you have used L is equal to M * H okay so I have got rid of L and I have written throughout the H okay so this is H to the sorry this is H to the 4 yeah H to the 4 okay now you see what is the good news that this term will cancel and finally you have something like this so c² this is also another term there will be so this minus this will simply give you K squ this minus this will give you K sare by m Square H Square so that will simply give you c² is equal to 2 e by m h² + K ² by m² H to the 4 okay now if you remember what was that c² so from B equation we had this C MH 2 * C by K is equal to e so m s c by K is equal to E m h² c by K we know we have okay we have this already and then I can simply write okay so c will be then I can simply write that or or I can write this uh I can do in both ways actually I can go for the here with a different color we have already this so e will be equal to M h² by K now c will be root over this with a plus minus I'm just taking the plus sign over here so root over 2 e by m h² + k s by m Square H to the 4 I have taken the plus sign only because from my this expression you see yeah from my this expression I have said that this will be a ratio which is in general positive okay and I will just take the plus over here so this will be the uh relation between the energy the angular momentum per unit mass the mass the force constant and the eccentricity and that is quite important okay now if you just take this relation a little bit more carefully you will see that e is equal to Simply if you just now insert I mean put this thing under the square root you will have root 1+ so it will go inside being an m² h² m² H to 4 by K sare which will totally be one over this term and that's why you will have a one plus so you will have 2 e by m h² * m² H 4 by K Square okay and you have the very fundamental result 1 + 2 e MH squar by K squ so you see now that is the expression you already have encountered l sare in the last lecture at the end of the last lecture I said that 1 + 2 e l s by m k s that must be greater than or equal to zero and that is because that and and and since that is greater than equal to zero I can call it that a square of some I mean some variable e or some some quantity e okay is not a variable sorry some quantity e and if you now think what they are so in terms of a l and m this is simply is equal to 1 + 2 e l s by m k s okay so you see this e is equal to this one okay and this e is nothing but the eccentricity okay and this is totally inconsistent with this result that your e plus this quantity must be greater than or equal to zero huh okay so in that way you are also crosschecking the con consistencies of this result so now finally you have you have found that e which came as the square root of 1+ 2 e l s by m² is nothing but the eccentricity of the conic section okay in the next lecture depending on the nature of the conic section we will actually try to differentiate different type of conic sections and uh analyze the corresponding motion okay thank you [Music] [Music] [Music]