Kepler's Laws: Why Are Planetary Orbits Elliptical?

we often take for granted that planets orbit around the Sun and that the reason why they do this is due to the force of gravity but this alone doesn't explain why orbits have the shape that they do where we notice that at one point in the planets year it can be closer to the Sun and that at other point in the year it can be further away from the Sun in what mathematically speaking we would call an elliptical orbit now hundreds of years ago a very smart chap by the name of Kepler first noticed that planetary orbits are ellipses but the question is why or today I'm going to use the laws of physics as formulated by Sir Isaac Newton to show you why plan tree orbits must be ellipses the question of how planets move around a star has become increasingly relevant in recent years with all of the discoveries of extrasolar planets so where better to start if we want to understand the dynamics of how planets move around the star then by looking at how the planets move within our own system so the first real major attempt in order to actually tackle this in a way which seems to match quite nicely with reality was by the German astronomer Johannes Kepler in the 17th century so the story goes that Kepler had access to a huge wealth a very precise astronomical measurements taken by the late astronomer Tycho Brahe he and from these measurements Kepler was able to spot a number of patterns and within these patterns he was able to tease out what later became known as his three laws of planetary motion so they can be summarized broadly by the first law being that planets orbit in ellipses and there's the added statement that the Sun in this elliptical system is at one of the focuses of the ellipse and I'll show you actually what this means just on the next slide after we've just stated what the Lord is so hold with me just for a moment the second law that he deduced is that let's say for instance here you have the Sun and here you have a planet draw a line between the Sun and the planet and then wait a certain amount of time and the planet will say move like this and then draw another line between the Sun and where the planet now is and if you shade in that area then wait until the plant is at a different point in its orbit say over here again draw a line between the star and the planet now wait the same amount of time for the planet to get into its second position as you waited this time around draw the line again shade in the area and if you compare the two areas for these two different arc lengths within the orbit you find that if you call this maybe a 1 and you call this area a 2 and the second law can be summarized as saying that a 1 is equal a 2 alternatively that planets rather will say that the radius vector of the planet sweeps out equal areas in equal times and this law may not seem that impressive at first glance there is actually some really hidden physics inside of there which I'll be illuminating in my next video and finally the third law which is incredibly useful and I use it all the time is in my own research in exoplanets is that the total amount of time it takes for a plant to go around a star notice the orbital period P if you square that you find that it is directly proportional to the semi-major axis of the orbit cubed now so major axis basically you can think of it as how far away the plant is from the star and in fact it's exactly that in the case of a circular orbit but I've shown you I'll show you precisely how to work it out in just a moment now the first two laws were deduced quite early on by Kepler actually in 1609 originally whilst the third law which took a little bit more effort to deduce you figured out in 1618 but these were just empirical laws he spotted the evidence for them amongst the data but he didn't know why these laws were true and an explanation of why they're true actually would have to wait for Sir Isaac Newton to come along now what I'm going to do in the rest of this video is I'm going to hone in on the first law of planetary motion firstly and explain what it actually means by showing you what an ellipse is and say how we can describe the orbits of planets and then for more advanced people interested in knowing why planets orbit ellipses I will then go through and present a modern version of Sir Isaac Newton's original deduction his original mathematical proof that planets orbit in ellipses due to his law of gravity so the end of the day an ellipse is just a fancy word for an oval which many of us will already be familiar with just from day-to-day life we can describe the shape of an ellipse by drawing two lines across it one from the cross from the two farthest ends and then one line intersecting the two closest parts and in the case of a circle these two will just be absolutely the same because in a circle you are always a constant distance away from the center when the case of an ellipse we call this longest axis the major axis and the shortest one predictably the minor axis and the distance halfway across the major axis is called the semi-major axis denoted by the letter a conventionally whilst the distance halfway across the minor axis is conventionally denoted by B which is the semi minor axis so when Kepler said that planets orbit in ellipses with the Sun at one of the foci what he means if we draw this diagram is let's say you have a star here which lies along the major axis but it's not quite in the middle of the major axis it's actually displaced from the middle by a distance a times by some number which is between zero and one so if this number E was just zero then the Sun would be at the center and if it was one then the Sun would be touching the outside edge of the ellipse and then the planet it could be the earth it could be Jupiter could be Mars then travels along this Green Line and that is what the orbit of the planet looks like so we can draw a line between the Sun and the planet and call this R and that's just the distance between the Sun and the planet as a function of time and we can also if you want to describe mathematically speaking define an angle theta between the major axis and the radial vector so for instance when this angle theta is equal to zero degrees the planet will be at its closest approach to the Sun and we call this point of closest approach the parry apps and when this angle theta is equal to 180 degrees when the plyatt is over here it will be at the furthest distance away from the Sun and we call this point the appo apps and we can figure out expressions for how far away the plant is from the Sun at the perry apps and the Apple apps just by looking at the geometry in this case so for instance when is that the perry apps the distance will just be this line here and let's call it RP for our perry apps and so we can see that from the center of the ellipse to the outside edge here by definition is just a and then if we subtract a times by e this little bit of distance here will then get what our P is so our P is just a minus AE which we can simplify as a times by 1 minus E and similarly for the appo apps the distance is just going to be so when the plight is here our Apple apps will be this distance and so it's just going to be a plus this small a times by E and again we can simplify that by taking out a factor of a we get a one plus e so this gives you the minimum distance of a planet from the Sun and this here gives you the maximum distance in terms of the quantity a the semi-major axis and the quantity e which is called the eccentricity it tells you basically how much the orbit deviates from being a circle so this here is just the eccentricity so notice from these two expressions that if the eccentricity is equal to zero then the minimum distance and the maximum distance are both just equal to a which is what defines a circular orbit where you're always at a constant distance away from the star in question so for ellipses the quantity e the eccentricity can vary between 0 and 1 where if it's zero you have a circular orbit and the case of it approaching one then actually the planet would then be unbound and it will be able to escape from the Sun so some comets for instance can have an eccentricity which exceeds one so now let's turn to ask why planets orbit in ellipses and to do this we need to understand how gravity works and you can explain why planets orbit in ellipses in the framework of Newtonian gravity as formulated by Sir Isaac Newton in his Principia Mathematica in 1687 and what's really significant about this is that by unifying the laws of how everyday objects move here on the earth with how the planets and the moons were orbiting up in the heavens Isaac Newton showed the laws of physics are universal so this is a very neat and very historically relevant trick that I'm going to show you here of how to prove Kepler's laws so what is his law of gravity how does gravity work at least according to Newton so what he stated is that say you take two objects each of which contains atoms and so they have mass Newton stated that these two objects attract each other due to gravity there's no reports of gravity and that the force of gravity depends on how much stuff there is in each of the objects so say if I double the amount of stuff in this object the force of gravity won't double he also stated that if you take the two objects and move them two times closer then the force of gravity becomes four times stronger so we say that gravity follows an inverse square law so both these observations can be summarized mathematically speaking by the force being proportional to the product of the two masses m1 and m2 and that also the force is inversely proportional to the distance between the objects squared and you can combine these two observations together in that the force will be equal to some constant of proportionality it's just a number here it's called Newton's constant well Newton's constant of universal gravitation so it's what six point six seven times by ten to the minus eleven if you want its actual value times by the product of the masses divided by the distance between them squared and this deceptively simple expression describes how the force of gravity depends both on mass and on distance and it's relevant whether you're measuring the force gravity say between myself and I don't know the books behind me or between the Earth and the moon for instance it's a universal law so this is the first ingredient we need in order to do this proof this is one other ingredient we need which is one of Newton's three laws of motion it's that if you exert a force F on an object then the relation between the force and the acceleration it feels is just given by the force is equal to the mass times by the acceleration F equals MA which is just Newton's second law of motion and these two results the one in the circle and the one in the square are all we need in order to prove Kepler's laws which is quite nice now the proof does get a little bit involved though so in order to do this efficiently speaking I will need a little bit of calculus and some vectors if you're not familiar of those then feel free to drop me a message down below and I'm happy to recommend materials so that you can learn how this all works it is worth noting though that in Newton's original proof he didn't actually use calculus because he'd only recently invented it but he used a very convoluted geometric proof which is actually even harder to understand than using calculus and so hence why I'm going to go through this more modern proof so to prove Kepler's first law of planetary motion what we want to solve is a classical problem in orbital mechanics called the two-body problem it's just basically how do two gravitating objects move as a function of time so set this up let's imagine we have the Sun which is over here for instance and over here we will have a planet the Sun will have a mass let's call it M star and the planet will have a mass M P from mass of the planet we can define some reference origin that we'll use to define all our coordinates from and in this coordinate system the distance between the origin and the Sun will be a vector that we will call r1 and the distance from the origin to the planet we're similarly going to call r2 and these are vector quantities meaning that they have a magnitude and a direction and equally the distance between the Sun and the planet we can also draw and we just could call this distance R now due to gravity the planet is going to experience a force pulling it towards the star which will be directed in this direction so let's call this f2 and equally due to the gravity of the planet the star will also experience attractive force that we shall call f1 now we can use our first key ingredient which is Newton's law of universal gravitation to figure out what f1 and f2 are going to be because he tells us that the force of gravity between two objects its magnitude is just given by G times by the product of the masses in this case M star times by M P divided by the distance between them R squared and the direction will be along the direction of the vector R between the two bodies so if we make F a vector then we have to include a vector of length 1 in the direction between them which is a unit vector so this little hat up here just means a vector in the direction between the planet and the star with total length one so for instance the vector R is just given by the length of the vector R which is just a scalar it's a number like say five meters or five million kilometres times by the unit vector so the unit vectors say would be like that and when you multiply by the distance between them are you get the total vector between them okay so this is how we can measure the force between them and notice though that f1 is in the opposite direction to f2 they're of the same magnitude klitz a mutually attractive force but this is actually a manifestation of Newton's third law which tells us that F 2 is equal to minus F 1 just because they're directed in opposite directions great so now we can use our second key ingredient which is Newton's second law of motion or just F is equal MA and we can use this to figure out the acceleration on the two bodies from the force of gravity acting on them so the acceleration is given in reference to the origin point which I'll call oh here for instance so the force on the first object this star would just be given by the mass of the star times by the acceleration it's experiencing and we'll have the same for the planet f2 will equal the mass of the planet times by its acceleration now how do we get the acceleration so the definition of acceleration as always is just the rate of change of velocity which we can write as DV by DT the time derivative of velocity and then remember that velocity is just the time derivative of position so a1 for instance is the derivative of V one which would just be the second derivative of R 1 because V 1 is d R 1 by DT so we can substitute this relationship into here in order to write this as M star times by D 2 R 1 by DT squared and this is still a vector and similarly for the force on the planet we have the mass of the planet times by D 2 R 2 over d t squared and then the question is how does this help us well the really key trick to note is that just from this diagram which we have here we can see that if you start here and want to travel from the star to the planet along the vector R you can also go against the vector r1 and then add the vector r2 or otherwise speaking the vector R is exactly equivalent to r2 minus r1 and so if we take the second derivative of both sides of this expression with respect to time then you can start to see here that we can find a way in order to use these two results down here in order to get expression involving the distance between the planet and the star okay so bear with me for just one moment because we also know the expression for f1 from up here so we can write that this here is just equal to G times by the mass of the star Ty's by the mass of the planet divided by R squared in the direction of R and this is exactly the same just with the aforementioned minus sign from Newton's third law that's the star mass the planet over R squared in the r direction right okay so now what we want to do now that we have these two expressions is that if we take this and substitute it into this expression at the top and do the same using this value of D to R by DT squared up here then we can figure out what D to our vector between the star in the planet divided by DT squared is equal to so let's do that on the next page right so here are two expressions for the force on the planet and the force on the star but what we want is an equation describing the time evolution of the distance between the planet and the star so remember that we had that the second derivative of the disk vector between the plant and the star is just given by the second derivative of the distance vector from the origin to the planet minus the second derivative with respect a time of the distance vector between the star and the origin and notice that we've almost got these two from these two equations up here which I can call equation 1 and equation 2 one way to do to get D 2 R 1 by DT squared is just cancel this M star here and here and for D 2 R 2 by DT squared just cancel the MP here and the MP here then all we do to get this is just take equation 2 and then subtract equation 1 from it which will then tell us that D 2 R by DT squared will just be given by minus G times by the mass of the star divided by R squared in the r direction from Equation 2 minus G times by the mass of the planet divided by R squared in also the r direction and so if we take these two quantities and bring them over to the left hand side we can get a differential equation that we want to solve that D 2 R vector by DT squared plus G times by the sum of the mass of the planet plus the mass of the star divided by R squared multiplied by the unit vector in the r direction is equal to zero and this is the equation that we need to solve if we want to figure out the time evolution of the distance vector between the planet and the star so let's go on and solve it in order to get a handle on what this equation is actually telling us about how the planet will orbit around the star we can use something called the cross product that you might have encountered before in just these vectors but for a brief reminder if we have two vectors say a and B the cross product of those two vectors is just defined as the magnitude or length of the vector a times by the magnitude of B times by the sine of the angle theta between them so say if a was here and B was here this would be the angle theta and the cross product of two vectors spits out a vector as well which is directed at 90 degrees to both the vector a and B so this is a unit vector Susheela noted by n because it's normal to both a and B right so how we going to use this resort here if we look at the equation that we're interested in right up here here's an idea why don't we take the vector R and cross it with this entire equation like this are crossed with all of this that would be what you would get on the left hand side whilst on the right hand side you have R cross with zero and zero times by anything always gives zero so we have R cross all of this up here and we can break this apart term by term because we have two terms which we're summing here and so we can take the cross product with each of them in turn so let's see what we'll get the first term gives us R cross D - R by DT squared we then add the second term if I take out the constant out the front we have G times by the mass of the star plus the mass of the planet divided by R squared just a number so I can take it out the front then we have our crossed with our unit vector and all of this is equal to zero but from the definition of the cross product here if we do R which is just directed in the vector of in the direction of R and then cross it with a vector in the same direction then the angle theta would actually be zero and so the cross product of any two vectors in the same direction always gives zero so this is just going to be zero and this term will die and hence what we are learning is that the vector R crossed with D to R by DT squared must equal the right hand side which is just zero now we can extract slightly more information from this equation by noting that this equation is exactly equivalent to the time derivative of the vector R crossed with dr by dt so this is just one time derivative here you can check this just by differentiating it term by term via the product rule the first term would give you dr by dt cross dr by dt which will give you zero and the second term from differentiating this then returns this expression and so this must equal zero and remember that if you ever have an expression of the form say DF by say dt is equal to zero then by integrating this expression it tells you that f must equal a constant because when you differentiate a constant you get zero so what this expression tells us is that the cross product of the vector R and this single time derivative of the vector R must equal a constant and because the cross-product always gives you a vector it must be a constant vector which we will denote by H so the conclusion from doing this is that are crossed with dr by dt is equal a constant vector h as it will turn out and we'll see shortly this is actually the angular momentum per unit mass but I'll be showing why that's true later but let's examine now the physical content of this nice simple equation we've deduced which actually tells us a lot about how planets are going to orbit so what is this equation mean well the key is buried in the fact that this vector H is a constant vector and because it's defined in terms of the cross product it must be at right angles to both the vector R between the planet and the star and dr by dt which you can think of just as a velocity so if we draw what this looks like we'll have the star will be here we have the vector R and then the planet is living out here will have the velocity which at some time may be say say it's like this for instance so this is the direction that R is changing the planet is then moving along let's say like this and now the vector H is perpendicular to both these two vectors so it will actually be sticking outwards like this so it's at a right angle to both of these and the key is because the vector H is constant meaning it has a constant direction and a constant magnitude this actually defines a plane in three-dimensional space because if H was not a constant say it was rotating and the plane would be rotating but because H is constant it forces our and the derivative of our to lie within a plane so we've deduced one key resort already in that planets orbit in a plane and this is good because it means that we can reduce this three-dimensional vector problem that we have just to a two-dimensional problem defined inside of this individual plane and to do this we're now going to introduce plane polar coordinates within this two-dimensional plane so let's orient ourselves such that we're looking down along the vector H looking down on the plane that the plant is orbiting about the star in so we'll see is the star say here and the planet just over here and we'll describe the system not in terms of x coordinates and Y coordinates but in terms of the distance R between the planet and the star and the angle theta then it makes to some reference direction we can though figure out the relation between the traditional x and y coordinate system and this polar coordinate system by noting just from trigonometry that the distance in the X Direction is R times by the cosine of theta and the distance in the Y Direction is just R times by the sine of theta so we can write the vector R between here and here in column vector notation as just being R times by cosine of theta for the X component and R times by sine of theta for the Y component we can then peel out a factor of R the magnitude of this vector and we just get R times by cosine of theta sine of theta but remember that our is just given by the magnitude or the length of the vector times by a unit vector in that direction so we've actually figured out here an expression for the unit vector in the r direction now the planet is going to live somewhere within this plane following some orbit that we don't know the shape of yet but if we want to define a two-dimensional corner system we also define a direction at right angles to this vector R so that we have well-defined coordinates so we can also define a unit vector theta at right angles to the vector R which will have components that you can figure out just from geometry for instance that's 90 and so this angle up here will be 90 minus theta and you can work through that to figure out that the vector at right angles of unit length is just minus sine of theta cosine of theta and when I say these are unit vectors what I mean is that you square each of the vectors say by dotting it with itself you get one because cosine theta squared plus sine theta squared just gives you one so they are indeed unit length okay so now that we know the two well the two basic vectors of our plane polar coordinate system we can now use these to figure out what H actually is because we want an expression for dr by dt so for instance since R is just given by the magnitude times by the unit vector we can work out the derivative of R with respect to time just by differentiating this product which we can do using the product rule where you differentiate the first term times by the second and then you add the second term times by the first one differentiated so sorry that should be we differentiate the unive actor and then times by the first term undifferentiated right well how do we get the derivative of a unit vector well we can actually use this expression down here so the derivative of the unit vector can just be written as the derivative with respect to time of cosine theta sine of theta which we can just write using the chain rule as D theta by DT times by D over D theta of this column vector cosine theta sine of theta and remember that cos differentiates to minus sine and sine differentiates to cos so this here will just be minus sine of theta cosine of theta which I'll look that looks a lot like the unit vector in the theta direction so we worked out that the derivative of the unit vector R with respect to time is just given by D theta by DT times by the unit vector in the theta direction so we now have this part and we can now substitute the first term and the second term which is just equal to this up into this expression and so what we will get down the bottom is that H is equal to R crossed with our unit vector dr by dt plus r times by d theta DT multiplied by the unit vector in the theta direction notice immediately though that the first term will die because we've got the cross product of two vectors in the same direction the second term will be fine though we'll get H being equal to R times by D theta by DT times by are crossed with theta in the unit direction and we could actually peel out a unit of the magnitude of R from here to make this be a unit vector and write this as R squared times by the unit vector in the direction R and the reason why I've done this is because by definition of the cross product our unit vector crossed with theta unit vector must be perpendicular to both of them and hence stick out of the plane and this just defines the vector H because these are unit vectors this will then give the unit vector in the H direction so at the end of the day what we figured out is that the vector H can just be written as R squared D theta by DT times by a unit vector in the X direction alternatively this is the magnitude of the vector H so H is just R squared D theta by DT so if we have a way to get H and we all know say R for instance you could figure out how the angle varies as a function of time great so this is our first key resort now what we're going to try and do is continue using this coordinate system to figure out how to solve our initial equation which we derive from Newton's law of gravitation right so this is the equation that fundamentally we're trying to solve we've already made some good progress in that we know we orbit within a plane and that we've also deduced that H is equal R squared D theta by DT and then working out hate we've already made some headway towards evaluating the second derivative in that we figured out that the first derivative of R with respect to T is just given at the end of the day by dr by dt in the r direction plus our D theta by DT in the theta direction so to get the second derivative we just continue doing this with the product rule we differentiate term by term the above expression and so we'll get D to our magnitude by DT squared times by our we then differentiate the second term here multiply the first dr by dt times by D our unit vector with respect to time and then for this term we just differentiate each of the three terms term by term multiplied by the two and differentiated using the pro roller game so we're going to get dr by dt x by d theta by dt in the theta Direction plus our d2 theta by dt squared in the theta direction and the final term our D theta by DT ties by the derivative of the theta unit vector now we have all the ingredients to works out apart from this bit we can figure that out just from this expression in that D theta unit vector respective time again using the chain rule is just D theta by DT times by D over D theta of this minus sine theta cos theta and because sine differentiates to cause and cause differentiates to minus sign this here just gives us for this part minus the I unit vector because we have an extra minus sign out the front from this derivative relative to just the our unit vector itself so this entire thing here it is equal to minus our unit vector D theta with respect to time although that's not a unit vector D theta right let's simplify this somewhat horrendous expression slightly so we also notice that we remember that the derivative of the our unit vector with respect to time is just given by D theta by DT in the theta direction so let's collect our terms in the our direction we get D to R by DT squared and then we get a minus R D theta by DT or squared so this is just the second term here multiplying these two D theta by DT s and then all the theta terms we also get as well which at the end of the day to the terms will combine and so we'll end up getting in the theta direction two times by dr DT multiplied by D theta by DT plus R times by the second derivative of theta with respect to time so we're not actually going to be that interested in the theta expression you can certainly use it to look at interesting effects like the courrier is false for instance but it's going to turn out that actually all we need is the r compound to this expression to figure out how planets are going to orbit so let's put all this together back into our original equation way up at the top and figure out what this is going to tell us okay so here's what we've just figured out or looking neat and so what we can do now that we have this expression for what the second derivative of the vector R is we can substitute this expression into the equation that we're trying to solve just up here and collecting the terms we get that D to R by DT squared minus our D theta by DT all squared and then I'm going to add this extra term here which is also in the r direction plus g times by the mass of the star plus the mass of the planet divided by r squared all of this in the r direction plus exactly what we have here in the theta direction must equal zero and because this is a vector equation we require each of the components to be exactly zero so this bracket here will equal zero and also this bracket here will equal zero and the key thing to note is that what we have in the square brackets there an equation involving our Thetas and time doesn't involve vectors anymore we've actually reduced the problem to a second order differential equation involving scalars which will be much more ready to solve and we can also use the relation which we figured out before if you want to eliminate the theta that H is equal to R squared D theta by DT to see that D theta by DT squared which we would rather not be in our equation is just going to be equal to H squared divided by R to the power 4 I've just divided by R squared here and then square the resulting expression so if we substitute this into this expression we'll get an equation just involving R as a function of time so we have D to R by DT squared where this R is no longer a vector it's just the distance between the planet and the star which is what ultimately we're interested in solving so we have this minus H squared over R to the power three we had one R up here which cancelled with one of the R's down here plus G times by the mass of the star plus the mass of the planet all divided by R squared is equal to zero solving this equation as it is in its current form as a function of time turns out not really something we can do so in order to solve it we're going to make a unit transformation to solve for R as a function of the angle theta specifically what we're going to do is define a new variable U as being equal to 1 over R such that R is just 1 divided by U and in terms of this new variable will have that dr by dt is just equal via the chain rule as dr by d u x by d u by D T and what is the R by D u well we can figure it out from this expression here just by differentiating this with respect to time so i with respect to u we get dr by d u is equal minus 1 divided by u squared so this could then just be written as say minus r squared x by d u by dt and then d by dt itself can just be written as d u by D theta times by D theta by DT okay and remember call also that H is equal to R squared D theta by DT so we can actually eliminate D theta by DT over here and replace this just with H divided by R squared and so this is just equal to minus H times by D u by D theta and then we can get the second derivative of R with respect to time D 2 R by D T squared we just differentiate that once more and we're going to get minus H times by D over DT of D u by D theta and again we can use the chain rule and we're going to get minus H D theta by DT times by D 2 u by D theta squared let's eliminate this D theta T again with a hey CH over R squared so all of this is just going to be equal to minus H squared and 1 over R squared is just U squared so we get minus h squared u squared times by D two u by D theta squared so we can use this over here in order to eliminate this up here at the top so if we do all of this and also say we can replace this R cubed with one over R cubed with u cubed on the top sing with the R squared over here at the end of the day what we'll get is that D two u by D theta squared plus U is just going to be equal to G times by the mass of the star plus the mass of the planet and then the expression is just divided by H squared so I've just taken that factor of G times mass of the star etc over to the right hand side and so crucially this expression here for u as a function of theta can actually be solved so let's go ahead and solve this and the idea being that once we know what u as a function of theta is we can get our just by one overing it working out the reciprocal okay so we're nearly there now we have a really simple second order differential equation to solve and so what we need effectively is some function U that when we differentiate it twice and then add it to itself we get a constant so I can immediately think of two functions that would do that so if u of theta was equal something times by sine of theta plus a constant then when you differentiate sine twice so d to you by D theta squared the first time you get cosign the second time you get minus sign and the constant dies so you will get minus a times by sine of theta and so you can see that when you add this to the second derivative you would just get a constant which is what we have on the right hand side so this expression would do it and likewise if you had an expression involving say cosine of theta that would also do it and we can generally combine these two expressions together using a little trig identity in order to get the general solution to this equation would be U of theta is equal to du to do some constant let's call it alpha times by cosine of the angle theta minus a reference angle Omega just a constant at this point and then the constant we have to add is what we have on the right-hand side here G times the mass of the star plus the mass of the planet all of this divided by H squared and you can verify this is the solution relatively straightforwardly just by plugging it into the differential equation at the top and just differentiating it twice and then adding it which will make this term vanish and you'll just be left with this so the question now is is there a slightly neater way that we can write this and well how about we say take out a factor of well G times the mass of the start with mass apply it over h squared and rewrite this as G times by the mass of the star plus the mass of the planet all of this divided by H squared so since we took out this factor here we're just going to get a 1 and then what I'm going to define since this is just an arbitrary constant alpha I will say that alpha divided by all this stuff here is just another constant that I'll call e and then we get the cosine of theta minus Omega so all they've done here is just redefined constants so I can write it as G times by the mass of the star plus the mass of the planet over H squared times by a constant e is what I defined to be equal to alpha so now that we have what u is equal to it's time for the moment of truth so we know that U is G times the mass of the star plus the mass of the planet divided by H squared multiplied by 1 plus e cosine of theta minus Omega and this i mean sometimes people define this theta minus Omega to be an angle cause the true anomaly and it's useful in exoplanets but for our purposes here we're going to define the reference direction Omega to be equal to 0 which just means that we define our reference to be along the periapsis direction but don't worry about that for the moment what we're interested in are how the distance between the planet and the star varies as a function of the angle and since R is just 1 divided by u we therefore have that R as a function of theta is just given by H squared divided by G times by the mass of the star plus the mass of the planet times by 1 divided by 1 plus e cosine of theta so we're practically there now we can just use one more resort in order to simplify this slightly in that remember that for the equation of an ellipse what we are looking at earlier we have say that the distance to Parry apps is given by a times by 1 minus e and that perry apps occurred when theta is equal to 0 so if you remember the diagram it looked a little like this here is the star here is the planet this is the angle theta this is R so when theta is equals zero you're at your periapsis position and in fact and the equation in this form here is already the equation of an ellipse so long as this constant down here E is between zero and one this mathematically is just the equation of an ellipse so if we plug in theta is equal 0 in which case recall that when theta is 0 cosine of theta is equal to 1 then what we'll find is that a times by 1 minus e is equal to H squared over G times the mass of the star plus the mass of the planet times by 1 divided by 1 plus e we can then say multiplied by this and we'll probably this to get that h squared is just going to be a times by 1 minus e squared times by G and this mass of the star plus the mass of the planet so now that we know an expression for H squared in terms of the semi-major axis we can write what our final result is going to be for the equation of a planetary orbit so putting everything together we finally have that the distance between the planet and the star is given by the semi-major axis a times by 1 minus the eccentricities squared all of this divided by 1 plus e cosine of theta and if you plot what r looks like as a function of theta then you find an elliptical orbit exactly as Kepler deduced all those years ago well that's all we have time for here today next time I will be doing the much simpler you'll be pleased to hear tasks of deriving Kepler's 2nd and 3rd laws of planetary motion please let me know down below what you thought about this somewhat experiment or video and I hope you didn't get lost too much in the technical details but thanks for watching and bye for now I hope you enjoyed this video which is laying the groundwork for a new series on the science of exoplanets but I'll be producing later in the year this week's feature video is an examination of the scientific objectives and instruments onboard the European Space Agency's ExoMars 2016 mission which successfully launched earlier this week next time I'll be concluding a tour of kepi floors and be looking at some applications of them in astrophysics feel free to subscribe so that you don't miss it and if you do have any questions or comments about this video and by all means are drop them down below and I'll be happy to answer

Channel: Martian Colonist

Views: 47,664

Rating: 4.7617555 out of 5

Keywords: Kepler's Laws, Orbits, Orbital Mechanics, Elliptical Orbits, Space, Astrophysics, Newton's laws, Newtonian Gravity, Kepler, Kepler's 1st law, Exoplanets

Id: DurLVHPc1Iw

Length: 58min 28sec (3508 seconds)

Published: Sun Mar 20 2016