Physics 320: Orbital Mechanics (Lecture 7) Dale Gary NJIT Physics Department

Reminder of Kepler’s Laws Kepler's Three Laws (quantitative version) First Law: Planets travel in elliptical orbits with the Sun at one focus, and obey the equation 𝑟=𝑐/(1+𝑒 cos 𝜃 ), where 𝑐=𝑎(1− 𝑒 2 ) for 0 < e < 1. (Comets and other bodies can have hyperbolic orbits, where 𝑐=𝑎( 𝑒 2 −1), for e > 1.) Second Law: The radius vector of a planet sweeps out equal areas in equal times (planet travels fastest when near perihelion). Third Law: The square of the orbital period of a planet is proportional to the cube of its semi-major axis: 𝑃 2 =𝑘 𝑎 3 , where k is a constant We also showed how Newton found the value of the constant 𝑘= 4 𝜋 2 𝐺𝑀 . So 𝑃 2 = 4𝜋 2 𝐺𝑀 𝑎 3 . Today, we will give a refinement of this result. September 25, 2018

Center of Mass Reference Frame We are now going to discuss the notion of center of mass, with which you are certainly already familiar. Think of a system of N particles a = 1, …, N, with masses ma and positions ra. The center of mass (or CM) is defined to be the position Like any vector equation, this represents separate equations for each of the components (X, Y, Z): You can think of the center of mass as a weighted average of the positions of each mass element, i.e. weighted by the mass of that element, or equivalently it is the vector sum of the ra, each multiplied by the fraction of mass at that location. To get a feeling for CM, let’s look at the center of mass for a two particle system, which might, for example, represent the Sun and Earth, or two stars in orbit around each other. where 𝑀= 𝛼=1 𝑁 𝑚 𝛼 September 25, 2018

Center of Mass and Equation of Motion In this case, , which can be seen in the figure. CM m1 m2 O r1 r2 R It is easy to show that the distance of the CM from m1 and m2 is in the ratio m2/m1. The figure shows the case where m1  4m2. In particular, if m1 >> m2, then the CM will be very close to m1. Note that the time derivative of the center of mass for N particles is just the CM velocity so the momentum of an N-particle system is related to its CM by . Differentiating this expression, we get the very useful relation for the Newton’s second law, the equation of motion of a system: This says that the CM of a collection of particles moves as if the external forces on all of the individual particles were concentrated at the CM. This is why we can treat extended objects (e.g. a planet) as a point mass. Here I am using “dot-notation” for time derivative. 𝑑𝑟 𝑑𝑡 = 𝑟 September 25, 2018

Reduced Mass If we now move our coordinate system to the center of mass (i.e. set R = 0), Then we can write r1 and r2 in terms of the vector r = r2 – r1 between the two masses: We now introduce the concept of reduced mass: so 𝐑= 𝑚 1 𝐫 1 + 𝑚 2 𝐫 2 𝑚 1 + 𝑚 2 =0 𝐫 1 =− 𝑚 2 𝑚 1 + 𝑚 2 𝐫 𝐫 2 = 𝑚 1 𝑚 1 + 𝑚 2 𝐫 𝐫 1 =− 𝜇 𝑚 1 𝐫 𝐫 2 = 𝜇 𝑚 2 𝐫 𝜇= 𝑚 1 𝑚 2 𝑚 1 + 𝑚 2 September 25, 2018

Reduced Mass and the CM Reference Frame When two objects of similar mass orbit each other, they both move around the common center of mass as in the figure below, left. We can reduce this complicated looking problem to an equivalent problem, where there is a single body of reduced mass m, orbiting a central body of mass M = m1 + m2, with a separation . r = |r1- r2| Note, as 𝑚 1 ∞, 𝜇 𝑚 2 and the problem corresponds to a relatively small mass planet orbiting a massive star, as we have been implicitly assuming up to now. In many of our expressions from now on, we will be using reduced mass instead of the mass of one body, since it is more general. r = |r1- r2| m m2 m1 r1 r2 O m2 m1 r1 r2 O r1 r2 CM M Arbitrary origin Origin at CM Equivalent one-dimensional problem Path relative to CM September 25, 2018

Kepler’s First Law Derived To derive the fact that a planet has to orbit in an ellipse, we have to begin with the orbital angular momentum (in the center of mass and using the reduced mass): An important fact about the gravitational force is that it is a central force, meaning it only acts radially, and cannot produce any change in angular momentum, i.e. 𝑑𝐋/𝑑𝑡=0. You can show this directly: The text starts with the expression for angular momentum, and goes through a purely mathematical derivation of Kepler’s first law, which is not at all obvious and uses several obscure mathematical tricks. I encourage you to go through the steps, but we will not do so here. We just write the final result: The equation for an ellipse in terms of L, Note that this means , which we will make use of later. 𝐋=𝐫×𝐩=𝐫×𝜇𝐯 v and p = mv are parallel, so the cross product is zero. r and F are parallel, so the cross product is zero. 𝑑𝐋 𝑑𝑡 = 𝑑𝐫 𝑑𝑡 ×𝐩+𝐫× 𝑑𝐩 𝑑𝑡 =𝐯×𝐩+𝐫×𝐅 𝑟= 𝐿 2 / 𝜇 2 𝐺𝑀(1+𝑒 cos 𝜃 ) cf. our earlier polar equation for an ellipse Kepler’s 1st Law 𝑟= 𝑎(1− 𝑒 2 ) 1+𝑒 cos 𝜃 𝐿=𝜇 𝐺𝑀𝑎(1− 𝑒 2 ) September 25, 2018

Kepler’s 2nd Law Derived Kepler’s 2nd law, about the radius sweeping out equal areas in equal times, can be derived by considering the area swept. The radius sweeps out an area given by half of this rectangle, or 𝑑𝐴= 1 2 𝑟 2 𝑑𝜃. Taking the derivative wrt time, 𝑑𝐴 𝑑𝑡 = 1 2 𝑟 2 𝑑𝜃 𝑑𝑡 = 1 2 𝑟 𝑣 𝜃 . But angular momentum is 𝐿=𝜇 𝐫×𝐯 =𝜇𝑟 𝑣 𝜃 , so the rate of change of area can be expressed in terms of L So Kepler’s finding, that this rate of change of area is a constant, is merely an expression of the fact that the angular momentum is a constant. r rdq dq orbit Sun 𝑑𝐴 𝑑𝑡 = 𝐿 2𝜇 Kepler’s 2nd Law September 25, 2018

Velocities in the Orbit At two particular locations in the orbit, namely perihelion (when the planet is closest to the Sun) and aphelion (when farthest from the Sun), there is no radial part of the velocity—the planet is moving only in the theta direction. At perihelion, 𝜃=0, so our earlier equation for the ellipse becomes, But also here, the angular momentum is just 𝐿=𝜇 𝑟 𝑝 𝑣 𝑝 , so A couple of lectures ago, we showed that 𝑟 𝑝 =𝑎(1−𝑒), so solving for 𝑣 𝑝 , we have and likewise 𝑟 𝑝 = 𝐿 2 / 𝜇 2 𝐺𝑀(1+𝑒) 𝑟 𝑝 = (𝜇 𝑟 𝑝 𝑣 𝑝 ) 2 / 𝜇 2 𝐺𝑀(1+𝑒) = ( 𝑟 𝑝 𝑣 𝑝 ) 2 𝐺𝑀(1+𝑒) 𝑣 𝑝 = 𝐺𝑀 𝑎 1+𝑒 1−𝑒 1/2 𝑣 𝑎 = 𝐺𝑀 𝑎 1−𝑒 1+𝑒 1/2 Perihelion speed Aphelion speed September 25, 2018

Kepler’s Third Law Derived Kepler’s third law can easily be obtained from his second law, by noting that the area swept out in a given time is constant, and the area swept out over one period is just the entire ellipse, which has an area 𝐴=𝜋𝑎𝑏. Thus, integrating the second law equation over an entire period, So Kepler’s third law becomes But we have an expression for b in terms of a, i.e., 𝑏 2 = 𝑎 2 (1+ 𝑒 2 ), and we also have our expression for L, which when squared is Putting these both into the above, we have our final form of Kepler’s third law: 𝐴= 𝐿 2𝜇 𝑃= 𝐿 2𝜇 =𝜋𝑎𝑏 𝑃 2 = 4 𝜋 2 𝑎 2 𝑏 2 𝜇 2 𝐿 2 𝐿 2 = 𝜇 2 𝐺𝑀𝑎(1− 𝑒 2 ). 𝑃 2 = 4 𝜋 2 𝐺( 𝑚 2 + 𝑚 1 ) 𝑎 3 Kepler’s 3rd Law September 25, 2018

Potential Energy We learned earlier that Newton’s law of universal gravitation provided the gravitational force equation: You may also recall that energy is force through a distance, and that potential energy is the negative of work done, i.e. Inserting the gravitational force into this equation (and being careful with signs!): Evaluating the integral: I stated earlier that we take the zero of potential energy at infinity, so we have the final result for any r: 𝐅=−𝐺 𝑀𝑚 𝑟 2 𝐫 𝑈 𝑓 − 𝑈 𝑖 =∆𝑈=− 𝑟 𝑖 𝑟 𝑓 𝐅∙𝑑𝐫 ∆𝑈= 𝑟 𝑖 𝑟 𝑓 𝐺 𝑀𝑚 𝑟 2 𝑑𝑟 A massive object M creates a gravity well around it 𝑈 𝑓 − 𝑈 𝑖 =−𝐺𝑀𝑚 1 𝑟 𝑓 − 1 𝑟 𝑖 𝑈=−𝐺 𝑀𝑚 𝑟 September 25, 2018

Escape Speed As a side note, since the potential energy at infinity is zero, an object falling from infinity will lose potential energy, which is everywhere negative inside the gravity well, and consequently the object will gain a kinetic energy given by the well-known K = ½ mv2. But it does this in such a way that its total energy remains zero! It simply converts its lost potential energy into kinetic energy, in exactly the same way that a ball dropped from a height on Earth does. Of course, it works the other way for an object moving outward. It will lose kinetic energy as it moves toward infinity (climbs out of the gravity well), and provided it has just enough energy to get to infinity, it will arrive there with 0 total energy. The speed that an object needs to go from some point in a gravity well (say, the surface of a planet) to infinity with zero energy is called the escape speed, derived from:𝐸=𝐾+𝑈= 1 2 𝑚 𝑣 2 −𝐺 𝑀𝑚 𝑟 =0, so 𝑣 esc = 2𝐺𝑀/𝑟 September 25, 2018

Orbital Energy It was mentioned a few lectures ago that any bound orbit has a total energy of less than 0. Let’s write the energy of a planet when it is at perihelion: Using our values for perihelion distance and velocity: We have If you calculate the same thing at aphelion, you will find the same expression. In fact, the energy is everywhere constant (and negative) over the orbit, and has this value: 𝐸= 1 2 𝜇 𝑣 𝑝 2 −𝐺 𝑀𝜇 𝑟 𝑝 𝑣 𝑝 = 𝐺𝑀 𝑎 1+𝑒 1−𝑒 1/2 𝑟 𝑝 =𝑎(1−𝑒) 𝐸= 1 2 𝜇𝐺 𝑀 𝑎 1+𝑒 1−𝑒 −𝐺 𝑀𝜇 𝑎 1−𝑒 = 1 2 𝜇𝐺 𝑀 𝑎 1−𝑒 𝑒−1 =−𝐺 𝑀𝜇 2𝑎 𝐸=−𝐺 𝑀𝜇 2𝑎 September 25, 2018

Velocities in the Orbit Again Since the energy is everywhere constant, we can equate it to the kinetic plus potential energy at any point: We can then get a general expression for velocity in an orbit In the above, we have used the fact that M is the total mass of the system, M = m1 + m2. This equation has a special name, the vis-viva equation. 𝐸=−𝐺 𝑀𝜇 2𝑎 = 1 2 𝜇 𝑣 2 −𝐺 𝑀𝜇 𝑟 𝑣 2 =𝐺( 𝑚 1 + 𝑚 2 ) 2 𝑟 − 1 𝑎 September 25, 2018

What We’ve Learned We learned that Kepler’s laws are just a consequence of the properties of a central force law, and that angular momentum is conserved in such a system. You should know that a star+planet system has a rather complex orbital relationship when the two masses are similar, but that this two-body system is equivalent to a simpler system with a planet of reduced mass 𝜇= 𝑚 1 𝑚 2 𝑚 1 + 𝑚 2 orbiting a central object of mass 𝑀= 𝑚 1 + 𝑚 2 ., at a distance 𝑟= 𝑟 1 + 𝑟 2 , where r1 is the distance of m1 from the CM, and r2 is the distance of m2 from the CM. The equations governing Kepler’s three laws are: We also have the orbital energy equation, constant (and negative) for any bound orbit. We derived the useful equation for speed in an orbit 𝑟= 𝐿 2 / 𝜇 2 𝐺𝑀(1+𝑒 cos 𝜃 ) Kepler’s 1st Law (equation for an ellipse) 𝑑𝐴 𝑑𝑡 = 𝐿 2𝜇 Kepler’s 2nd Law (conservation of energy) 𝑃 2 = 4 𝜋 2 𝐺( 𝑚 2 + 𝑚 1 ) 𝑎 3 Kepler’s 3rd Law (period-distance law) 𝐸=−𝐺 𝑀𝜇 2𝑎 𝑣 2 =𝐺( 𝑚 1 + 𝑚 2 ) 2 𝑟 − 1 𝑎 September 25, 2018