What are Kepler’s Laws? Notes written by Boming Jia 07/17/2014 [This page is intentionally left as blank] Kepler's laws of planetary motion are three scientific laws describing the motion of planets around the Sun. They are 1. The planets move in elliptical orbits, with the Sun at one focus point. 2. The radius vector to a planet sweeps out area at a rate that is independent of its position in the orbits. 3. The square of the period of an orbit , is proportional to the cube of the semi-major-axis length 푇 푎 in equation: = 2 3 2 4휋 푎 Here and throughout the note 푇 퐺푀 stands for the Gravitational Constant which퐺 is about 6.67 × 10 ( / ) ) −11 2 is the mass of the Sun.푁 ∙ 푚 푘푔 푀We assume the mass of the star is very large compared to the mass of the planets so we can treat the position of the star to be a fixed point in the space. Without this assumption we need to use the idea of reduced mass and modify this equation a little bit. In 1609 Johannes Kepler published his first two laws about planetary motion, by analyzing the astronomical observations of Tycho Brahe. And Kepler's Third Law was published in 1619 (only in the form of proportionality of and ). In 2 3 1687 Isaac Newton showed that his own laws of motion and law푇 of universal푎 gravitation implies Kepler's laws, which is the beginning of all mathematical formulations of Laws of Physics. In this note, I am going sketch the derivation of all three of Kepler's Laws from classical Newtonian mechanics. Also we will see how much information about gravity we can get from Kepler’s Laws. Newton’s laws of physics implies the Kepler’s Laws Newtonian Mechanics Laws tells us essentially the relation = where is the total external force acted on a point particle with mass퐹 푚푎 and is 퐹the acceleration of this particle. Since the sizes of the planets and the푚 Sun 푎are very small compared to the distance between the planet and the star, we can treat them as point particle. Another reason that we can treat them as mass points is because they are essentially spherical. And we can show that the Newton’s gravity between two solid balls and with mass and whose centers are distance apart is the same as the퐵1 gravity퐵2 between two푀 point1 mass푀2 and with distance 푟. Newton’s Gravity Law describes the gravity between푀1 푀 two2 point mass (푟 and ) with distance to each other. The relation is = where 푀is the 퐺푀푚 2 푚magnitude of the gravity푟 between them. However this퐹 is 푟not everything퐹 that Newton’s Gravity Law says. We know that force, acceleration and position can all be described by vectors. So we can put an arrow above all the vector quantities which has not only a magnitude but also a direction. Thus the Newtonian Mechanics Laws states = which also tells us the particle accelerates in exactly the direction of 퐹⃑the 푚total푎⃑ external force on it. Now we can state the Newton’s Gravity Law in the vector form which is as following: = | | | | 퐺푀푚 푟⃑ 퐹⃑ − 2 Here is the position vector of (from 푟⃑) and푟⃑ is the gravity acted on by . And actually푟⃑ it is also true if we 푚switch “ 푀” and “퐹⃑ ” in the last sentence. 푚 푀 푚 푀 We see this formula is a little more complicated than the formula for magnitude. The minus sign “ ” tells us that gravity is always attractive. And the direction of the force is exactly the opposite of which is the direction of the − | | 푟⃑ position vector from M to m. 푟⃑ In this note we adopt a useful notation for the time derivative as following: = = 2 for any quantity 푑푦 푑 푦 2 So by definition we have푦̇ the푑푡 velocity푎푛푑 푦 ̈ =푑푡 and the acceleration푦 = . Another piece of important ingredient푣⃑ 푟⃑̇ for the derivation푎⃑ is 푟⃑thë angular momentum. For a mass particle with position vector and velocity (both of which depends on the reference frame),푚 the angular momentum푟⃑ of it is 푣a⃑ vector defined by = × and here “×” is the usual cross product in three퐿�⃑ dimension. By퐿�⃑ taking푟⃑ derivative푚푣⃑ of = × and using the relation = we will get = × where is the퐿�⃑ gravitational푟⃑ 푚푣⃑ force acted on the planet.퐹⃑ 푚 Since푎⃑ Newton’s퐿�⃑ ̇ Gravity푟⃑ 퐹 ⃑Theory tells퐹⃑ us that is collinear with , then we have = 0, hence is a constant function of time퐹⃑ (in physics we 푟say⃑ such a quantity퐿�⃑̇ is “conserved”퐿�⃑ ). The conservation of angular momentum for a planet tells us two things, one is that | | is a constant with respect to time,퐿�⃑ the other is the planet always moves in the 퐿�⃑plane spanned by the velocity of the planet and the position vector 푣⃑ If we푟⃑ let = | | and set up a coordinate system in the plane of the span of and then we푟 can 푟express⃑ as ( , ) for some [0,2 ). Now we can푣⃑ see that푟⃑ the magnitude of the푟⃑ angular푟 푐표푠휃 momentum푟 푠푖푛휃 of our planet휃 ∈ is 휋= = 2 which is conserved. Also we have | | = + ( ) . 퐿 �퐿�⃑� 푚푟 휃̇ 2 2 2 푣⃑ 푟̇ 푟휃̇ We know that the kinetic energy for a particle with mass and velocity is | | (by integrating = with respect to spatial 푚displacement 푣⃑ ). 1 2 Therefore2 푚 푣⃑ the expression 퐹for⃑ the푚푎 ⃑kinetic energy is + ( ) . And푑 t푥⃑he 1 2 1 2 potential energy for Newton’s Gravitational Force is2 푚푟̇ 2(by푚 푟integrating휃̇ the 퐺푀푚 gravity equation = as a function of ). Let = − 푟 (throughout this note 퐺푀푚 2 we adopt this convention),퐹 푟 then the potential푟 energy훼 is퐺푀푚 . Also if we use E to 훼 denote the total energy of the planet, then we have − 푟 1 1 + ( ) = 2 2 2 2 훼 Since = , we can 푚rewrite푟̇ the푚 above푟휃̇ −formula퐸 as 푟 2 ̇ 1 퐿 푚푟 휃 + = 2 2 2 2 퐿 훼 A bit of algebra applied to 푚=푟̇ and2 − this formula퐸 we can get 푚푟 푟 1 21 2 2 ( 퐿) =푚푟 휃̇ + + 푟̇ 2 푚훼 푚퐸 2 − 2 2 2 We know that = 푟so 휃thė above푟 equation푟퐿 can퐿 be treated as a differential 푟̇ 푑푟 equation of the function휃̇ 푑휃 ( ). We can solve it by making a series of substitutions: = , = , = 푟 1휃+ , then we will get the general solution is 2 1 푚훼 2퐸퐿 2 2 푟 퐿 푚훼 푦 = 푧 푦 − ( 휀 ) whe� re is an arbitrary constant. We can take care of the 푚훼 2 axis푧 such퐿 휀 푐표푠that 휃 −=휃0 , hence we휃0 get = and in terms of and we 푚훼 0 2 have = 휃 . Let = , then푧 we퐿 have휀 푐표푠 휃 푟 휃 2 2 퐿 1 퐿 푚훼 1−휀 푐표푠휃 푚훼 푟 푘 2 = = = 1 + (1) 1 2 2 푘 퐿 퐸퐿 푟 푤ℎ푒푟푒 푘 푎푛푑 휀 � 2 − 휀 푐표푠휃 푚훼 푚훼 This equation describes the trajectory of the planet, and if we assume < 1 then calculate the equation in Cartesian coordinate by substitute = 휀 and = , with some algebra we can get 푥 푟 푐표푠휃 ( + ) 푦 푟 푠푖푛휃 + = 0 2 2 푥 푐 푦 2 2 = 푎, = 푏 = 1 1 1 푘 푘 푘휀 With some analytic푤ℎ푒푟푒 푎geometry, 2this푏 equation says2 푎푛푑 that푐 the traject2 ory is an ellipse − 휀 √ − 휀 − 휀 with one focus at the origin. Hence we have demonstrated Kepler’s First Law. (Question: what will happen if = 1 or > 1 ? In what cases these will happen?) Now we are going to prove 휀Kepler’s휀 Second Law. First let ( ) be the area that the radius vector of the planet sweeps in the time interval 0 to퐴 , 푡then the claim of ( ) the second law is just is constant. The proof is as following:푡 ′ 퐴 푡 ( ) = ( ) = = 2 2 ′ ′ 푟 휃̇ 퐿 퐴 푡 퐴 휃 휃̇ And is a constant. The second equality holds푚 because the shape that 퐿 sweeps in2 푚a very small angle is approximately a triangle with area . 푟⃑ 1 2 Now we demonstrate Kepler’s푑휃 Third Law. Let be the period of 2the푟 orbit,푑휃 i.e. the time it takes the planet to finish a full cycle of the푇 orbit. And for simplicity here we let = ( ). Since the rate between sweeping area and time is the constant , then퐴 we퐴 have푇 = . Also from the property of the ellipse we know that 퐿 퐿푇 2푚= = 퐴 , 2푚= = = where , 2 and . With a bit algebra we 푘 푘 퐿 2 2 퐴get A,휋푎푏 L, b, k, all 1cancelled−휀 푏 √and1−휀 end푘 up with푚훼 훼 퐺푀푚 4 훼 = 2 3 2 휋 푎 Which is exactly what Kepler’s 푇Third Law states. 퐺푀 Kepler’s Laws indicate Newton’s Gravity Law First we state Kepler’s laws in a more mathematical way. First Law: = (1) where , are constants, is the length of the radius vector and is the - 푟 푘 − 휖 푥 coordinate 푘of휖 the planet in the푟 Cartesian coordinate with the Sun at the푥 origin and푥 major axis as -axis. Second Law:푥 = ( ) (2) 푥 푥푦 where D is a constant, y is∫ 0the푦 y-푑푥coordinate− 2 of퐷 the푡 −planet푡0 in the Cartesian coordinate and will be different constants whenever the planet crosses the x-axis. (Draw a picture,푡 0then you will understand this better) Third Law: = (3) 2 퐷 where is a constant independent푘 퐻 of time It might퐻 be a little bit hard to see why the third law is in this form. But if you imitate the proof of the third law from Newtonian Mechanics in the last session, you can show that = 2 3 2 휋 푎 푘 푇 2 Then it is clear that indeed the Third Law tells퐷 us that 2 is a constant.