Gravitation and Kepler’s Laws Newton’s Law of Universal Gravitation in vectorial form: Gm1m2 F12 F21 r̂12 2 r12 where the hat (ˆ) denotes a unit vector as usual. Gravity obeys the superposition principle, so if particle 1 is attracted by particles 2 and 3, the total force on 1 is F12 F13. Central forces are conservative, we can define gravitational potential energy: r r Gm1m2 Gm1m2 dr V r F dr r r 2 ∞ ∞ Define also gravitational potential (aka gravitational potential energy per unit mass) Gm Φ r r (set m1 m and replace m2 with 1). Likewise, gravitational field g as gravitational force per unit mass: Gm g r r̂ r2 Field and potential are related in the usual way: g ∇Φ 2 Gravity From A Spherical Shell: Direct Calculation Consider a thin spherical shell of radius a, mass per unit area ρ and total mass m 4πρa2 Superposition principle leads to element of mass: m dm ρ 2π a sin θ adθ sinθ dθ 2 Contribution to the potential from annulus is: dΦ G dm Gm sin θ dθ R 2 R Integrate over θ from 0 to π. Change the integration variable from θ to R, via R2 r 2 hence a2 2ar cosθ sinθ dθ R dR ar 3 If r a integration limits are r a and r r a they are a r and a r: a; if Gm r a Gm r for r a Φ r dR Gm a for r a 2ar r a Gravitational field by differentiation: Gm r̂ r2 for r a g r 0 for r a 1. Outside the shell, the potential is just that of a point mass at the centre. 2. Inside the shell, the potential is constant and so the force vanishes. 4 Now Using Analogy With Coulomb Force Apply integral form of Gauss’ Law to gravitational case: S g dS 4πG ρm dV V That is: Surface integral of normal component of gravitational field over given surface S is equal to 4πG times the mass contained within surface, with mass obtained by integrating mass density ρm over volume V contained by S. From spherical symmetry gravitational field g must be radial: g g r̂ Choose a concentric spherical surface with radius r a: mass enclosed is just shell mass m and Gauss’ Law says 4πr2g 4πGm giving g Gm r̂ 2 r for r a. Likewise, choose concentric spherical surface inside shell: mass enclosed is zero and g vanish. 5 Orbits: Preliminaries Two-body Problem: Reduced Mass Express position ri as CM location R plus displacement ρi relative to it: r1 R ρ1 r2 R ρ 2 Change variables from r1 and r2 to R and r r1 r2 (recall F F12 F21, internal forces only): m1r̈1 F m2r̈2 Set M m1 m2, thus M R̈ m1r̈1 F m2r̈2 0 i.e., CM velocity is constant. Consider relative displacement r, r̈ r̈1 r̈2 1 1 F m1 m2 i.e., F µr̈ and re-encounter reduced mass µ m1 m2 m1 m2 m1 m2 F m1 m2 6 For conservative force F there is potential energy V r , hence total energy is: E 1 1 2 M Ṙ2 µṙ V r 2 2 Likewise, when F is central, total angular momentum is L M R Ṙ µ r ṙ Since CM velocity is constant, choose inertial frame with origin at R, i.e.: R 0 Hence: E L 1 2 µ ṙ V r 2 µ r ṙ Rationale: two-body problem reduces to equivalent single body one of mass µ at position r relative to fixed centre, acted upon by force F ∂V ∂r r̂ 7 If m2 m1 : µ m1m2 m1 m2 m1 m1 r 1 m2 r 2 r2 R m1 m2 (‘fixed Sun and moving Planet approximation’). Commentary: 1. Approximation valid for Kepler’s Laws (m1 mPlanet and m2 mSun). 2. Can ignore interactions between Planets in comparison to gravitational attraction Planet-Sun. 8 Two-body Problem: Conserved Quantities Gravity is central force: gravitational attraction between two bodies acts along line joining them. Gravitational force on mass µ acts in direction r and no torque is exerted about fixed centre: L constant Both magnitude and direction of L are fixed ! Since L r p r µṙ L is perpendicular to plane defined by position and momentum of µ. Conversely: r and p must always lie in fixed plane of all directions perpendicular to L. Can therefore describe motion using plane polar coordinates r θ , with origin at fixed centre ! Radial and angular equations of motion become: k radial equation r̈ rθ̇2 mr2 1d 2 r θ̇ 0 angular equation r dt Gravitational force is F kr̂ r2 k GMm wherein µ m mPlanet and M mSun . 9 Angular equation expresses conservation of angular momentum: L mr2θ̇ Other conserved quantity is total energy: E 1 2 1 2 2 mṙ mr θ̇ k r 2 2 wherein gravitational potential energy is V r k r Commentary: 1. Gravitational potential energy will help deducing shape of planetary orbits ! 10 Two-body Problem: Two Problems Comet 1. Comet approaching Sun in plane of Earth’s orbit (assumed circular) crosses orbit at angle of 60 travelling at 50 km s 1. 2. Closest approach to Sun is 1 10 of Earth’s orbital radius (re). 3. Ignore attraction of comet to Earth compared to Sun (i.e., reduced mass µ m mComet). 4. Aim: compute comet’s speed at point of closest approach. Solution Key: angular momentum conservation L r p r mv of comet about Sun At point of closest approach comet’s velocity must be tangential only: r v rmin vmax At crossing point: r v rev sin30 Equate two expressions: rmin vmax 0 1 revmax 1 rev 2 11 Finally, vmax 5v 250 kms 1 12 Cygnus X1 1. Cygnus X1 is a binary system of a supergiant star of 25 solar masses and a black hole of 10 solar masses, each in a circular orbit about CM with period 5 6 days. 2. Aim: Determine distance between supergiant and black hole, given solar mass 1 99 1030 kg. Solution Key: 2-body equation of motion in polar coord.s: (m1 2 Gm1m2 m1 m2 2 r ω r2 m1 m2 mass, r distance, ω ang. velocity). Where is RHS 2nd term coming from ? Introduce period: T 2π ω Extract distance: G m1 r 3 m2 T 2 4π2 That is, 27 5 1030 m3 r 3 1010 m 13 Kepler’s Laws State Kepler’s Laws: 1. The orbits of the planets are ellipses with the Sun at one focus. 2. The radius vector from the Sun to a planet sweeps out equal areas in equal times. The square of the orbital period of a planet is pro3. portional to the cube of the semimajor axis of the planet’s orbit (T 2 ∝ a3). Next lecture: their derivation 14 Kepler’s 2nd Law This is statement of angular momentum conservation under action of central gravitational force. Angular equation of motion gives: r2θ̇ L const m Leads to: dA dt 1 2 r θ̇ 2 L const 2m 15 Orbit equation Ellipses are specific to inverse square law for force, hence first and third laws are specific to inverse square law force. Study radial equation of motion (k GMm) r̈ rθ̇2 k ! mr2 (i) Remove θ̇ using angular momentum conservation, θ̇ L mr2 get r̈ L2 m2 r 3 k mr2 (ii) Use relation d d θ̇ dt dθ L d mr2 dθ (differential equation for r in terms of θ). (iii) Substitute u 1 r to obtain orbit equation: d 2u u 2 dθ mk L2 16 Kepler’s 1st Law Solution of orbit equation is 1 r mk 1 e cosθ L2 First law: for 0 e 1 is an ellipse, with semi latus rectum l L2 mk. Kepler’s 3rd Law Start with 2nd law for rate of area: dA L dt 2m Integrate over complete orbital period T : T 2mA L A πab is area of ellipse ! Substituting for b in terms of a gives third law: T 2 4π2 3 a GM 17 Scaling Argument for Kepler’s 3rd Law Suppose you found a solution to orbit equation r̈ rθ̇2 k mr2 i.e., r and θ as functions of t. Scale radial and variables by constants α and β: r αr t βt In terms of r and t , LHS of orbit equation is: dθ 2 d 2r r dt 2 dt α θ̇ 2 αr r̈ β2 β α 2 r̈ r θ̇ β2 RHS becomes: k mr 2 1 k α2 mr2 Compare two sides, new solution in terms of r and t provided β2 α3 That is, T 2 ∝ a3 1. Need solving orbit equation for proportionality constant. 2. Scaling argument makes clear third law based on inverse-square force law. 18 Problem Sheet 6 Section B. Earth’s speed in circular motion about Sun !? Orbit equation in polar coordinates: l 1 r e 0 ! That is, re l L2 mk m2v2e re2 m2MSunG Inverting, v2e re GMSun re2 v2e GMSun re Finally, Everything can be expressed in terms of ve ! 19 Energy Considerations: Effective Potential Gravitational force is conservative, hence total energy E of orbiting body is conserved: 1 2 1 2 2 mṙ mr θ̇ V r 2 2 E Angular momentum is also conserved (force is central), hence use r2θ̇ L m to remove θ̇2: 1 2 E mṙ 2 L2 V r 2mr2 Formally, energy equation of particle in linear motion under effective potential U r L2 V r 2mr2 Effective potential contains centrifugal term, L2 2mr2 arising because angular momentum is conserved. Replace V r U r k r and use l L2 mk: kl k 2r2 r Fig 20 Interpret U r as a function of r for given E By definition ṙ2 0, implying kl k E U r 2r2 r Draw a horizontal line for E, U r lies below it ! Circular Orbit At minimum U r E, r is constant at rc l L2 mk hence orbit is circular and total energy is E k 2l mk2 2L2 Elliptic Orbit If k 2l E 0, motion is allowed for r p r ra perihelion r p and aphelion ra given by roots of E kl 2r2 k r Parabolic Orbit If E 0, there is always minimum value for r but escape to infinity is just possible. Hyperbolic Orbit For E 0, escape to infinity is possible with finite kinetic energy at infinite separation. 21 Orbits in a Yukawa Potential Consider Yukawa potential αe V r r κr α 0 κ 0 Describes, e.g., attractive force between nucleons in an atomic nuclei. Neglect quantum-mechanics and use classical dynamics. Effective potential becomes U r L2 αe κr 2mr2 r Trajectories are more complicated: Fig. 22 Interpret U r as a function of r for given E E 0 but greater than Umin Rosette orbit, i.e., ellipse with rotating orientation, aka precession of perihelion. Typical of small (κ 0) perturbations of planetary orbits, e.g., due to other planets (irregularities in Uranus’ motion led to discovery of Neptune, 1846). Large r limit Term L2 2mr2 dominates exponentially falling Yukawa term, so U r becomes positive ! If Umax E 0, two possible orbits classically distinct. In quantum mechanics, ‘tunnelling’ becomes possible (e.g., alpha decay) !