Short Version : 8. Gravity

Retrograde Motion Retrograde motion of Mars. As explained by Ptolemy.

Ptolemaic (Geo-Centric) System epicycle equant  deferent  swf

Cassini Apparent Sun Venus

8.1. Toward a Law of Gravity 1543: Copernicus – Helio-centric theory. 1593: Tycho Brahe – Planetary obs. 1592-1610: Galileo – Jupiter’s moons, sunspots, phases of Venus. 1609-19: Kepler’s Laws 1687: Newton – Universal gravitation. Phases of Venus: Size would be constant in a geocentric system.

Kepler’s Laws Explains retrograde motion Mathematica

8.2. Universal Gravitation Newton’s law of universal gravitation: m1 & m2 are 2 point masses. r12 = position vector from 1 to 2. F12 = force of 1 on 2. G = Constant of universal gravitation = 6.67  1011 N m2 / kg2 . F12 m2 r12 m1 Law also applies to spherical masses.

Example 8.1. Acceleration of Gravity Use the law of gravitation to find the acceleration of gravity at Earth’s surface. at the 380-km altitude of the International Space Station. on the surface of Mars.  (a) (b) (c) see App.E

Cavendish Experiment: Weighing the Earth ME can be calculated if g, G, & RE are known. Cavendish: G determined using two 5 cm & two 30 cm diameter lead spheres.

8.3. Orbital Motion g = 0 projectiles orbit Orbital motion: Motion of object due to gravity from another larger body. E.g. Sun orbits the center of our galaxy with a period of ~200 million yrs. Newton’s “thought experiment” g = 0 Condition for circular orbit Speed for circular orbit projectiles orbit Orbital period Kepler’s 3rd law

Example 8.3. Geosynchronous Orbit What altitude is required for geosynchronous orbits? Altitude = r  RE Earth circumference = Earth not perfect sphere  orbital correction required every few weeks.

Elliptical Orbits Perihelion: closest point to sun. Projectile trajectory is parabolic only if curvature of Earth is neglected. ellipse Orbits of most known comets, are highly elliptical. Perihelion: closest point to sun. Aphelion: furthest point from sun.

Open Orbits Open (hyperbola) Closed (circle) Borderline (parabola) (ellipse) Mathematica

8.4. Gravitational Energy U = 0 on this path How much energy is required to boost a satellite to geosynchronous orbit? U = 0 on this path U12 depends only on radial positions. … so U12 is the same as if we start here.

Zero of Potential Energy  Gravitational potential energy E > 0, open orbit Open Closed E < 0, closed orbit Bounded motion Turning point

Energy in Circular Orbits  r > r K K  E K E U K U Higher K or v  Lower E & orbit (r) . To catch the satellite, the shuttle needs to lose energy. It does so by turning to fire its engine opposite its direction of motion. It drops lower, turns again , and fires its engine to achieve a circular orbit, now faster and lower than before. Mathematica

energy Altitude  K K > K E = K+U = U / 2 h h < h E = K+U = U / 2 E = K+U < E ( K < K ) U U < U UG

8.5. The Gravitational Field Two descriptions of gravity: body attracts another body (action-at-a-distance) Body creates gravitational field. Field acts on another body. near earth Near Earth: Large scale: Action-at-a-distance  instantaneous messages Field theory  finite propagation of information Only field theory agrees with relativity. Great advantage of the field approach: No need to know how the field is produced. in space

Application: Tide Two tidal bulges Moon’s tidal (differential) force field at Earth’s surface Moon’s tidal (differential) force field near Earth Two tidal bulges Mathematica Sun + Moon  tides with varying strength. Tidal forces cause internal heating of Jupiter’s moons. They also contribute to formation of planetary rings.