1. Short Version : 8. Gravity

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

3. Ptolemaic (Geo-Centric) System
epicycle
equant 
deferent 
swf

4. Cassini Apparent
Sun
Venus

5. 8.1. Toward a Law of Gravity 1543: Copernicus – Helio-centric theory.
1593: Tycho Brahe – Planetary obs.
: Galileo – Jupiter’s moons, sunspots, phases of Venus.
: Kepler’s Laws
1687: Newton – Universal gravitation.
Phases of Venus:
Size would be constant in a geocentric system.

6. Kepler’s Laws
Explains retrograde motion
Mathematica

7. 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.

8. 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

9. 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.

10. 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

11. 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.

12. 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.

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

14. 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.

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

16. 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

17. 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

18. 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

19. 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.