2. Sample
Determine the net gravitational force acting on the Earth during a total lunar eclipse.
msun = 1.99 x1030 kg
mmoon = 7.36 x 1022 kg
rsun = 1.5 x 108 km
rmoon = km

3. Kepler ( )
Used Tycho Brahe's precise data on apparent planet motions and relative distances.
Deduced three laws of planetary motion.
Took him the last 30 years of his life.

4. Kepler’s First Law
The orbits of the planets are elliptical (not circular) with the Sun at one focus of the ellipse.
'a' = semi-major axis: Avg. distance between sun and planet

5. Kepler’s First Law Perihelion – close to sun (perigee)
Aphelion – furthest from sun (apogee)
Eccentricity – how not a circle are you?
circle e = 0
parabola e = 1

6. Kepler’s First Law Examples: Earth: e = 0.0167 Mercury: e = 0.2056
Venus: e =

7. Kepler’s First Law
a = semi-major axis

8. Translation: planets move faster when closer to the Sun.
Kepler's Second Law
A line connecting the Sun and a planet sweeps out equal areas in equal times.
slower
faster
Translation: planets move faster when closer to the Sun.

9. Kepler's Second Law
A line connecting the Sun and a planet sweeps out equal areas in equal times.
The speed of the planet in orbit is dependent on its distance from the sun
voro = vf rf
slower
faster

10. Sample
If the Earth has an orbital speed of 29.5 km/sec at apogee, determine the orbital speed at apogee.
ra = 1.52 x 108 km
rp = 1.47 x 108 km

11. Kepler’s Second Law
voro = vf rf
Note where the highest speeds of tornados and hurricanes are.

12. Kepler’s Third Law
The square of a planet’s orbital period is proportional to the cube of its semi-major axis .
Translation: the further the planet is from the sun, the longer it will take to go around

13. Why Does it Work?
Newton discovers that ellipses are pretty close to circular, just with the sun offset
Fc = Fg

14. Sample
How fast must a satellite move to maintain an orbit of 500 km over the Earth’s surface?
What is the period of rotation?

15. Kepler’s Third Law
Newton took the idea of centripetal force and applied it to the Kepler problem
Fc = Fg

16. Solution . Where M is the mass being orbited (as opposed to orbiting)
T is the period of the orbit
r is the radius of the orbit