Sample Determine the net gravitational force acting on the Earth during a total lunar eclipse. m sun = 1.99 x10 30 kg m moon = 7.36 x 10 22 kg r sun =

SampleSampleDetermine 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 = 384 400 km

Kepler (1571-1630)

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.

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

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

Kepler’s First LawExamples:

Earth: e = 0.0167Mercury: e =

0.2056Venus: e =

0.00657

Kepler’s First Law

a = semi-major axis

Kepler's Second Law

A line connecting the Sun and a planet sweeps out equal areas in equal times.

Translation: planets move faster when closer to the Sun.

slower faster

Kepler's Second Law

A line connecting the Sun and a planet sweeps out equal areas in equal times.

slower faster

The speed of the planet in orbit is dependent on its distance from the sun

voro = vf rf

SampleSampleIf 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

Kepler’s Second LawKepler’s Second Law

voro = vf rf

Note where the highest speeds of tornados

and hurricanes are.

Kepler’s Third LawKepler’s Third LawThe 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

Why Does it Work?Why Does it Work?

Newton discovers that ellipses are pretty close to circular, just with the sun offset

Fc = Fg

SampleSampleHow fast must a satellite move to maintain an orbit of 500 km over the Earth’s surface?

What is the period of rotation?

Kepler’s Third LawKepler’s Third Law

Newton took the idea of centripetal force and applied it to the Kepler problem

Fc = Fg

SolutionSolution

.

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