 # 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.

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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

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.

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 Law Examples: Earth: e = 0.0167 Mercury: e = 0.2056
Venus: e =

Kepler’s First Law a = semi-major axis

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.

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

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

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

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

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

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?

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

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

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