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Published byColin Casey Modified over 7 years ago

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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

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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.

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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

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**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

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**Kepler’s First Law Examples: Earth: e = 0.0167 Mercury: e = 0.2056**

Venus: e =

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Kepler’s First Law a = semi-major axis

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**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.

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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

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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

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

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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

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

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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?

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

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**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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