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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 =

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Presentation on theme: "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 ="— Presentation transcript:

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2 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 kg r sun = 1.5 x 10 8 km r moon = 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 Keplers 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 Keplers 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 Keplers First Law Examples: Earth: e = Mercury: e = Venus: e =

7 Keplers First Law a = semi-major axis

8 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

9 Kepler's Second Law A line connecting the Sun and a planet sweeps out equal areas in equal times. slowerfaster The speed of the planet in orbit is dependent on its distance from the sun v o r o = v f r f

10 Sample If the Earth has an orbital speed of 29.5 km/sec at apogee, determine the orbital speed at apogee. r a = 1.52 x 10 8 km r p = 1.47 x 10 8 km

11 Keplers Second Law v o r o = v f r f Note where the highest speeds of tornados and hurricanes are.

12 Keplers Third Law The square of a planets 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 Earths surface? What is the period of rotation?

15 Keplers 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


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