Sample Determine the net gravitational force acting on the Earth during a total lunar eclipse. m sun...
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![Page 1: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/1.jpg)
![Page 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 10 22 kg r sun =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/2.jpg)
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
![Page 3: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/3.jpg)
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.
![Page 4: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/4.jpg)
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
![Page 5: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/5.jpg)
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
![Page 6: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/6.jpg)
Kepler’s First LawExamples:
Earth: e = 0.0167Mercury: e =
0.2056Venus: e =
0.00657
![Page 7: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/7.jpg)
Kepler’s First Law
a = semi-major axis
![Page 8: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/8.jpg)
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
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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
![Page 10: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/10.jpg)
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
![Page 11: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/11.jpg)
Kepler’s Second LawKepler’s Second Law
voro = vf rf
Note where the highest speeds of tornados
and hurricanes are.
![Page 12: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/12.jpg)
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
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Why Does it Work?Why Does it Work?
Newton discovers that ellipses are pretty close to circular, just with the sun offset
Fc = Fg
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SampleSampleHow fast must a satellite move to maintain an orbit of 500 km over the Earth’s surface?
What is the period of rotation?
![Page 15: 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 =](https://reader035.fdocuments.in/reader035/viewer/2022062417/5514dcdb55034693478b557a/html5/thumbnails/15.jpg)
Kepler’s Third LawKepler’s Third Law
Newton took the idea of centripetal force and applied it to the Kepler problem
Fc = Fg
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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