kepler’s laws of planetary motion © david hoult 2009
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Kepler’s Laws of Planetary Motion
© David Hoult 2009
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© David Hoult 2009
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© David Hoult 2009
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© David Hoult 2009
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© David Hoult 2009
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The eccentricity of an ellipse gives an indication of the difference between its major and minor axes
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The eccentricity depends on the distance between the two points, f (compared with the length of the piece of string)
The eccentricity of an ellipse gives an indication of the difference between its major and minor axes
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eccentricity = distance between foci / major axis
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The eccentricity of the orbits of the planets is low; their orbits are very nearly circular orbits.
eccentricity = distance between foci / major axis
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Law 1
Each planet orbits the sun in an elliptical path with the sun at one focus of the ellipse.
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Mercury 0.206
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Mercury 0.206 Venus 0.0068
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Mercury 0.206 Venus 0.0068 Earth 0.0167
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Mercury 0.206 Venus 0.0068 Earth 0.0167 Mars 0.0934
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Mercury 0.206 Venus 0.0068 Earth 0.0167 Mars 0.0934 Jupiter 0.0485
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Mercury 0.206 Venus 0.0068 Earth 0.0167 Mars 0.0934 Jupiter 0.0485 Saturn 0.0556
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Mercury 0.206 Venus 0.0068 Earth 0.0167 Mars 0.0934 Jupiter 0.0485 Saturn 0.0556 Uranus 0.0472
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Mercury 0.206 Venus 0.0068 Earth 0.0167 Mars 0.0934 Jupiter 0.0485 Saturn 0.0556 Uranus 0.0472 Neptune 0.0086
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Mercury 0.206 Venus 0.0068 Earth 0.0167 Mars 0.0934 Jupiter 0.0485 Saturn 0.0556 Uranus 0.0472 Neptune 0.0086 Pluto 0.25
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...it can be shown that...
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minor axismajor axis
= 1 - e2
where e is the eccentricity of the ellipse
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minor axismajor axis
= 1 - e2
where e is the eccentricity of the ellipse
which means that even for the planet (?) with the most eccentric orbit, the ratio of minor to major axis is only about:
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minor axismajor axis
= 1 - e2
where e is the eccentricity of the ellipse
which means that even for the planet (?) with the most eccentric orbit, the ratio of minor to major axes is only about:
0.97
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In calculations we will consider the orbits to be circular
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Eccentricity of ellipse much exaggerated© David Hoult 2009
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Law 2
A line from the sun to a planet sweeps out equal areas in equal times.
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Law 3
The square of the time period of a planet’s orbit is directly proportional to the cube of its mean distance from the sun.
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T2
r3= a constant
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F = Gr2
Mm
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F = m r 2
F = Gr2
Mm
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F = Gr2
Mm F = m r 2
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F = Gr2
Mm F = m r 2
r2G M m m r 2
=
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F = Gr2
Mm F = m r 2
= T
2
r2G M m m r 2
=
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T2
r3= 42
GM
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T2
r3= 42
GM
in which we see Kepler’s third law
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