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Does Kepler's 2nd mean that, for instance, both Pluto and Mercury sweep out an equal area over 1 hour? My gut reaction, without calculating anything, is "yes".

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- #1

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Does Kepler's 2nd mean that, for instance, both Pluto and Mercury sweep out an equal area over 1 hour? My gut reaction, without calculating anything, is "yes".

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marcus

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Actually not.

The law is about one planetary orbit. It doesn't equate sweep-rates of different planets.

You could imagine a highly elliptical comet orbit with perihelion equal to Mercury's and aphelion equal to Plutos. (nearest and farthest points in orbit)

At farthest, the comet would be sweeping slower than Pluto, because destined to fall in towards Sun.

At nearest it would be sweeping faster than Mercury because destined to swing out away from Sun.

The comet's two rates would be equal (and slower than Pluto's but faster than Mercury's).

So Pluto > Mercury.

Also think of the area sweep rate as expressing angular momentum of unit mass in the given orbit.

Pluto has more angular momentum (per unit mass)

The law is about one planetary orbit. It doesn't equate sweep-rates of different planets.

You could imagine a highly elliptical comet orbit with perihelion equal to Mercury's and aphelion equal to Plutos. (nearest and farthest points in orbit)

At farthest, the comet would be sweeping slower than Pluto, because destined to fall in towards Sun.

At nearest it would be sweeping faster than Mercury because destined to swing out away from Sun.

The comet's two rates would be equal (and slower than Pluto's but faster than Mercury's).

So Pluto > Mercury.

Also think of the area sweep rate as expressing angular momentum of unit mass in the given orbit.

Pluto has more angular momentum (per unit mass)

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for circular orbits, this is found by:

[tex]A = \sqrt{GMa}[/tex]

where a is the semi-major radius of the orbit (mean orbital distance)

for an elliptical orbit, it is found by:

[tex] \sqrt{GMa \frac{1+e}{1-e}}[/tex]

Where e is the eccentricity of the orbit.

Note that as the mean orbital distance goes up, so does the Areal velocity (However, since it also goes up with eccentricity, it would be possible for an object with a high eccentricity to have a greater Areal velocity than a object with a higher mean orbital distance but a lower eccentricity.)

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Yup, makes perfect sense.At farthest, the comet would be sweeping slower than Pluto, because destined to fall in towards Sun.

At nearest it would be sweeping faster than Mercury because destined to swing out away from Sun.

The comet's two rates would be equal (and slower than Pluto's but faster than Mercury's).

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Indeed.A little learning is a dangerous thing;

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There shallow draughts intoxicate the brain,

And drinking largely sobers us again. -- Alexander Pope

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