OPENING QUESTION: A red dwarf star is a tiny star that isn't much bigger than Earth. Nonetheless nuclear fusion (nucleosynthesis!) occurs there so it is very definitely a star.
Consider the case of such a star with a mass of 5.97 × 1029 kg.
A planet circles that star with a period of 11.86 years (the same as Jupite's orbital period here in our own Solar System), what is a necessary FIRST step in using the basic form of Kepler's Laws (See below) to determine that planet's average distance from that red dwarf star?
Now please find that planet's average distance to that red dwarf star.
LEARNING OBJECTIVES: I will be able to describe Kepler's 3 laws to an articulate high school freshman during today's class
WORDS O' THE DAY:
Period = T = Time for one orbit around the sun measured in years
A = Average distance to the sun measured in AUs
Eccentricity = a measure of the 'roundness' of an ellipse (or in this case an elliptical orbit). An eccentricity of 0.00 = a circle
FORMULAE OBJECTUS:
T2 ∝ A3: The square of the period (in years) of an object orbiting the sun is proportional to the cube of the average distance to the sun (in years). This is kind of archaic in that we rarely see the proportional symbol anymore.
T2 = a3: The square of the period (in years) of an object orbiting the sun is approximately equal to the cube of the average distance to the sun (in years). We MUST keep in mind this is an observational relationship. Although it gets us close (in most instances) it is not an exact value so the "=" sign isn't really appropriate although it is widely used.
MsT2 = a3: This version is still approximate but it allows us to substitute in the mass of *other* stars as long as we measure the mass of the other star in terms of the mass of the sun being 1.00
T2 = (4π2/GM)(a)3 = This version is much more accurate is often referred to as Newton's version of Kepler's Law
WORK O' THE DAY:
Kepler's three laws of planetary motion can be described as follows:
- The path of the planets about the sun is elliptical in shape, with the center of the sun being located at one focus. (The Law of Ellipses)
- An imaginary line drawn from the center of the sun to the center of the planet will sweep out equal areas in equal intervals of time. (The Law of Equal Areas)
- The ratio of the squares of the periods of any two planets is equal to the ratio of the cubes of their average distances from the sun. (The Law of Harmonies)
Mr W Addendums:
First Law: Regardless of the eccentricity of the orbit, the sun exists at one of the foci of each planet's orbit.
Second Law: Each arc of a planet's orbit sweeps out the same areas in the same amount of time.
Third Law: T2 = A3
Where T is the period of the planet around the sun and A is the average distance to the sun in Astronomical Units. 1 AU is equal to the average distance from the Earth to the sun and equal 93,000,000 miles which is about 150,000,000 km
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I 'built out' yesterday's lesson plan a bit more and uploaded that in case you need to refer to it later. I also included those bits in today's lesson plan.
With that in mind plealse review words of the the day and formulae as well
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