31

Kepler's account of how the third law came to be is as follows (Caspar p.286; emphasis mine): On the 8th of March of this year 1618, if exact information about the time is desired, it appeared in my head. But I was unlucky when I inserted it into the calculation, and rejected it as false. Finally, on May 15, it came again and with a new onset conquered ...


26

Kepler's third law is trivial (in my opinion) compared to his first law. I am quite impressed that he was able to deduce that the orbits were ellipses. To get that, he had to go back and forth plotting Mars' direction from Earth and Earth's direction from Mars. He knew the length of both planets' years, so observations taken one Mars year apart would differ ...


8

Short Answer: Kepler expressed his laws with the sun at a foci rather than a barycenter. Long Answer: In Astronomia Nova (pub. 1609) Kepler presents the first version of something we can recognize as Kepler’s first law: On page 285 of Astronomia Nova: CAPVT LIX. Demonftratio, quod orbita MARTIS, librati in diametro epicycli, fiat perfecta ellipfis: Et ...


6

I know that quoting Wikipedia is frowned upon here, but as there has been no other answer posted, this is what the Wikipedia article on Kepler has to say about the matter: He then set about calculating the entire orbit of Mars, using the geometrical rate law and assuming an egg-shaped ovoid orbit. After approximately 40 failed attempts, in early 1605 he ...


5

It is easy enough to do the calculations, formulae for the in and circum radii of the Platonic solids can be found here which give ratios of circ to in radii of (note formulea for the radii have dropped common factor of the side length, which we don't need as we are interested in the ratios): >ri4=sqrt(6)/12,rc4=sqrt(6)/4 0.204124 0.612372 &...


5

If you plot the log of the period against the log of the semi-major axis then it is obvious that $P^2 \propto a^3$. Any other power law relationship simply wouldn't fit. The following passage (from https://www.mathpages.com/rr/s8-01/8-01.htm ) seems relevant: Is it just coincidental that John Napier's "Mirifici Logarithmorum Canonis Descripto" (...


4

The year length depends on the distance between the planet's centre & the Sun's centre, not the Sun's surface. So if the Sun merely expands, Earth at 1 AU will still take a standard year to perform 1 orbit. However, when the Sun becomes a red giant, it won't just expand. As Wikipedia mentions, red giants shed a considerable amount of mass in the form of ...


4

The AAS Nova article cites Rawls et al. 2016, who analyze light curves and spectra of the eclipsing binary KIC 9246715 to estimate its physical properties. Besides the stellar masses and radii in the article, their Table 2 lists orbital semimajor axis a = 211 R☉ = 147 million km = 0.98 au, and orbital eccentricity e = 0.356. During the 171-day period, ...


4

Wasn't it pretty obvious to a careful astrologer a thousand years ago, that the Moon does not have a circular orbit and does not describe epi-cycles? The ancient Greek model of the motion of planetary bodies remained unchallenged for almost two millennia, so obviously not. Hipparchus' model did a fairly good job dealing with the elliptical motion of the ...


4

That is not correct. The area is the total area between the two radius lines, so there is a curved side. Imagine you have two points almost 180 degrees from each other. Using just a triangle, the area is close to zero. Now, two points placed closely together can have the same area in between them. Then you have an equal are, but not equal time, so your ...


2

With copious research and thinking, I managed to answer my own question, but I am going to share the answer seeing that no one has answered it satsifactorily. Q1: the position vector $\vec{r}$ of the reduced mass is also the relative distance between the 2 masses, therefore if the reduced mass is undergoing elliptical motion about its centre of mass, it is ...


2

I must say that I agree with you, this way of derivate the Kepler law is not the most intuitive, it's maybe for this reason they specify: "revisited". R1: The reason is stated at the very begin of the chapter, where you derive the relation between $r$ and the angle to the perihelion $\theta$ for the general case of an ellipse (Eq 3 in your book in ...


2

The Earth moves in an elliptic orbit around the sun (or around the barycenter). If, in helocentric coordinates the Earth is at position (x,y), then in Geocentric coordinates the position of the sun is in position (-x,-y) So the locus of the Sun in Geocentric coordinates exactly matches the locus of the Earth in Heliocentric. The path of the sun in ...


2

I'll say yes here, but only if the epicycles are used as approximations. Here's how they would be useful: The eccentricity of the celestial body's orbit is low, i.e. $e<<0$. This means that it would be nearly circular, and epicycles could therefore be reasonably accurate. For example, it would not be a good idea to use epicycles to model Mercury's ...


2

Hohmannfan's answer is correct, but as I understand your question, you do understand the general idea correctly and the answer to your question is yes, no matter how eccentric the orbit, the planet spans equal areas over equal time, at least nearly perfectly. (more on that later) You're mistake is in calling the sectors "triangles". They're ...


2

Though the question may be closed as a duplicate, here's an answer that shows how to calculate periods using Kepler's $\color{blue}{\text{3rd}}$ law but with all the constants and units explained. Here's the equation from Wikipedia's Orbital_period; Small body orbiting a central body. The idea here is that the the size difference is so big that we can ...


1

On the risk of not fully answering your question: Trajectories of a system through some solution space, are in physics always solutions of some set of differential equations. However using the right formalism, one can often derive properties of the solution, without actually knowing it. For the Kepler-problem, there is such a trick. Let us consider the ...


1

None of these things are epicycles. Epicycles are perfect circles. These orbits are ellipses: it's a fundamental difference and so the answer is a very firm 'no'.


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