Astronomy in a Simple Solar System

[zbikraw]

Let's face it, the easiest way to determine that the Earth orbits the Sun is to appreciate that the stars will transition from left to right of the central/stationary Sun and parallel to the orbital plane. I personally feel that it is from the most unappreciated satellite out there.

The stars in the region of space known as the constellation Scorpius are now transitioning to the right of the central Sun due to the orbital motion of the Earth just as anyone who drives around a traffic circle will see background objects change position from one side of the roundabout to the other as the car travels/orbits the centre.

https://sol24.net/data/html/SOHO/C3/96H/VIDEO/

https://theskylive.com/planetarium

The old geocentric astronomers marked the transition of the stars from an evening to morning appearance as heliacal risings but, with a satellite out there free of daily rotational influences, that transition can now be described in heliocentric or Sun-centred perspectives as the Earth's orbital motion is responsible for that transition.

It is a matter of a more expansive view so try not to be caught between two stools.

 

[sophiecentaur]

Without an appreciation of Gravity, what’s to say the stars don’t go round us but just at a different rate from the Sun? When the Earth is clearly the largest and most important body then why challenge that model?
Partial ignorance is not too hard to mimic but total ignorance is pretty much impossible.

 

[saddlestone-man]

As the originator of the thread, I was happy to let it diverge and drift to some extent (don't most threads on most Forums?) as all the contributions were constructive and interesting.

I modified the question halfway through the thread's life to something like "if you now allow all the planets to exist, is there a simple experiment a child could conduct which would prove that the Sun is at the centre of the Solar System, and not the Earth". Bearing in mind the computing power, almost instantaneous worldwide voice and data communication, GPS receivers at our fingertips, reasonable quality cameras built into our phones, etc, we have today, I thought that a child has considerably more resource at his/her fingertips than any scientist in the past.

The conclusion so far seems to be that "no, there is no such experiment".

 

[zbikraw]

Historically, the European discovery of heliocentricity was a result of astrological activity. Before Newton both "sciences" evolved parallelly. In astrology there was some incosistency impossible to solve in geocentric model. Because of Christianity rule, open discussion of these problems was impossible (astrology was forbidden officially, but in reality practicized under rich feodals shield). The most strinking problem was "retrograde movement of planets", now called "pseudo-reversal of their course". We know that it results from changing wiewpoint because of simultaneous movements of Earth and other planets. In geocentric model Earth stands in place and retrogradation is impossible.
Of course, early astronomers registered positions of all visible planets (fast moving objects) and stars (slow moving objects). By digital simulations it was simple to prove that stars have one-day period of movement, and return to previous position. Cycles of planets was more complex, as they do not return to the same position next day. Correlation of digital coefficients describing movements of celestial objects fits quite well to heliocentric model and not geocentric-at all.
Of course this was discovered before Copernicus, but early European astronomers lived relatively nearby to Inquisition centers, and Copernicus relatively far, in northern Poland. He had also powerful protection of Polish kings from Jagiellonian dynasty. So, after publishing its theory, the Vaticane preferred to ignore it and prosecute followers of Copernicus in italy.
Galileo, Kepler and other Renaissance astronomers have had "perspectives", which makes impossible for honest observer to support geocentric theory. Relatively slow changing details on Moon and planets surfaces made possible identifications of their movements without correlations.

Without an appreciation of Gravity, what’s to say the stars don’t go round us but just at a different rate from the Sun? When the Earth is clearly the largest and most important body then why challenge that model?
Partial ignorance is not too hard to mimic but total ignorance is pretty much impossible.

These astronomers and astrologs were not scientists, rather the crafsmans of instruments and ideas. Astronomic observations were important for sailors, astrology was behind many political decisions. So, the consequences of error might be hard. Both disciplines calculate future positions of stars and planets: to accurately fins sailor's position on ocean and to find best day for many human activities. Details of calculation were hidden as the intellectual properties of "scientists", but some informations about such activity were written. In both disciplines gravity was unimportant, so mechanical dynamics was not needed. The results of calculations and observations were only positions of stars and planets. The science of astronomy starts from Newton, his predecessors were generally craftsmans, not scientists. Nota bene, Newton was astrologer, too.
Regards,
zbikraw

 

[Ibix]

The conclusion so far seems to be that "no, there is no such experiment".

Some things are made easier with modern technology, like you say. For example, an experiment to disprove the flat Earth is pretty much trivial with modern communication technology. It makes long distance coordination quite straightforward, and quite crude measurements with a wide geographic spread are good enough.

The problem with heliocentric versus geocentric models is that the differences are only detectable on longer timescales. Those measurements are possible for a sufficiently committed amateur to make, but aren't made much easier by modern technology. The only thing I think you could do is a cheap and cheerful simulation to show that you expect both prograde and retrograde motion given a heliocentric model (and ask why planets would reverse course in a geocentric model). Then use published tables of planetary positions to show that they have this behaviour. Then do a couple of observations to show that the tables accurately predict the positions of the planets on a few nights.

 

[DrStupid]

The problem with heliocentric versus geocentric models is that the differences are only detectable on longer timescales.

Some geocentric models even result in the same astronomical observations (e.g. a simple transformation of the heliocentric model into the rest frame or Earth). Additional information would be required in order to ruel them out (e.g. conservation of momentum). Maybe the best approach is to demonstrate how to predict the astronomical observations with a heliocentric model and than to ask for a pure geocentric model with at least the same accuracy. It would be a surprise if the latter turns out to be more convenient.

 

[sophiecentaur]

It would be a surprise if the latter turns out to be more convenient.

We tend to take for granted that the 'right' model would be the simplest model. It's in the DNA of Scientists to go for the most basic model. Should Convenience really be the touchstone? After all, every one of those 'Laws' ends up with codicils and those codicils tend to be more huge than the original Law.
One person's law of nature can be another person's quick fix for working out what to do.

 

[DrStupid]

Should Convenience really be the touchstone?

Physics shall be useful. Thus, convenience should matter. However, it depends on the praticular application. A model that is convenient for one application (e.g. to provide a better understanding for humans) might not be suitable for another (e.g. for computer simulations). Therefore it is not an objective criterion.

In regard to the current topic this means that from different models (e.g. barycentric, heliocentric or geocentric) resulting in the same predictions (within a given accuracy), the model that is less painful to use should be preferred. That would be the barycentric model for Newtonean dynamics or the heliocentric model for Keplers laws. I am not aware of an application that would cry for a geocentric model.

 

[Vanadium 50]

What is this thread about now? It seems to be all over the place.

 

[zbikraw]

Some things are made easier with modern technology, like you say. For example, an experiment to disprove the flat Earth is pretty much trivial with modern communication technology. It makes long distance coordination quite straightforward, and quite crude measurements with a wide geographic spread are good enough.

The problem with heliocentric versus geocentric models is that the differences are only detectable on longer timescales. Those measurements are possible for a sufficiently committed amateur to make, but aren't made much easier by modern technology. The only thing I think you could do is a cheap and cheerful simulation to show that you expect both prograde and retrograde motion given a heliocentric model (and ask why planets would reverse course in a geocentric model). Then use published tables of planetary positions to show that they have this behaviour. Then do a couple of observations to show that the tables accurately predict the positions of the planets on a few nights.

Remains of stronomical observatories in China, India and pre-Incan Andean countries contains stone tools (parts of the buildings) making possible (with recording materials like paper, pergamine, wood, etc.) recording of observation through years. In a medieval Europe mobile instruments and notebooks were much easier to hidden such observation from Inquisition. Both systems were capable of centurial observations. Unfortunately we don't know if these capabilities were exploited in reality.
zbikraw

 

[sophiecentaur]

Physics shall be useful.

Yes, I agree and that's a good pragmatic approach. Eventually, a simple law proves to be inadequate, though and the new law has to be less and less simple. In the example of the OP, I'm wondering what would make those Scientists look for anything other than a simple 'planetarium' model which doesn't consider actual distances or masses.

 

[saddlestone-man]

Let's say I allow the child (or maybe its parent) to ring or e-mail a scientist on the ISS. Would they be able to make any useful observations that the child on Earth couldn't make.

Maybe I'm hung up on the observations being made by a child. How about it's a group of year 10/11 students under the guidance of their science teacher?

PS: I'm not a science teacher looking for a project for my students.

 

[DrStupid]

Let's say I allow the child (or maybe its parent) to ring or e-mail a scientist on the ISS. Would they be able to make any useful observations that the child on Earth couldn't make.

Maybe I'm hung up on the observations being made by a child. How about it's a group of year 10/11 students under the guidance of their science teacher?

The only experiment that I could think of is making several pictures of the same regions of the sky over the year and than to measure the angle between neighboring stars. With sufficient accuracy this should reveal changes of the positions (in the images) of nearby stars relative to distant stars. Within the ecliptic the appearent motion should be linear and in the polar regions it should be circular. That should be an acceptable indication for the motion of Earth. This has already been mentioned above (parallax method).

Astronomical observations within the solar system could distinguish between different geocentric models (e.g. Sun and all other planets orbit Earth vs. Sun orbits Earth and all other plantes orbit Sun) but not between a heliocentric model and geocentric models in general (because some of them are equivalent to the heliocentric model in regard to the predicted astronomical observations).

 

[Ibix]

Some geocentric models even result in the same astronomical observations (e.g. a simple transformation of the heliocentric model into the rest frame or Earth).

A model that is only a choice of origin away from the modern standard Newtonian (or post-Newtonian) model is not a distinct model. I would describe that as a disguised heliocentric model; a simpler description is available in the Sun frame (or center of mass frame). A geocentric model would be one such as that sketched in #28, where the simplest description is available in the Earth-centered frame. Obviously, no experiment can differentiate a heliocentric model from a coordinate transformed version of itself.

Both systems were capable of centurial observations. Unfortunately we don't know if these capabilities were exploited in reality.

Indeed. And Tycho Brahe's observatory was a naked eye observatory. But, again, the OP was asking for an experiment that kids could do. Sure you can make sufficient measurements over the course of a few months to convince yourself, but I don't think there's any one-or-two night measurements they could make without borrowing data from some other source.

 

[sophiecentaur]

The only experiment that I could think of is making several pictures of the same regions of the sky over the year and than to measure the angle between neighboring stars. With sufficient accuracy this should reveal changes of the positions (in the images) of nearby stars relative to distant stars.

It's far from trivial to make parallax measurements and those measurements rely on long history of astronomical techniques. I am not sure that our fictitious situation would allow that. It seems that we are picking and choosing about what we are allowed to use out present knowledge to do. If we're in a position to do parallax measurements then we would also be in a position to measure the 'proper motion' of stars as they move about the galaxy.

Way before parallax measurements were possible, the distances to and between the bodies in the Solar System had been measured (amazingly, without the use of telescopes in the early cases). I think that we'd hard put to deduce those distances accurately without using a heliocentric model.

 

[sophiecentaur]

Maybe I'm hung up on the observations being made by a child.

I agree. Too hard and too extended for all but the exceptional child with lots of available time.

I think you may be overestimating what simple astronomical observations reveal. I was rather taken up with the process measuring astronomical distances and tried to communicate the ideas to two sets of adults. It was fun.

Bear in mind that all you can ever measure is angles and knowledge of one or more distances. A working knowledge of school geometry and assumptions about circular orbits can take you from the Moon's distance (using a baseline on Earth) to the Sun's distance and then to the distances to the planets. But just the angles tell a very limited story.

Hats off to the old astronomers with their basic instruments.

 

[PeterDonis]

We do that because we know the physics behind it.

Actually, the physics says the Moon orbits the Sun, not the Earth. More precisely, the Sun's gravitational force on the Moon is always stronger than the Earth's, so the Moon's orbit is always concave towards the Sun.

 

[sophiecentaur]

Is this correct?

Yes. That is what we 'see'. Any particular star changes apparent position but, after one year, it's got round to where it was, after 366 days (approx). Parallax amongst the stars is not visible from a baseline within the Earth and relative (proper) motion is very subtle, year on year.

 

[sophiecentaur]

so the Moon's orbit is always concave towards the Sun.

I wonder what the limiting case for this is - i.e. what's the lowest Earth orbit that's still concave wrt the Sun?

 

[Ibix]

I wonder what the limiting case for this is - i.e. what's the lowest Earth orbit that's still concave wrt the Sun?

Does concave just mean "no retrograde motion"? If so, the orbital radius whose speed is the same as the Earth's orbital speed, so that the satellite is instantaneously stationary with respect to the Sun.

If Earth's orbital speed is $v$ and its orbital radius is $R$ from the Sun of mass $M_S$, then $v^2=GM_S/R$. Similarly, if a satellite has orbital speed $u$ and orbital radius $r$ from an Earth of mass $M_E$, then $u^2=GM_E/r$. Equating $u$ and $v$ I get $r=R(M_E/M_S)$, or a bit less than 45km from the center of the Earth(!).

Edit: obviously that's nonsense because you can't treat the Earth as a point mass at the center once you are inside it. So there probably is no such orbit.

 

[PeterDonis]

Does concave just mean "no retrograde motion"?

No. It means the net acceleration is always towards the Sun, never away from the Sun.

 

[PeterDonis]

I wonder what the limiting case for this is - i.e. what's the lowest Earth orbit that's still concave wrt the Sun?

For an object directly on a line between the Earth and the Sun, the accelerations due to the Sun's and Earth's gravity are equal and opposite at about 257,000 km from the Earth (the Moon's orbit is at an average distance of about 400,000 km). So that altitude is the limiting altitude for an orbit to always be concave towards the Sun.

 

[sophiecentaur]

For an object directly on a line between the Earth and the Sun, the accelerations due to the Sun's and Earth's gravity are equal and opposite at about 257,000 km from the Earth (the Moon's orbit is at an average distance of about 400,000 km). So that altitude is the limiting altitude for an orbit to always be concave towards the Sun.

So not a lot lower than the Moon's orbit then? Below that, the orbital speed of the satellite around the Earth would be greater than its orbital speed round the Sun. (?)

 

[PeterDonis]

So not a lot lower than the Moon's orbit then?

Yes.

Below that, the orbital speed of the satellite around the Earth would be greater than its orbital speed round the Sun. (?)

No, that's not what I said. Please re-read what you quoted from my post, and note that "orbital speed" does not appear at all. I stated what the criterion was explicitly in that quote, and it has nothing to do with orbital speed.

 

[Ibix]

So not a lot lower than the Moon's orbit then? Below that, the orbital speed of the satellite around the Earth would be greater than its orbital speed round the Sun. (?)

No, that's the calculation I did. A bit of Googling turns up this paper, which models a moon moving in an Earth-centered circle superimposed on the Earth's Sun-centered circle. It shows that the instantaneous radius of curvature of the orbit can change sign for close-in orbits, but does not for the Moon's real orbit.

 

[DrStupid]

Actually, the physics says the Moon orbits the Sun, not the Earth. More precisely, the Sun's gravitational force on the Moon is always stronger than the Earth's, so the Moon's orbit is always concave towards the Sun.

The force doesn't matter. You need to look at the energy. Moon is gravitationally bound to Earth and not only to the Sun. It would still orbit Earth when the Sun would be removed.

 

[PeterDonis]

The force doesn't matter.

It does if you're looking at the shape of the orbit and whether or not it is always concave towards the Sun.

Moon is gravitationally bound to Earth and not only to the Sun.

In other words, it is bound to both. Yes, I never said it wasn't.

It would still orbit Earth when the Sun would be removed.

And it would still orbit the Sun if the Earth were removed. None of that contradicts what I said.

 

[DrStupid]

In other words, it is bound to both. Yes, I never said it wasn't.

You claimed that Moon does not orbit Earth.

 

[PeterDonis]

You claimed that Moon does not orbit Earth.

In the sense that its orbit is not always concave towards the Earth, but is always concave towards the Sun. I said nothing about the Moon not being "bound" to the Earth.

 

[DrStupid]

In the sense that its orbit is not always concave towards the Earth, but is always concave towards the Sun.

In the rest frame of Earth it is not always concave towards the Sun, but is always concave towards the Earth. And no, there is no reason to prefer the rest frame of Sun. The world lines of both, Earth and Sun are geodesics.

Long story short: You are wrong. The Moon orbits Earth - even according to your own definition.

 

[PeterDonis]

In the rest frame of Earth

Which is not inertial over the relevant time scale.

there is no reason to prefer the rest frame of Sun

Yes, there is: that it is inertial over the relevant time scale (a month), while the rest frame of the Earth is not.

More precisely, the inertial frame in which the barycenter of the solar system is at rest is the frame in which the orbit of the Moon has the property I stated, and it is inertial over the relevant time scale. Strictly speaking, the rest frame of the Sun is not, because of the Sun's movement relative to the barycenter, but it is still much closer to being inertial over a time scale of a month than the Earth's rest frame, because the Sun is so much closer to the barycenter than the Earth is (so the non-inertial terms are much, much smaller).

The Moon orbits Earth - even according to your own definition.

No, it doesn't, because my definition requires an inertial frame.

 

[PeterDonis]

The world lines of both, Earth and Sun are geodesics.

This implies a GR viewpoint, rather than the Newtonian viewpoint I have been taking up to now. From a GR viewpoint, I'm not sure what the relevant criterion would be and whether, or how, it would depend on our choice of frame. The first criterion that occurs to me is to look for (approximate) Killing vector fields in the spacetime geometry. The second is to compare the metric perturbations due to the Sun and Earth in the vicinity of the Moon.

 

[DrStupid]

Which is not inertial over the relevant time scale.

Earth is free falling around the Sun. Thus, the rest frame of Earth is a local inertial frame (you don't need GR to see that). There are tidal forces from the Sun but for the Moon they are orders of magnitude below the tidal forces from Earth and therefore negligible in good approximation.

No, it doesn't, because my definition requires an inertial frame.

As the rest frame of Earth is sufficiently inertial for the Moon you are still wrong.

 

[sophiecentaur]

In the sense that its orbit is not always concave towards the Earth

It is always that way from my viewpoint. It's not a good idea to try to describe what things 'really' are because it's all relative.

Imagine you are in a craft, orbiting the Moon. How would you describe things - assuming you knew nothing about modern astronomy? And what would the orbits of the ISS look like from there`?

 

[PeterDonis]

Earth is free falling around the Sun. Thus, the rest frame of Earth is a local inertial frame (you don't need GR to see that)

Wrong. A local inertial frame is local in time as well as in space in a curved spacetime. That is true even if the worldline of an object at rest in the frame is a geodesic. A local inertial frame centered on Earth now and a local inertial frame centered on Earth a month from now are not the same frame; they are moving in different directions. A local inertial frame centered on the barycenter of the solar system does not have that problem; the time scale for it to be "moving in a different direction" to a large enough degree to matter is millions of years.

You could set up Fermi normal coordinates centered on the Earth's worldline, and those would cover the Earth now and a month from now (or indeed as far into the future as you like); however, those would not extend far enough in space to cover the Sun, so it is impossible to describe an "orbit" for the Sun in such a frame.