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Determine the relative distances of each of the planets from the Sun given [ ].

  1. Jan 4, 2013 #1

    s3a

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    "Determine the relative distances of each of the planets from the Sun given [...]."

    1. The problem statement, all variables and given/known data
    Devise methods to determine the relative distances of each of the planets from the Sun given the information available to Copernicus (observable angles between the planets and the Sun, orbital configurations, and synodic periods).

    The solution is attached as TheSolution.jpg.

    2. Relevant equations
    Trigonometry.

    3. The attempt at a solution
    1) Is the case with the inferior planet also dealt with using sidereal periods?

    2) How is the person answering the question supposed to know to consider the specific situation where the inferior planet is at greatest elongation and the specific situation where the superior planet goes from opposition to quadrature especially since, if different situations are chosen, the relative distances differ? (I'm asking this in case I am missing some important insight.)

    Any input would be greatly appreciated!
     

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  2. jcsd
  3. Jan 5, 2013 #2

    SammyS

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    Re: "Determine the relative distances of each of the planets from the Sun given [...]

    No sidereal information is needed. All that is needed is observing what is the maximum angle between the line of sight to the Sun and the line of sight to the inferior planet.

    At greatest elongation, Earth would be at position P2 in figure S1.1 and the inferior planet would be at position E2, assuming that angle P2E2S is 90°. (The labels in the figure were chosen for the case of observing a superior planet.)

    attachment.php?attachmentid=54464&d=1357357656.jpg
    The superior planet is at quadrature when the line of sight to the planet makes an angle of 90° with respect to the line of sight to the Sun.
     
  4. Jan 6, 2013 #3

    s3a

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    Re: "Determine the relative distances of each of the planets from the Sun given [...]

    So, the availability of the synodic/sidereal periods are computationaly useless in solving this problem, right? In other words, they serve only for qualitatively stating that we can obtain ∠P_1 S P_2, right?

    To be more specific, you meant the following, right?:
    The superior planet is at quadrature when the line of sight to the superior planet from the inferior planet makes an angle of 90° with respect to the line of sight from the inferior planet to the Sun.

    I'm assuming you're saying the following.:
    For the statement quoted above, what about the question would indicate that we are looking for the maximum angle between the line of sight from the inferior planet to the superior planet with respect to the line of sight from the inferior planet to the Sun? Did you just assume that/figure that out based on the given solution to the problem or what?

    Also, to confirm, in the attached Position_astronomy.jpg image, the circle labeled with “Greatest western elongation” is at quadrature, right? Furthermore, why is it called the greatest WESTERN elongation if it's on the right of the image? Is it because the superior planet is to the left (=west) of the inferior planet?

    Another thing I'd like to know is if points E_1 and E_2 in Figure S1.1 are at greatest elongation with the superior planet as a reference point. Are they?

    Yet another thing I'd like to know is, in real life, how is one supposed to know that P_2 is at quadrature? Put differently, how is the astronomer (or whatever person) supposed to know when to stop calculating how much time it took for P_1 to go from opposition to quadrature?
     

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  5. Jan 6, 2013 #4

    SammyS

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    Re: "Determine the relative distances of each of the planets from the Sun given [...]

    Well, it's going to take some effort to respond to you last post ...

    When I stated the following:
    I meant only what I said, not what you assumed that I meant.

    You see, the observations are made from earth. The observations of an inferior planet are totally independent (not simultaneous with) the observations of a superior planet.

    In the following quote (also from post #2) I merely pointed out how to use figure S1.1 to understand the relative positions of the earth, the sun, an an inferior planet for the case of greatest elongation.
    From your original post, I gathered that ' observable angles between the planets and the Sun, orbital configurations, and synodic periods ' were assumed to be available, as they would have been available to Copernicus.

    For the case of an inferior planet, such as Venus, you could consult a table with a daily listing of the angle between the line of sight to the Sun and the line of sight to Venus. When this angle is at a maximum, on that day, Venus is at greatest elongation. This does require that you understand what is mean by "greatest elongation", or for that matter, what is mean by "elongation", of an inferior planet.

    From the standpoint of more casual observation, the greatest elongation of Venus occurs either when the time between Venus rising and sunrise is greatest, or when the time between Venus setting and sunset is greatest.

    All the above is in regards to finding the distance of an inferior planet from the Sun, relative to earth's distance from the sun. All that is needed for this calculation is the angle observed between Venus and the Sun at the time of greatest elongation.

    The following quote from Post #2 is in regards to the case of a superior planet. It states the condition for quadrature of a superior planet from the point of view of an observer on Earth.


    All the rest of the quoted material is from you in Post #3. (Some deletions ... not all of Post #3 is quoted here.)
    The availability of sidereal periods is only important for the case of superior planets, in which case it is crucial. In this case it's not only important in obtaining ∠P1 S P2, it's also important in finding ∠E1 S E2. Both of these are needed for finding ∠P2 S E2.

    I do not see this figure.

    Hopefully, after reading the above, you will see that this question makes no sense.

    This question has also been answered above, but to repeat:
    The superior planet is at quadrature when the line of sight to the planet makes an angle of 90° with respect to the line of sight to the Sun.
    This information should have been available to Copernicus. It occurs when the planet rises or sets at Solar midnight (midway between sunset and sunrise).


    Of course, all of this assumes circular orbits, and in the case of superior planets, assumes that the orbits are in the same plane.
     
    Last edited: Jan 6, 2013
  6. Jan 7, 2013 #5

    s3a

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    Re: "Determine the relative distances of each of the planets from the Sun given [...]

    Sorry about that.

    I understand that the observations from different celestial bodies differ but, I'd like to make the statement as explicit as possible since, I am confused by these concepts.

    So, would it be correct to say the following?:
    As for your quote from post #2, is it still correct with this modification?:
    I currently understand what is going on with this. :)

    I converted what I found on Wikipedia ( http://upload.wikimedia.org/wikipedia/commons/f/f6/Positional_astronomy.svg ) from an svg to jpg and then attached it here. I don't know why you can't see it but, hopefully, the Wikipedia link should work out.

    I think I just phrased it badly (but, I am not sure).

    What I meant, here, is that the superior planet is Earth and points E_1 and E_2 are dealth with one at a time (in two different cases). Tell me if what I just said is also confusing.

    So, specifically, one can deduce the existence of the 90 degree angle by using a telescope (from Earth) to view the planet rising/setting and, if that happens while the Sun is setting/rising (as viewed from Earth), then, that means that the superior planet is at quadrature?
     
  7. Jan 7, 2013 #6

    SammyS

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    Re: "Determine the relative distances of each of the planets from the Sun given [...]

    Yes. That's correct.
    Yes. That's also correct.
    I don't know why
    I couldn't/didh't see it either. I can see it now.

    attachment.php?attachmentid=54491&d=1357490555.jpg
    The 90° angle is correct, but this occurs when the superior planet rises or sets at midnight (solar midnight) or at solar high noon. It's quite difficult to observe the planet rising or setting with the Sun high in the sky.

    If the planet rises at sunset or sets at sunrise, then the planet is at opposition.



    As for that other question ...

    You wrote in post #3:
    which is then referred to next.
    I'm still not sure what you are referring to here, but here's another try at using figure S1.1 to show greatest elongation - which is only interesting for an inferior planet.
    First of all, positions P1 and E1 are irrelevant regarding greatest elongation.

    If we take the outer circle to represent Earth's orbit and the inner circle to represent the orbit of the inferior planet, then the triangle P2 E2 S corresponds to greatest elongation provided that the triangle is a right triangle with the inferior planet sitting at the vertex of the right angle.

    If you were to observe this situation from the inferior planet, then you might say that Earth was at quadrature.​
     
  8. May 11, 2013 #7

    s3a

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    Sorry for the very late reply. I was overwhelmed with school (and this is not part of my school curriculum).

    1) Initially, I was thinking that these equations were to describe the relative distances between the planets at any point in time but, I'm now thinking that these relative distances are merely the relative distances at a "snapshot" in time/fixed time. Am I correct now or was I correct before?

    2) Is the inferior planet being at elongation the same thing as the superior planet being at quadrature (except that they're each from a different perspective)?

    3) So, let's say I wanted to draw a diagram/figure for the part of the problem where we aim to get an equation with relative distances of the planets for an inferior planet at greatest elongation (=the first part), do I just erase/delete/remove ##P_1##, ##E_1##, place ##E_2## where ##P_2## is and place ##P_2## where ##E_2## is in Figure S1.1 and that's it?

    4) Assuming I am correct now about what is stated in my point #1, why was the decision made to "freeze time" at a particular instant with this geometric configuration?

    If something I said does not make perfect sense, tell me and I will attempt to improve my phrasing.
     
  9. May 11, 2013 #8

    SammyS

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    Yes, it's been a long time.

    Whoa! Which equations? We must be more specific.

    This discussion began with the rather confusing situation of using one figure (image) to describe two quite different procedures for calculating the orbital radii of the planets. One procedure for determining the radius of an inferior planet. A different procedure for a superior planet.

    For an inferior planet: At the moment that the planet is at greatest elongation, the equation ##\displaystyle \ \cos(\angle PES)=\frac{EP}{ES} \ ## does give the distance of the planet from the Sun, relative to Earth's distance from the Sun. That is valid for that moment in time.

    It is in the following sense. At the moment that the superior planet is at quadrature, an observer on that planet would consider Earth to be at maximum elongation.

    That will work. The equation then becomes ##\displaystyle \ \cos(\angle P_2E_2S)=\frac{E_2P_2}{E_2S} \ ##

    As far as being correct now or correct before, --- or a decision being made to "freeze time", you will need to be way more specific regarding which statements are being given as a choice.
     
    Last edited: May 11, 2013
  10. May 14, 2013 #9

    s3a

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    The first equation is the one you listed [cos(∠PES) = ##EP / ES##]. The second equation is cos(∠##P_2####S####E_2##) = ##E_2####S##/##P_2####S## (or the reciprocal of both sides of the equal sign - as given in the solution - which is the attachment in the first post).

    Thanks. :)

    Thanks. :)

    I think we're miscommunicating. What I was asking is why did the author choose to get these equations from when the superior planet is at quadratureinferior planet is at greatest elongation instead of any other point in time? The equations would likely differ from those currently given in the solutions manual but, if I'm correct, they would still represent relative distances as the question asked for such that the answer would still be correct ... basically, how is the person doing this problem supposed to know to consider the particular time that the superior planet is at quadratureinferior planet is at greatest elongation instead of any other "snapshot" of time? Is there any constraint in the wording of the problem which makes this the only possible solution?
     
  11. May 14, 2013 #10

    SammyS

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    Yes, I agree.

    That's why I tried to be very specific about which situation I was referring in my posts, as well as which situation each of your questions referred to.

    Again, I will give an answer as unambiguously as I can, and hope it addresses your question.

    As we have established the geometry for an inferior planet at maximum elongation is the same as for a superior planet at quadrature, provided that you switch the labels properly. The helpful aspect of this configuration is that the Sun, the planet, and Earth are each at a vertex of a right triangle.

    In the case of an inferior planet: One of the acute angles is formed at the Earth. This angle can be observed fairly directly. Basic trigonometry then yields the relative distances of interest.

    You could observe the inferior planet for other configurations, but you would need to combine information from more than one observation/configuration as well as information regarding sidereal periods.



    The case of an superior planet is different: In this case Earth sits at the vertex of the right angle of the right triangle. To measure one of the acute angles directly, would require measuring the acute angle from either the location of the planet, or the location of the Sun. To determine the acute angle at the vertex occupied by by the planet, at the time of quadrature, you need to use the sidereal periods of Earth and the planet and use the time between opposition and quadrature. And, perhaps to finally answer your question: Using the acute angle thus determined, does give the relative distances at the moment of quadrature (as in a "snapshot") provided that the planet and Earth have circular (or nearly circular) orbits.​
     
  12. May 18, 2013 #11

    s3a

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    Edit: I double posted. This forum should implement something where responses less than five seconds apart are not posted or something of the sort.
     
  13. May 18, 2013 #12

    s3a

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    Sorry for still not getting it but, I'm still not understanding why one cannot answer this differently.

    Can't someone just observe the celestial bodies in a different geometric configuration such as one where the bodies are located at vertices of a hypothetical non-right-angled triangle and find relative distances using the law of sines, for example?
     
  14. May 18, 2013 #13

    SammyS

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    Yes, that can happen.

    It is possible to delete your own post, provided you do it promptly enough so that you can still edit it.

    (I think that time is something like 700 minutes.)
     
  15. May 18, 2013 #14

    SammyS

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    Looking back at the OP, the available observations are rather limited. We're talking about data available to Copernicus .


    If you're asking, if it's possible to determine relative distances for situations other than those corresponding to right triangles, I suppose it is. The problem is: how do you determine those angles using observations made purely from Earth?

    For instance: Suppose we send an observer to Mars, and have that observer determine the angle between the line of sight to Earth and the line of sight to the Sun and simultaneously have an observer on Earth determine the angle between the line of sight to Mars and the line of sight to the Sun . Then we would not need to use a right triangle configuration. However, I suppose if we were able to send an observer to Mars, we would know with great precision, the actual distances involved.
     
  16. May 23, 2013 #15

    s3a

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    Take a look at the PotentialBetterWay.jpg attachment. (The logic in the PotentialBetterWay.jpg assumes that there is only an observer on Earth - wherever Earth is located.) Could that have been done by Copernicus in addition to what the solution in the book says? If so, is it a better solution than what the book shows, in your opinion?

    [Also, if the PotentialBetterWay.jpg is correct and could have been done by Copernicus, that would imply that there was more than one way that Copernicus could have solved this problem despite the solution of the book (understandably) only giving one method of solving the problem.]

    If there is something wrong with my logic in the PotentialBetterWay.jpg attachment, please point out what it is.
     

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  17. May 25, 2013 #16

    SammyS

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    The observations required for for your PotentialBetterWay.jpg certainly could have been made by someone at the time of Copernicus. The method should work as well as the methods used in the solutions given in your book.

    Looking back at the original question: The main information available to Copernicus would likely not include the observations needed for your PotentialBetterWay.jpg . The information available would likely be: synodic periods, times at which superior planets were at opposition, conjunction, and either quadrature, times at which inferior planets were at either conjunction or at greatest elongation and the measure of the greatest elongation.

    Any sidereal periods would have to be calculated by Copernicus, using something like the Eq. 1.1 mentioned in your first figure.


    In reviewing the posts in this thread, I did notice an inconsistency. The problem asks you to find "the relative distances of each of the planets from the Sun". The solution they give for an inferior planet gives ## \displaystyle \ \cos(\angle PES)=\frac{EP}{ES} \,, \ ## which is the distance the planet is from Earth relative to Earth's distance from the Sun. To get the relative distance of an inferior planet from the Sun, one would need to combine this result with the Pythagorean Theorem, or use the sine in place of the cosine : ## \displaystyle \ \sin(\angle PES)=\frac{PS}{ES} \ . ##
     
    Last edited: May 25, 2013
  18. May 26, 2013 #17

    s3a

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    You told me what was available to Copernicus but, what about my PotentialBetterWay.jpg file was NOT available to Copernicus, specifically?

    Are you saying sidereal periods were not available to him? If so, that would imply that we are defining "available to Copernicus" differently. To me, "available to Copernicus" means information that he is given directly and indirectly where indirectly-given information is information that he derives through calculations, etc.

    So, if you are indeed saying that sidereal periods were not available to him, wouldn't that then mean, by my logic, that Copernicus could have done what my PotentialBetterWay.jpg file shows?

    Thanks for pointing that out to me. :)
     
  19. May 27, 2013 #18

    SammyS

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    It seems to me that to derive a formula for converting synodic periods to sidereal periods would require the use of a heliocentric model. Therefore, I doubt that sidereal periods for the planets were available to Copernicus. He would have had to develop such a formula based on a heliocentric model of the Solar System.

    Based on the instructions for your exercise,
    It seems to me that the methods described in your "PotentialBetterWay.jpg" file could be used.
     
  20. May 29, 2013 #19

    s3a

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    I'm not sure if I misunderstood what was said but, my interpretation of your answers seems contradictory.

    To "freshen the conversation", we know that (1) the solution in the book uses sidereal periods (which would not make sense without acknowledging the existence of the heliocentric model), (2) Copernicus was the one who created the heliocentric model ( http://en.wikipedia.org/wiki/Copernican_heliocentrism ), (3) my PotentialBetterWay.jpg requires, at least, the heliocentric model in order to use sidereal periods so, finally, like you had said at the end of your previous post, he actually could have done it the way shown in my PotentialBetterWay.jpg, right?

    I think I have come to the realization that "my way" is, in fact, not a better way (but, it is still a correct way) of solving this problem if we assume the orbits to be fully/perfectly circular because, to think of this problem from a slightly different perspective, we are essentially being asked to find a relationship between the radii of the orbits of the two planets around the Sun (assuming that the book's solution made a mistake when it said part of the answer was cos(∠PES) = EP/ES instead of sin(∠PES) = PS/ES, assuming that Figure S1.1 had no E_1 and P_1 and that it had E where P_2 currently lies and P where E_2 currently lies - also, based on the way the question was phrased and other "traditional academic" stuff I found online, I'm pretty sure this is the case). If we were thinking about ellipses, the law-of-sines (or law-of-cosines) way would actually be better, assuming the angles could be accurately determined (which they can) since it would give the "instantaneous radius" but, given the largely-but-not-fully realistic, close-to-circular elliptical motion, we can assume theoretical perfectly-circular motion in which case, the radii remain the same at all times so it is far better to just use a right-angled triangle in the quadrature/greatest elongation geometric configuration. That and, Copernicus didn't know that the orbits were elliptical at all (and the question doesn't even want us to consider Kepler's accomplishments). It was Kepler that discovered that the orbits were elliptical at some amount of time after Copernicus died, right? Another noteworthy statement is that, assuming perfectly circular orbits, it would always be at least (or, usually, be exactly) two of the three sides of the triangle that would be equal to each other, when comparing "my way" with the "right-angled-triangle way".

    Sorry for the long post and for being repetitive but, given that I felt certain statements were contradicting each other, I wanted to clarify the situation. I think I now have a very good handle on this question but, I just need a final confirmation (assuming I do, in fact, understand it correctly).

    Is everything I said in this latest post correct?
     
  21. Jun 24, 2013 #20

    s3a

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    Sorry for bothering you again but, have you forgotten about this thread?

    I feel that I understand this problem because, I feel like I have no unanswered questions/confusions left about it but, are the explanations that I gave in my previous post correct explanations (as far as you can tell)?
     
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