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2013-Dec-16, 03:24 AM
Hello everyone,

I am trying to calculate the sizes that various bodies would appear in the sky as viewed from a fictional planet that I have created, and would appreciate the assitance of someone who knows the necessary math.

The planet from which the sky is being viewed is the third and outermost in the system.
The bodies being viewed are:

Host Star - Diametre: 0.837 Sol, Distance: 0.633 AU

Planet a - Diametre: 12,410 km, Semi-major axis: 0.252 AU
Planet b - Diametre: 9830 km, Semi-major axis: 0.399 AU

In the case of the two inner planets, I would also like to determine how far from the host star they would appear throughout their orbital periods, and how this compares to the visual properties of Mercury and Venus as viewed from the Earth.

glappkaeft
2013-Dec-16, 03:38 PM
That is not enough information by far to calculate this (see http://en.wikipedia.org/wiki/Keplerian_orbit ).

You only give the Semimajor axis but Eccentricity, Inclination, Longitude of the ascending node, Argument of periapsis and the True anomaly are also needed. If we assume these are all zero (everything orbits in the ecliptic and perfectly circular) the distance between to the third planet is the only important part.

The apparent size (using the small angle approximation) is just diameter/current distance = apparent size in radians (1 radian = 180/PI degrees = 10800/PI arcminutes).

chornedsnorkack
2013-Dec-16, 04:00 PM
Angles can come from distance ratios.
b/c = 0.399/0.633=0.630
a/c=0.252/0.633=0.398
In Solar system
Venus/Earth=0.723
Mercury/Earth=0.387

2013-Dec-17, 02:46 AM
That is not enough information by far to calculate this (see http://en.wikipedia.org/wiki/Keplerian_orbit ).

You only give the Semimajor axis but Eccentricity, Inclination, Longitude of the ascending node, Argument of periapsis and the True anomaly are also needed.

For all of the planets in the system, eccentricity: <0.01 and Inclination: <0.1°

Are the other properties going to be important when the orbits are so straightforward?

Angles can come from distance ratios.
b/c = 0.399/0.633=0.630
a/c=0.252/0.633=0.398
In Solar system
Venus/Earth=0.723
Mercury/Earth=0.387

So, what I take from this is that the innermost planet will appear a little larger than Mercury, and the second will appear a bit smaller than Venus?

How can I determine whether the orbits of these two will take them far enough away from the star from the viewer's perspective that they won't be lost in the glare?

chornedsnorkack
2013-Dec-17, 08:53 AM
So, what I take from this is that the innermost planet will appear a little larger than Mercury, and the second will appear a bit smaller than Venus?

How can I determine whether the orbits of these two will take them far enough away from the star from the viewer's perspective that they won't be lost in the glare?

NO. You should have understood that my calculations addressed exactly how far away they get from the star - not the size.
b gets slightly not as far as Venus. a is slightly further than Mercury on average, but since Mercury is on an eccentric orbit and a is not, a does not get as far as Mercury at furthest.

glappkaeft
2013-Dec-18, 02:25 AM
For all of the planets in the system, eccentricity: <0.01 and Inclination: <0.1°

Are the other properties going to be important when the orbits are so straightforward?

If and only if (AKA IFF) Eccentricity = Inclination = 0 then the "Longitude of the ascending node" and "Argument of periapsis" have no effect (they determine the orientation of the eccentricity and inclination) but the "True anomaly" determines where in the orbit the planet is and therefore it's always important.

If one just wants the extremes the angular size/apparent diameter (the formula I provided) will be greatest/smallest when the inner planets are in front of/behind the star and the separation (the formula chornedsnorkrak used although I would instead use tan(a/c) and tan(b/c) as this will give you angle instead of a ratio) will be the greatest when the planets are orthogonal to each other.

Draw it and you'll see that all of these questions are really very simple geometry problems using simple triangles.

How can I determine whether the orbits of these two will take them far enough away from the star from the viewer's perspective that they won't be lost in the glare?

Like most things involving vision it is seriously non-trivial and requires much more data and work plus enough different variables and scientific disciplines that I'm not certain I can answer it.

chornedsnorkack
2013-Dec-18, 08:36 AM
If and only if (AKA IFF) Eccentricity = Inclination = 0 then the "Longitude of the ascending node" and "Argument of periapsis" have no effect (they determine the orientation of the eccentricity and inclination) but the "True anomaly" determines where in the orbit the planet is and therefore it's always important.

No, because it goes through full circle. It is only important if the planets are in resonance, and if both are in circular orbits in same plane then it is unimportant even in case of resonance.

If one just wants the extremes the angular size/apparent diameter (the formula I provided) will be greatest/smallest when the inner planets are in front of/behind the star and the separation (the formula chornedsnorkrak used although I would instead use tan(a/c) and tan(b/c) as this will give you angle instead of a ratio)
Yes, but you are taking arc functions of the ratios. It is unimportant to do the trigonometry if you already get your comparison from ratios.

Like most things involving vision it is seriously non-trivial and requires much more data and work plus enough different variables and scientific disciplines that I'm not certain I can answer it.
But you can simply compare with solar system. It is already clear that a is slightly further than mean Mercury, and much bigger. Do you want numbers?

George
2013-Dec-18, 10:42 PM
Host Star - Diametre: 0.837 Sol, Distance: 0.633 AU The Sun is about 32 arcminutes in apparent diameter as seen from 1 AU. Thus, this sun would appear 0.837 x 32 arc minutes in size if your planet was at 1 AU. Since it is at 0.633 AU, you must divide by this short distance to get your final apparent size (note chornedsnorkack's work), which yields about 42 arc minutes.

Planet a - Diametre: 12,410 km, Semi-major axis: 0.252 AU If you prefer, you can use basic trig...
arctan of 12,410/(.381*150000000). [0.633 - 0.252 = 0.381] which is for conjunction

glappkaeft
2013-Dec-19, 12:55 AM
No, because it goes through full circle. It is only important if the planets are in resonance, and if both are in circular orbits in same plane then it is unimportant even in case of resonance.

The relative positions of the planets are vital to many of the questions posed by the OP so I disagree.

Yes, but you are taking arc functions of the ratios. It is unimportant to do the trigonometry if you already get your comparison from ratios.

Sure both treatments are for most purposes equivalent, I only suggested that based on my experience as a substitute math teacher and tutor. IME most people have a better grasp of trigonometric functions than ratios.

Do you want numbers?

Yes, if I had the time available right now to put in the amount of work that required to get an accurate number to this question I'd like to know lots of more numbers. To start if like to know or calculate thing like the albedo and phase of planets a and b, the luminosity of the star, atmospheric composition/transparency/extinction of planet c, the air mass between the observer and the planet, etc. The numbers shown so far are not sufficient even for a basic BotE calculation.

chornedsnorkack
2013-Dec-19, 08:08 AM
The relative positions of the planets are vital to many of the questions posed by the OP so I disagree.

Repeat: Both inner planets go through full orbit.

Yes, if I had the time available right now to put in the amount of work that required to get an accurate number to this question I'd like to know lots of more numbers. To start if like to know or calculate thing like the albedo and phase of planets a and b,
Albedo is one thing I also miss.
My point is that the planets go through all phases.

Now IŽll show my calculations:
Assuming Earth and Venus were exactly circular orbits, 0,72 AU for Venus, the distances should be:
superior conjunction 1,72 AU (1+0,72) and full phase
inferior conjunction 0,28 AU (1-0,72) and new phase
Now maximum elongation?
The phase is half.
Understand that Venus is at the right angle of a right triangle. Distance Earth to Sun is hypotenuse, and Sun to Venus is one cathetus.
So the distance Earth-Venus is 0,69 AU.
Mercury is harder because it is so eccentric. On average 0,39 AU to Sun, the distance to Earth should be 0,92 AU.
Now going to the described system:
c to b at superior conjunction - 1,032 AU. (So bigger than Venus)
c to b at inferior conjunction - 0,234 AU
c to b at maximum elongation - 0,49 AU

c to a at superior conjunction - 0,885 AU
c to a at inferior conjunction - 0,381 AU
c to a at maximum elongation - 0,58 AU

Do you wish more details of angular size?

glappkaeft
2013-Dec-20, 08:57 PM
Repeat: Both inner planets go through full orbit.

Well obviously. Repeat: The true anomaly says where in the orbit the planet is, something you have implicitly relied on in your calculations. Sure now that the OP has specified near circular orbits the use Keplerian orbits is overkill and positions can be calculated assuming a position along a circle (i.e. True Anomoly). At this point I think that we are simply looking at this problem from different points and are in violent agreement. As is the case most of the time there is more than one way to calculate something.

Do you wish more details of angular size?

No, I have no interest in any of those numbers at all, if I did I would have calculated them long ago but IronRadiant might. My only interest in this thread was to try to point IronRadiant in the right direction.