# Background

In my story, HD 28185 star system harbors life. One of them is an earthlike planet called Aucafidus, located at L4 of HD 28185 d (Subralis). In-universe, Subralis is white-blue in colors with white-ish clouds due to water vapors on its atmosphere, its diameters is approximately 420,000 kilometers with around six times the mass of jupiter.

# Question

Assuming all of these (for Subralis's albedo and exact coloration of its atmosphere is yet to be determined, and might be posted as separate question), what would this configuration looks like as seen from a terrestrial trojan planet spaced around 1 AU from it?

• I'd assume it would look the same as orbiting a star at 1AU normally does except that there would be a very bright morning or evening star (depending on the direction of rotation) with a noticeable diameter. The gas giant would be a disk not a bright point. You can calculate the diameter from the diameter and distance. Large moons of the gas giant might also be visible. – Ville Niemi Jun 27 '15 at 12:47
• I'm dubious about a visible disk. Apparent size is about .16 degrees, or the equivalent of a 1/32" diameter ball at 10 feet. With any sort of optics, the permanent phase (slightly more than a half-moon) will be visible. – WhatRoughBeast Jun 27 '15 at 14:11
• Before looking at such details, you should see if such a situation even makes sense physically. A massive planet will clear out the area near its orbit. Planets will space themselves out to a stable configuration. A planet 1 AU from this giant would be ejected or captured, in less than a million years. – JDługosz Jun 27 '15 at 14:16
• The planet is on jovian's trojan point, smaller than earth but denser. Perhaps I'll post another question about in near future. – Hendrik Lie Jun 27 '15 at 17:35
• @WhatRoughBeast. Our moon subtends roughly 0.5 degrees. So 0.16 degrees would be a very clearly visible disk, around 1/3 the diameter of our moon. On a quick check, I agree with 0.16 degrees. – Neil Slater Jun 28 '15 at 18:26

One AU is roughy 150M kilometers. A quick check of the angle in radians subtended by a planet of diameter 420,000 kilometers is given simply by diameter/distance (very good approximations for small angles) = 0.0028 radians.

Astronomers usually give the angle in arc-minutes/arc-seconds. There are 60 arc-minutes in each full degree (with 360 degrees in a full circle). One radian is approximately 3,437 arc-minutes. So the gas giant would appear roughly 9.6 arc-minutes across in the sky.

A general formula for this is $θ^{arcmins} = 6875 * r / d$ where $r$ is radius of planet, and $d$ is the distance away in the same units. Provided you get a value below 600 or so then it should be reasonably accurate. The multiplier 6875 is $2 * 60 * 180 / \pi$

You would think this is quite small, but compare it to our moon (and our sun), which subtends 31 arc-minutes, and is clearly visible as a disk with different structures on the surface, then you can get a good sense of what it might look like.

So it will be roughly 1/3 diameter of Earth's moon. To the naked eye, large-scale structures (such as wide banding, or exceptionally large clouds/storms like Jupiter's red spot) should be visible. At night, there may be enough reflected light that the colour could be perceived (although that would also be influenced by any atmosphere it had to travel through).

As you have suggested that the Earth-like planet is in a trojan orbit, that should also set some other characteristics:

• The gas giant planet would always appear either 60 degrees ahead of or behind the sun in daily motion (depending on orbital configuration and direction of spin of the Earth-like). On a 24-hour clock that equates to rising (and setting) 4 hours ahead (or behind) the sun. The sun will be much brighter, so appearance during daylight hours is washed out, but even then you could reasonably expect to see the disk in the sky.

• The disk will not be a perfect circle, but instead will always resemble a gibbous moon. Unlike our moon, it would always have the same phase.

• Earth's Moon has an albedo in visible wavelengths of 7%, whilst Jupiter's is roughly 5 times more at 34%. Assuming something similar for your scenario, and accounting for the smaller size (which we need to square, and also allow for the gibbous shape), then the disk centre will seem brighter than the moon, but shed total light roughly 30% that of the full moon due to being smaller. That's still pretty bright, someone with good night vision should notice shadows from it.

• As a more flexible formula: Total brightness of this imagined planet in "Full Moons" $B = 0.66 * (Albedo/0.07) * (Arcmins/31)^2$ where Albedo and Arcmins are the angle of disk in the sky of the planet and visual reflectivity. The values 0.07 and 31 are the same values from our moon, and factor of 0.66 accounts roughly for the incomplete disk.
• Unlike our moon, there is no chance of any tidal locking at the given distance, so any visible features will be seen to rotate. Jupiter rotates at once every 10 hours, so it is possible for effects of the rotation to be very obvious even in a single night.

• If the gas giant has any large moons, they could be visible (as star-like dots, perhaps similar brightness to e.g. Mars for us), and it would be relatively easy to tell with only a few observations, that they are in motion around the gas giant, unlike any other stars or planets that could be observed which would have more complex motions. The orbital period of e.g. Io around Jupiter is 42 hours, so even during a single night you would notice significant shifts of position. I am not sure if it would be possible to see the shadow of a moon on the gas giant's clouds - maybe not with the naked eye, but it might be possible with basic equipment e.g. binoculars.

• Whoaa, that's pretty much a neat answer. – Hendrik Lie Jun 29 '15 at 12:16
• I'm sorry to bother you, but I would really be helped if you were to provide me with some formula (of results posted in your answer), in case the size of the planet were changed. It would be very nice of you if you were willing to complete your answer :) – Hendrik Lie Aug 29 '15 at 10:18
• @HendrikLie. That is in the first paragraph. Provided the planet is not so close or large that it would take up a big chunk of sky, then the formula is $\theta = 2 * r / d$ where $r$ is planet radius, $d$ is distance in same units, and $\theta$ is the angle in radians that this subtends. To get arc-minutes, for comparison with astronomy text-books etc, multiply $\theta * 60 * 180 / \pi$ – Neil Slater Aug 29 '15 at 10:25
• hey thanks. How about characteristics number three (albedo...)? How to calculate how bright is the planet? I'm sorry if I were to interrupt you again, but it is supposed to be my last question on your answer. – Hendrik Lie Aug 29 '15 at 12:55
• Total brightness of imagined planet in "Full Moons" $B = 0.66 * (Albedo/0.07) * (Arcmins}/31)^2$ where $Albedo$ and $Arcmins$ are the reflectivity and angle of disk in the sky of the planet, and the values 0.07 and 31 are the same values from our moon. The factor of 0.66 is because the planet is not showing a full disk. – Neil Slater Aug 29 '15 at 19:40